Kinematics in One Dimension Question Bank Set 1

PHYS 202 - GENERAL PHYSICS II Kinematics in One Dimension Question Bank - Set 1 Liberty University Question 1 Question 2 A car starts from rest and accelerates at a constant rate of 2 m/s . How far does the car travel in the first 5 seconds? Solution Step 1: We can use the kinematic equation d = vi t + 12 at2 , where d is the distance traveled, vi is the initial velocity, a is the acceleration, and t is the time. Step 2: Since the car starts from rest, the initial velocity vi = 0. Plugging 2 in the values vi = 0, a = 2 m/s , and t = 5 s into the equation, we get: d=0×5+ 1 × 2 × (5)2 2 Step 3: Simplifying the expression gives: d=0+ 1 × 2 × 25 2 d = 0 + 1 × 25 = 25 m Step 4: Therefore, the car travels a distance of 25 meters in the first 5 seconds under the given conditions. Question 2 Question A car is initially at rest and accelerates uniformly to a speed of 25 m/s in 5 seconds. What is the acceleration of the car? Solution Step 1: Identify the given values. The initial velocity of the car, u, is 0 m/s. The final velocity of the car, v, is 25 m/s. The time taken for the car to reach this final velocity, t, is 5 seconds. The acceleration of the car, a, is what we need to find. Step 2: Use the kinematic equation relating velocity, acceleration, and time. The kinematic equation we will use is: v = u + at Substitute the given values into the equation: 25 = 0 + a × 5 Step 3: Solve for acceleration. a= 25 5 a = 5 m/s 2 Step 4: Check the units. The unit of acceleration is in m/s², which is the correct unit for acceleration. Therefore, the acceleration of the car is 5 m/s². Question 3 Question A car is initially traveling at a speed of 20 m/s. It accelerates at a constant rate of 3 m/s2 for 8 seconds. What is the final velocity of the car at the end of the 8-second interval? Solution Step 1: Determine the acceleration of the car. Given that the car accelerates at a constant rate of 3 m/s2 , the acceleration a is 3 m/s2 . Step 2: Determine the change in velocity of the car. Using the kinematic equation v = u + at, where: - v is the final velocity, - u is the initial velocity, - a is the acceleration, and - t is the time interval, we can find the change in 2 velocity: ∆v = a · t = 3 m/s · 8 s = 24 m/s. Step 3: Determine the final velocity of the car. The final velocity v of the car at the end of the 8-second interval is given by: v = u + ∆v = 20 m/s + 24 m/s = 44 m/s. Therefore, the final velocity of the car at the end of the 8-second interval is 44 m/s. 2 Question 4 Question A particle moves along a straight line according to the equation of motion x(t) = 2t3 − 3t2 + 6t + 1, where x is in meters and t is in seconds. Determine the displacement and distance traveled by the particle during the time interval t = 0 to t = 3 seconds. Solution Step 1: To find the displacement of the particle during the time interval t = 0 to t = 3 seconds, we need to evaluate x(3) − x(0). Displacement = x(3)−x(0) = (2(3)3 −3(3)2 +6(3)+1)−(2(0)3 −3(0)2 +6(0)+1) = (54 − 27 + 18 + 1) − (0 − 0 + 0 + 1) = 46 m Step 2: To find the distance traveled by the particle during the time interval t = 0 to t = 3 seconds, we need to calculate the total length of the curve x(t) over this interval. We can do this by integrating the absolute value of the velocity function. dx = 6t2 − 6t + 6 v(t) = dt Speed = |v(t)| = |6t2 − 6t + 6| Step 3: To integrate the speed function over the interval t = 0 to t = 3 seconds, we need to break the integral into two parts where the velocity is positive and negative, and then sum them up. Z 3 Distance = |6t2 − 6t + 6| dt 0 Z Distance = 1 (6t2 − 6t + 6) dt + 0 Z 3 −(6t2 − 6t + 6) dt 1 = 9m Therefore, the displacement of the particle during t = 0 to t = 3 seconds is 46 meters, and the distance traveled by the particle during that time interval is 9 meters. Question 5 Question A car traveling at a constant velocity of 25 m/s passes a stopped truck. At the instant the car passes, the truck starts accelerating at a rate of 2 m/s2 . How long will it take for the truck to catch up with the car? 3 Solution Step 1: Define the variables. Let t be the time it takes for the truck to catch up with the car. Step 2: Determine the equations of motion for both the car and the truck. The position of the car as a function of time can be described by: xcar (t) = 25t The position of the truck as a function of time can be described by: xtruck (t) = 1 2 (2t ) 2 Step 3: Set up an equation to find the time at which the truck catches up with the car. This occurs when the positions of the car and the truck are equal: 25t = 2t2 Step 4: Solve for t by rearranging the equation: 2t2 − 25t = 0 2t(t − 12.5) = 0 Step 5: Find the positive value of t by solving for t in the equation t − 12.5 = 0. This gives t = 12.5 seconds. Answer: It will take 12.5 seconds for the truck to catch up with the car. Question 6 Question A car travels along a straight road with a constant velocity of 20 m/s. At time t = 0, the car is 50 m behind a truck that is moving along the same road in the same direction with a constant velocity of 15 m/s. At what time will the car overtake the truck, and how far from the starting point will this occur? Solution Step 1: Determine the relative velocity between the car and the truck. The relative velocity vrel of the car with respect to the truck is the difference between the car’s velocity and the truck’s velocity. vrel = vcar −vtruck = 20 m/s−15 m/s = 5 m/s. Step 2: Determine the time it takes for the car to overtake the truck. Let t be the time it takes for the car to overtake the truck. During this time, the car will have closed the initial distance of 50 meters and some additional distance from the truck. The additional distance closed by the car is equal to the relative 4 velocity multiplied by time: dadditional = vrel · t = 5t. Setting this equal to the initial distance of 50 meters, we have: 50 = 5t =⇒ t = 10 s. Step 3: Determine the distance from the starting point where the overtake occurs. The distance d traveled by the car when it overtakes the truck is given by: d = vcar · t = 20 m/s · 10 s = 200 m. Thus, the car will overtake the truck 200 meters from the starting point. Question 7 Question A car accelerates uniformly from rest at a rate of 3 m/s2 for a distance of 100 m. What is the speed of the car after it has traveled 100 m? Solution Step 1: Determine the final velocity using the equation for uniformly accelerated motion: q vf = vi2 + 2a · d where - vf is the final velocity, - vi is the initial velocity (0 m/s, as the car starts from rest), - a is the acceleration (3 m/s2 ), - d is the distance traveled (100 m). Step 2: Substitute the given values into the equation: p √ √ vf = 0 + 2(3)(100) = 0 + 600 = 600 ≈ 24.49 m/s Therefore, the speed of the car after it has traveled 100 m is approximately 24.49 m/s. Question 8 Question A car traveling at a constant velocity of 25 m/s passes a street lamp. A bicyclist traveling in the same direction has a constant velocity of 10 m/s and passes the same lamp 5 seconds after the car passes it. How far from the lamp will the bicyclist catch up to the car? Solution Step 1: Let’s denote the distance from the lamp as x. We will first find the distance traveled by the car in 5 seconds. Step 2: The distance traveled by the car can be calculated using the formula d = vt, where d is the distance, v is the velocity, and t is the time. 5 Substitute v = 25 m/s and t = 5 s into the formula to find the distance traveled by the car: dcar = (25 m/s) · (5 s) = 125 m Step 3: Now, let’s find the distance between the car and the bicyclist when the bicyclist starts after 5 seconds. This distance is given by the equation: x = (25 m/s − 10 m/s) · 5 s = 15 m/s · 5 s x = 75 m Step 4: At this point, the bicyclist is 75 meters behind the car. Now, both the car and the bicyclist are moving at the same velocity of 15 m/s (25 m/s 10 m/s). To calculate the time it takes for the bicyclist to catch up to the car, we can use the formula: x = vt 75 m = 15 m/s · t Step 5: Solve for t: t= 75 m = 5s 15 m/s Step 6: Finally, we can find how far from the lamp the bicyclist will catch up to the car by using the formula: Distance traveled by the bicyclist = 10 m/s · 5 s = 50 m Adding the 75 meters the bicyclist was behind the car initially, the total distance from the lamp where the bicyclist catches up to the car is: 75 m + 50 m = 125 m Therefore, the bicyclist will catch up to the car 125 meters from the lamp. Question 9 Question 2 A car accelerates from rest at a constant rate of 2.0 m/s for a distance of 100 m. After reaching this distance, the car continues to accelerate but at a lower 2 rate of 1.0 m/s . How long does it take for the car to reach a speed of 25 m/s? 6 Solution Step 1: Determine the time it takes for the car to reach the end of the first acceleration phase. vf = vi + at 2 25 m/s = 0 + 2.0 m/s · t 25 m/s = 2t t= 25 m/s 2 2 m/s t = 12.5 s Step 2: Calculate the distance traveled during the first acceleration phase. 1 d = vi t + at2 2 1 2 d = 0 · 12.5 + · 2.0 m/s · (12.5 s)2 2 d = 0 + 156.25 d = 156.25 m Step 3: Determine how far the car has left to travel after the first acceleration phase. ∆d = 100 m − 156.25 m ∆d = −56.25 m Step 4: Determine the time it takes to reach a speed of 25 m/s after the first acceleration phase. vf = vi + at 2 25 m/s = 0 + 1.0 m/s · t 25 m/s = t t = 25 s The total time for the car to reach a speed of 25 m/s is 12.5 s (first acceleration phase) + 25 s (second acceleration phase) = 37.5 s. Question 10 Question A car initially traveling at a speed of 25 m/s accelerates uniformly at 2 m/s2 for 10 seconds. What is the final velocity of the car? 7 Solution Let’s denote the initial velocity of the car as vi = 25 m/s, the acceleration as a = 2 m/s2 , and the time duration as t = 10 s. We need to find the final velocity vf . Step 1: Use the kinematic equation vf = vi + at to find the final velocity. 2 vf = 25 m/s + 2 m/s × 10 s vf = 25 m/s + 20 m/s vf = 45 m/s Step 2: Therefore, the final velocity of the car after accelerating uniformly at 2 m/s2 for 10 seconds is 45 m/s . Question 11 Question A car travels along a straight road. The car’s velocity as a function of time is given by v(t) = 8t2 − 2t, where v is in m/s and t is in seconds. Find the total distance the car travels between t = 0 and t = 2 seconds. Solution Step 1: To find the total distance traveled by the car, we need to integrate the absolute value of the velocity function over the interval [0, 2]. Z 2 Total distance = |v(t)|dt 0 Step 2: First, we need to find the critical points of v(t) where v(t) = 0. 8t2 − 2t = 0 2t(4t − 1) = 0 1 4 Step 3: Split the integral depending on the sign of v(t). t = 0, t = Z 2 Z 1/4 |v(t)|dt = 0 Z 2 2 −(8t − 2t)dt + 0 (8t2 − 2t)dt 1/4 Step 4: Evaluate the integral.  1/4  2 8 3 8 3 2 2 + t −t = − t +t 3 3 0 1/4 8      8 1 1 8 8 1 3 1 = − ( )3 + ( )2 + (2)3 − (2)2 − ( ) − ( )2 3 4 4 3 3 4 4 1 64 1 65 + −1− = m 3 3 3 3 Therefore, the total distance the car travels between t = 0 and t = 2 seconds is 65 3 meters. = Question 12 Question A car accelerates along a straight road, starting from rest at time t = 0. The 2 acceleration of the car is given by a(t) = 2t, where a is in m/s and t is in seconds. Find the expression for the velocity of the car as a function of time, v(t). Solution Step 1: To find the expression for the velocity of the car as a function of time, we need to integrate the acceleration function with respect to time to get the velocity function. Step 2: The acceleration is given as a(t) = 2t. Integrating with respect to time gives the velocity function: Z Z a(t) dt = 2t dt Z v(t) = 2t dt = t2 + C where C is the constant of integration. Step 3: To determine the value of the constant of integration C, we can use the initial condition that the car starts from rest, meaning v(0) = 0. Substituting t = 0 into the velocity function: v(0) = 02 + C = 0 C=0 Step 4: Therefore, the expression for the velocity of the car as a function of time is: v(t) = t2 9 Question 13 Question An object is thrown vertically upward from the ground. Its initial velocity is 20 m/s. How long does it take for the object to reach its maximum height? 2 (Assume the acceleration due to gravity is −9.81 m/s ) Solution Step 1: We start by finding the time it takes for the object to reach its maximum height using the kinematic equation for vertical motion: vf = v0 + at where: - vf is the final velocity (which is 0 at the maximum height), - v0 is the initial velocity (20 m/s in this case), - a is the acceleration due to gravity (-9.81 m/s2 ), - t is the time. Step 2: Substitute the known values into the equation: 0 = 20 − 9.81t Step 3: Solve for t: 9.81t = 20 20 t= 9.81 t ≈ 2.04 s Therefore, it takes approximately 2.04 seconds for the object to reach its maximum height. Question 14 Question A car initially at rest accelerates at a constant rate for 10 seconds until it reaches a speed of 30 m/s. It then maintains this speed for 20 seconds before coming to a stop with a constant deceleration. If the total distance traveled by the car is 750 meters, what is the magnitude of the deceleration? Solution Step 1: Find the acceleration during the first 10 seconds. The final velocity of the car, vf , is 30 m/s, the initial velocity, vi , is 0 m/s, and the time, t, is 10 seconds. We can use the equation for acceleration in one dimension: a= v f − vi t 10 30 m/s − 0 m/s 2 = 3 m/s 10 s Step 2: Find the distance covered during the acceleration phase. We can use the equation: 1 s = vi t + at2 2 1 s = 0 · 10 + · 3 · 102 = 150 m 2 Step 3: Find the distance covered during the constant velocity phase. During this phase, the car moves at a constant speed of 30 m/s for 20 seconds. Thus, the distance covered is: a= s = v · t = 30 · 20 = 600 m Step 4: Find the deceleration. The total distance traveled by the car is 750 meters, the distance covered during acceleration is 150 meters, and the distance covered during constant speed is 600 meters. The remaining distance covered during deceleration is: 750 − 150 − 600 = 0 Since the car comes to a stop, the final velocity is 0. Therefore, we can use the equation of motion for deceleration: vf2 = vi2 + 2as 0 = 302 + 2 · a · s 900 900 2 =− = −∞ m/s a=− 2s 2·0 Therefore, it appears there may have been a calculation error or assumption made in the problem statement, as the calculated deceleration is not physically possible. Question 15 Question A car is initially traveling at a speed of 25 m/s. It then accelerates uniformly at 2 m/s2 for 10 seconds. What is the final velocity of the car? Solution Step 1: Calculate the acceleration of the car using the formula a = ∆v ∆t where a is the acceleration, ∆v is the change in velocity, and ∆t is the change in time. Given: ∆t = 10 s, a = 2 m/s 11 2 ∆v 2 , we find: ∆v = a × ∆t = 2 m/s × 10 s = 20 m/s ∆t Step 2: Calculate the final velocity of the car using the formula vf = vi + ∆v where vf is the final velocity, vi is the initial velocity, and ∆v is the change in velocity. Given: vi = 25 m/s, ∆v = 20 m/s Using a = vf = vi + ∆v = 25 m/s + 20 m/s = 45 m/s Therefore, the final velocity of the car after accelerating uniformly at 2 m/s2 for 10 seconds is 45 m/s. Question 16 Question A car starts from rest and accelerates uniformly at 2.5 m/s2 for 10 seconds. After this time, the brakes are applied and the car decelerates uniformly at 3.0 m/s2 until it comes to a stop. Determine the total distance the car travels during the entire motion. Solution Step 1: Calculate the distance traveled during the acceleration phase. 2 ˆ Given acceleration, a = 2.5 m/s ˆ Time taken for acceleration, t1 = 10 s ˆ Initial velocity, u = 0 m/s (starting from rest) Using the kinematic equation s = ut + 21 at2 , we can find the distance traveled during acceleration. 1 s1 = 0 × 10 + × 2.5 × (10)2 2 s1 = 0 + 125 s1 = 125 m Step 2: Calculate the distance traveled during the deceleration phase. 2 ˆ Given deceleration, a = −3.0 m/s (negative because it’s deceleration) ˆ Final velocity, v = 0 m/s (the car comes to a stop) Again, using the kinematic equation v 2 = u2 + 2as, we can find the distance traveled during deceleration. 0 = 02 + 2 × (−3.0) × s2 12 0 −6.0 s2 = 0 m s2 = Step 3: Calculate the total distance traveled by summing up the distances from the acceleration and deceleration phases. Total distance = s1 + s2 Total distance = 125 + 0 Total distance = 125 m Therefore, the total distance the car travels during the entire motion is 125 meters. Question 17 Question An object is initially at rest. It undergoes acceleration for 5 seconds, then decelerates for the next 2 seconds until it comes to a stop. The total displacement of the object during this time is 90 meters. If the object moves in a straight line, what is its average velocity during the entire motion? Solution Step 1: Calculate the initial acceleration using the first phase of motion. Given that the object is initially at rest, the final velocity after 5 seconds can be calculated using the equation of motion: vf = u + at Where: - vf = 0 (as the object comes to rest) - u = 0 (initial velocity) - a = acceleration Plugging in the values and solving for a: 0 = 0 + 5a a=0 Therefore, the initial acceleration of the object is 0 m/s2 . Step 2: Calculate the final velocity of the object after the deceleration phase. Given that the final velocity is 0 m/s after the 2-second deceleration, the deceleration b can be calculated: vf = u + bt 0 = 0 + 2b 13 b=0 Therefore, the deceleration is also 0 m/s2 . Step 3: Calculate the total distance traveled during the acceleration phase. During the acceleration phase, the total distance traveled can be calculated using the kinematic equation: 1 s = ut + at2 2 Where: - s = total distance (90 meters) - u = 0 (initial velocity) - a = 0 (acceleration) - t = 5 seconds Plugging in the values: 1 × 0 × 52 2 90 = 0 + 90 = 0 This results in a contradiction, indicating an error in our assumptions or calculations. This suggests that the question may have been set up incorrectly, as the given values do not lead to a consistent solution. Therefore, the average velocity cannot be determined with the provided information due to the inconsistency in the calculations. Question 18 Question A car accelerates from rest at a constant rate of 2 m/s2 for 5 seconds, after which it maintains a constant velocity for 10 seconds, and then decelerates at a rate of 1 m/s2 until it stops. Find the total distance the car travels during this time period. Solution Step 1: Find the distance traveled during the acceleration phase. The distance traveled during the acceleration phase can be calculated using the equation: d= 1 2 at 2 2 where: a = 2 m/s (acceleration) t = 5 s (time) Substitute the given values into the equation: 1 × 2 × (5)2 2 1 d = × 2 × 25 2 d= 14 d = 25 m Therefore, the distance traveled during the acceleration phase is 25 meters. Step 2: Find the distance traveled during the constant velocity phase. During the constant velocity phase, the distance traveled can be calculated using the equation: d = vt where: v = 2 m/s (constant velocity) t = 10 s (time) Substitute the given values into the equation: d = 2 × 10 d = 20 m Therefore, the distance traveled during the constant velocity phase is 20 meters. Step 3: Find the distance traveled during the deceleration phase. The distance traveled during the deceleration phase can be calculated using the equation: 1 d = vi t + at2 2 2 where: vi = 2 m/s (initial velocity) a = −1 m/s (deceleration) t = 15 s (time) Substitute the given values into the equation: 1 × (−1) × (15)2 2 1 d = 30 − × 225 2 d = 30 − 112.5 d = 2 × 15 + d = −82.5 m Therefore, the distance traveled during the deceleration phase is -82.5 meters. Step 4: Find the total distance traveled. To find the total distance traveled, we add up the distances traveled in each phase: Total distance = 25 m (acceleration) + 20 m (constant velocity) + (-82.5) m (deceleration) Total distance = 25 m + 20 m - 82.5 m Total distance = -37.5 m Therefore, the total distance the car travels during this time period is 37.5 meters. 15 Question 19 Question A car is traveling along a straight road with an initial velocity of 25 m/s. The car then accelerates at a constant rate for 10 seconds, reaching a final velocity of 35 m/s. After that, the car decelerates at a constant rate of 2 m/s² until it comes to a stop. Determine the total distance traveled by the car during this journey. Solution Step 1: Calculate the distance traveled during acceleration. vf = vi + at Where: vf = 35 m/s (final velocity), vi = 25 m/s (initial velocity), a is the acceleration, t = 10 s (time taken to accelerate). Solving for the acceleration a: 35 = 25 + a(10)35 = 25 + 10a10 = 10aa = 1 m/s 2 Using the formula for distance traveled during acceleration: 1 d = vi t + at2 2 Substitute vi = 25 m/s, a = 1 m/s², and t = 10 s: 1 d = (25)(10) + (1)(10)2 2 d = 250 + 50 = 300 m Step 2: Calculate the distance traveled during deceleration. The final velocity during deceleration is 0 m/s, and the acceleration a = −2 m/s². Using the formula for distance traveled during deceleration: vf2 = vi2 + 2ad Since vf = 0, the equation simplifies to: 0 = (35)2 + 2(−2)d0 = 1225 − 4d4d = 1225d = 306.25 m Step 3: Calculate the total distance traveled by the car. Total distance = 300 + 306.25 = 606.25 m Therefore, the total distance traveled by the car during the journey is 606.25 meters. 16 Question 20 Question A particle starts from rest at t = 0 and moves along the x-axis with a constant acceleration. If the particle moves a distance of 120 m in the fourth second, what is the acceleration of the particle? Solution Step 1: Let’s denote the acceleration of the particle as a m/s2 . Since the particle starts from rest, its initial velocity v0 = 0 m/s. Step 2: We know that the distance traveled by a particle with constant acceleration is given by the equation: 1 x(t) = v0 t + at2 2 where x(t) is the position of the particle at time t, v0 is the initial velocity, a is the acceleration, and t is the time. Step 3: Given that the particle moves a distance of 120 m in the fourth second, we can write the position equation as: 1 x(4) = 0 · 4 + a · 42 = 120 2 8a = 120 120 2 a= = 15 m/s 8 Step 4: Therefore, the acceleration of the particle is 15 m/s2 . Question 21 Question 2 A car starts from rest and accelerates at a constant rate of 2.5 m/s . How long does it take for the car to reach a speed of 30 m/s? Solution Step 1: Let’s first identify the given values and the unknown in this problem. Initial velocity (u) = 0 m/s (car starts from rest) Acceleration (a) = 2.5 m/s2 Final velocity (v) = 30 m/s Time taken (t) = ? Step 2: We can use the kinematic equation relating final velocity, initial velocity, acceleration, and time: v = u + at 17 Substitute the known values into the equation: 30 = 0 + 2.5t Step 3: Solve for t: 30 = 12 s 2.5 Step 4: Therefore, it takes 12 seconds for the car to reach a speed of 30 m/s. t= Question 22 Question A particle moves along a straight line according to the equation of motion: x(t) = 3t2 − 2t. Determine the velocity and acceleration of the particle at time t = 2 s. Solution Step 1: To find the velocity of the particle, we need to take the derivative of the equation of motion with respect to time. Velocity v(t) = dx dt d(3t2 − 2t) dt v(t) = 6t − 2 v(t) = Step 2: Substitute t = 2 s into the velocity equation to find the velocity at t = 2 s. v(2) = 6(2) − 2 v(2) = 12 − 2 v(2) = 10 m/s Therefore, the velocity of the particle at t = 2 s is 10 m/s. Step 3: To find the acceleration of the particle, we need to take the derivative of the velocity equation with respect to time. Acceleration a(t) = dv dt d(6t − 2) dt a(t) = 6 a(t) = Step 4: Substitute t = 2 s into the acceleration equation to find the acceleration at t = 2 s. 2 a(2) = 6 m/s Therefore, the acceleration of the particle at t = 2 s is 6 m/s2 . 18 Question 23 Question 2 A car accelerates uniformly from rest at a rate of 3 m/s . How long does it take the car to reach a speed of 30 m/s? Solution Step 1: Let’s denote the initial velocity of the car as v0 = 0 m/s, the acceleration 2 as a = 3 m/s , and the final speed reached as v = 30 m/s. We want to find the time t it takes the car to reach this final speed. Step 2: We can use the kinematic equation that relates final velocity, initial velocity, acceleration, and time: v = v0 + at Step 3: Substituting the given values into the equation, we get: 2 30 m/s = 0 m/s + 3 m/s · t Step 4: Solving for t, we find: t= 30 m/s 2 3 m/s = 10 s Step 5: Therefore, it takes the car 10 seconds to reach a speed of 30 m/s. Question 24 Question 2 A car accelerates from rest at a constant rate of 3 m/s for a distance of 100 meters. After reaching a certain velocity, the car decelerates at a constant rate 2 of 2 m/s until it comes to a stop. Determine the total time it takes for the car to come to a stop. Solution Step 1: Let’s first calculate the time it takes for the car to reach a certain velocity before deceleration: The final velocity vf can be calculated using the equation: √ vf = 2a · d 2 where a is the acceleration and d is the distance. Plugging in a = 3 m/s and d = 100 m: √ √ vf = 2 · 3 · 100 = 600 ≈ 24.49 m/s 19 Step 2: Next, we calculate the time taken to reach the final velocity using the equation: v = at Rearranging the equation to solve for time, t: 24.49 v ≈ 8.16 s t= = a 3 2 Step 3: After reaching the final velocity, the car decelerates at 2 m/s until it comes to a stop. The time taken to stop can be calculated using the equation: vf = at 2 where vf = 0, the final velocity after deceleration. Substitute a = −2 m/s for deceleration: 0 = −2t t = 0s Step 4: The total time taken for the car to come to a stop is the sum of the times for acceleration and deceleration: Total time = 8.16 s + 0 s = 8.16 s Therefore, the total time it takes for the car to come to a stop is 8.16 seconds. Question 25 Question A car initially traveling at a speed of 20 m/s accelerates uniformly to a speed of 40 m/s in 5 seconds. What is the distance traveled by the car during this time? Solution vf − v i Step 1: Find the acceleration of the car using the formula a = , where a t is the acceleration, vf is the final velocity, vi is the initial velocity, and t is the time. Given: vi = 20 m/s, vf = 40 m/s, t = 5 s 40 m/s − 20 m/s 20 m/s 2 a= = = 4 m/s 5s 5s Step 2: Use the equation of motion d = vi t + 12 at2 to find the distance traveled by the car, where d is the distance, vi is the initial velocity, a is the acceleration, and t is the time. 2 Given: vi = 20 m/s, a = 4 m/s , t = 5 s 1 2 d = (20 m/s)(5 s) + (4 m/s )(5 s)2 2 d = 100 m + 50 m = 150 m Therefore, the distance traveled by the car during this time is 150 meters. 20 Question 26 Question A car initially at rest accelerates along a straight road. It covers a distance of 400 m in the first 20 s and a distance of 600 m in the next 30 s. Find the acceleration of the car. Solution Step 1: Calculate the average velocity of the car during the first 20 s. Average velocity = displacement 400 m = = 20 m/s time 20 s Step 2: Calculate the acceleration of the car during the first 20 s using the equation: initial velocity + final velocity Average velocity = 2 0 + v1 20 m/s = 2 v1 = 40 m/s Step 3: Calculate the acceleration during the first 20 s using the equation: a= v−u t 40 m/s − 0 2 = 2 m/s 20 s Step 4: Calculate the average velocity of the car during the next 30 s. a= Average velocity = displacement 600 m = = 20 m/s time 30 s Step 5: Calculate the acceleration of the car during the next 30 s using the equation: initial velocity + final velocity Average velocity = 2 v1 + v2 20 m/s = 2 v1 + v2 = 40 m/s Step 6: Calculate the acceleration during the next 30 s using the equation: a= v−u t v2 − 40 m/s 2 = 2 m/s 30 s Step 7: Since the acceleration is constant, the acceleration of the car is 2 2 m/s during both time intervals. a= 21 Question 27 Question A car initially traveling at 20 m/s accelerates uniformly at 2 m/s2 for 10 seconds. After this time, the brakes are applied, causing the car to decelerate at 3 m/s2 . Determine the total distance the car travels during this process. Solution Step 1: Let’s first find the distance traveled during the acceleration phase using the equation: 1 d = vi t + at2 2 where d is the distance traveled, vi is the initial velocity, a is the acceleration, and t is the time. Plugging in the values, we get: 1 2 d = (20 m/s)(10 s) + (2 m/s )(10 s)2 2 d = 200 m + (0.5)(2)(100) m d = 200 m + 100 m = 300 m Step 2: Next, let’s find the distance traveled during the deceleration phase. We can use the same equation, but this time the acceleration is negative: 1 d = vi t + at2 2 Plugging in the values, we get: 1 2 d = (20 m/s)(10 s) + (−3 m/s )(10 s)2 2 d = 200 m + (0.5)(−3)(100) m d = 200 m − 150 m = 50 m Step 3: Finally, to find the total distance traveled, we add the distances traveled during acceleration and deceleration: Total distance = 300 m + 50 m = 350 m Therefore, the car travels a total distance of 350 meters during this process. Question 28 Question A car starts from rest and accelerates at a constant rate of 3.0 m/s2 for 10.0 seconds. After this time, the car maintains a constant velocity for another 20.0 seconds. Finally, the car decelerates at a rate of 2.0 m/s2 until it comes to a stop. Determine the total distance traveled by the car during this entire motion. 22 Solution Step 1: Calculate the distance traveled during the acceleration phase. The distance traveled during the acceleration phase can be calculated using the equation: 1 d = at2 2 2 where a is the acceleration and t is the time. Substitute a = 3.0 m/s and t = 10.0 s into the equation: d= 1 (3.0)(10.0)2 = 150.0 m 2 Step 2: Calculate the distance traveled during the constant velocity phase. The distance traveled during the constant velocity phase can be calculated using the equation: d = vt where v is the constant velocity and t is the time. Since the car maintains a constant velocity, the distance traveled is simply the product of the velocity and time. Substitute v = 3.0 × 10.0 = 30 m Step 3: Calculate the distance traveled during the deceleration phase. The distance traveled during the deceleration phase can be calculated using the equation: 1 d = vi t + at2 2 where vi is the initial velocity (in this case, the constant velocity) and a is the 2 acceleration. Substitute vi = 30.0 m/s and a = −2.0 m/s (negative because it’s decelerating) and t = 10.0 into the equation: 1 d = 30.0 × 10.0 + (−2.0)(10.0)2 = 300.0 − 100.0 = 200.0 m 2 Step 4: Calculate the total distance traveled by the car. The total distance traveled by the car is the sum of the distances traveled during each phase. Total distance = 150.0 + 30.0 + 200.0 = 380.0 m Therefore, the total distance traveled by the car during this entire motion is 380.0 meters. Question 29 Question An object is launched vertically upward from the ground. The initial velocity 2 of the object is 12 m/s and the acceleration due to gravity is −9.81 m/s . How long will it take for the object to reach its maximum height? 23 Solution Step 1: We know that at the maximum height of the object, the final velocity will be 0 m/s since the object momentarily stops before falling back down. We can use the kinematic equation vf = vi + at where vf = 0, vi = 12 m/s, a = 2 −9.81 m/s , and we want to solve for t. 0 = 12 − 9.81t 9.81t = 12 12 t= 9.81 t ≈ 1.22 s Therefore, it will take approximately 1.22 seconds for the object to reach its maximum height. Question 30 Question A particle moves along a straight line according to the equation of motion x(t) = 2 3 (6.0 m/s )t2 − (2.0 m/s )t3 , where x is in meters and t is in seconds. Determine the velocity of the particle at t = 2.0 s. Solution Step 1: Find the expression for the velocity of the particle. The velocity of the particle is given by the derivative of the position function x(t) with respect to time: dx v(t) = dt Step 2: Calculate the derivative of the position function. Differentiating x(t) = 6.0t2 − 2.0t3 with respect to t gives: v(t) = dx = 12.0t − 6.0t2 dt Step 3: Find the velocity at t = 2.0 s. Substitute t = 2.0 s into the expression for velocity: v(2.0) = 12.0(2.0) − 6.0(2.0)2 v(2.0) = 24.0 − 24.0 v(2.0) = 0 m/s Therefore, the velocity of the particle at t = 2.0 s is 0 m/s. 24 Question 31 Question A stone is thrown vertically upward with an initial speed of 15 m/s. How high does it go above its starting point? Take the acceleration due to gravity as 2 9.81 m/s . Solution Step 1: Identify the knowns and unknowns. Let’s denote the initial velocity of the stone as vi = 15 m/s, the acceleration 2 due to gravity as a = −9.81 m/s (where the negative sign indicates that the acceleration is directed downward), and the final velocity when the stone reaches its highest point as vf = 0 m/s. We need to find the height the stone reaches, denoted as h. Step 2: Choose the appropriate kinematic equation. In this particular scenario, the stone is thrown vertically upward and comes to a stop at its highest point before falling back down. Therefore, we can use the kinematic equation: vf2 = vi2 + 2ah Step 3: Substitute the known values into the kinematic equation. 2 Substitute vi = 15 m/s, vf = 0 m/s, and a = −9.81 m/s into the equation to solve for h: (0)2 = (15)2 + 2(−9.81)h Step 4: Solve for the height h. Simplify the equation: 0 = 225 − 19.62h 19.62h = 225 225 h= 19.62 h ≈ 11.47 m Therefore, the stone reaches a height of approximately 11.47 meters above its starting point. Question 32 Question A car initially traveling at 25 m/s accelerates uniformly at 3 m/s2 for 10 seconds. What is the final velocity of the car? 25 Solution Step 1: Identify the given variables and the acceleration equation. The initial velocity of the car, vi , is 25 m/s. The acceleration of the car, a, is 3 m/s2 , and the time interval, t, is 10 seconds. We can use the kinematic equation: vf = vi + a · t where vf is the final velocity. Step 2: Substitute the known values into the equation. 2 vf = 25 m/s + 3 m/s · 10 s Step 3: Calculate the final velocity of the car. vf = 25 m/s + 30 m/s = 55 m/s Therefore, the final velocity of the car is 55 m/s . Question 33 Question A car is initially at rest. It then accelerates at a constant rate of 2 m/s2 for 10 seconds. After this period of acceleration, the car continues to move at a constant velocity for 20 seconds. If the total distance travelled by the car during this time is 300 meters, what is the total displacement of the car? Solution Step 1: Calculate the distance travelled during acceleration phase. The distance travelled during acceleration phase can be calculated using the equation s = ut + 12 at2 , where s is the distance travelled, u is the initial velocity, a is the acceleration, and t is the time. Given that the car starts from rest, u = 0. Therefore, s = 0 + 12 × 2 × (10)2 . s = 0 + 21 × 2 × 100 = 100 meters. Step 2: Calculate the distance travelled during constant velocity phase. During the constant velocity phase, the car moves at a constant speed for 20 seconds. Since speed is constant, we can use the formula s = vt to calculate the distance traveled. The speed v during this phase is the final velocity after acceleration. Using the equation of motion, v = u + at, where u = 0 and a = 2 m/s2 , we have v = 0 + 2 × 10 = 20 m/s. Therefore, s = 20 × 20 = 400 meters. Step 3: Calculate the total displacement of the car. Displacement is a vector quantity that accounts for the change in position from the initial to the final position. Since the car started and ended at the same point, the total displacement is the sum of the displacements during the two phases. The displacement during acceleration is 100 meters in the positive direction, and during constant velocity is 400 meters in the positive direction. Therefore, the total displacement is 100 + 400 = 500 meters. 26 Question 34 Question A car starts from rest and accelerates in a straight line with a constant accel2 eration of 4 m/s for 10 seconds. After this time, the car maintains a constant velocity for another 5 seconds before coming to a stop with a constant deceler2 ation of 2 m/s . Determine the total distance the car travels during this entire motion. Solution Step 1: Find the distance traveled during acceleration. ˆ The initial velocity vinitial is 0 m/s. ˆ The acceleration a is 4 m/s2 . ˆ The time t is 10 s. The distance x traveled during acceleration can be found using the kinematic equation: 1 x = vinitial t + at2 2 1 x = 0 × 10 + × 4 × (10)2 2 x = 0 + 2 × 100 x = 200 m Step 2: Find the distance traveled during constant velocity. ˆ The velocity v is constant during this time. ˆ The time t is 5 s. The distance x traveled during constant velocity is given by: x = vt x=4×5 x = 20 m Step 3: Find the distance traveled during deceleration. ˆ The final velocity vfinal is 0 m/s. ˆ The deceleration a is −2 m/s2 (opposite direction to motion). ˆ The time t is 5 s. 27 The distance x traveled during deceleration can be found using the kinematic equation: 1 x = vfinal t + at2 2 1 x = 0 × 5 + × (−2) × (5)2 2 x = 0 + (−1) × 25 x = −25 m Step 4: Find the total distance traveled. The total distance traveled by the car is the sum of the distances traveled during acceleration, constant velocity, and deceleration. Total distance = 200 m + 20 m + (−25) m Total distance = 195 m Therefore, the total distance the car travels during this entire motion is 195 meters. Question 35 Question A car is traveling along a straight road at a constant speed of 25 m/s. Suddenly, the driver observes an obstruction 120 m ahead. The driver applies the brakes, causing the car to decelerate at a rate of 4 m/s2 . Will the car stop before reaching the obstruction? If not, how far from the obstruction will the car stop? Solution Step 1: Identify the given variables. The initial velocity of the car (vi ) is 25 m/s. The acceleration of the car (a) due to braking is -4 m/s2 (negative because it is deceleration). The distance to the obstruction (d) is 120 m. Step 2: Determine if the car will stop before reaching the obstruction. We will use the equation of motion: vf2 = vi2 + 2ad where vf is the final velocity of the car, and since the car is stopping, vf = 0. Substitute the known values into the equation: 0 = (25)2 + 2(−4)(d) 28 Solution Step 1: Identify the given values. The initial velocity of the car, u, is 0 m/s. The final velocity of the car, v, is 25 m/s. The time taken for the car to reach this final velocity, t, is 5 seconds. The acceleration of the car, a, is what we need to find. Step 2: Use the kinematic equation relating velocity, acceleration, and time. The kinematic equation we will use is: v = u + at Substitute the given values into the equation: 25 = 0 + a × 5 Step 3: Solve for acceleration. a= 25 5 a = 5 m/s 2 Step 4: Check the units. The unit of acceleration is in m/s², which is the correct unit for acceleration. Therefore, the acceleration of the car is 5 m/s². Question 3 Question A car is initially traveling at a speed of 20 m/s. It accelerates at a constant rate of 3 m/s2 for 8 seconds. What is the final velocity of the car at the end of the 8-second interval? Solution Step 1: Determine the acceleration of the car. Given that the car accelerates at a constant rate of 3 m/s2 , the acceleration a is 3 m/s2 . Step 2: Determine the change in velocity of the car. Using the kinematic equation v = u + at, where: - v is the final velocity, - u is the initial velocity, - a is the acceleration, and - t is the time interval, we can find the change in 2 velocity: ∆v = a · t = 3 m/s · 8 s = 24 m/s. Step 3: Determine the final velocity of the car. The final velocity v of the car at the end of the 8-second interval is given by: v = u + ∆v = 20 m/s + 24 m/s = 44 m/s. Therefore, the final velocity of the car at the end of the 8-second interval is 44 m/s. 2 Question 4 Question A particle moves along a straight line according to the equation of motion x(t) = 2t3 − 3t2 + 6t + 1, where x is in meters and t is in seconds. Determine the displacement and distance traveled by the particle during the time interval t = 0 to t = 3 seconds. Solution Step 1: To find the displacement of the particle during the time interval t = 0 to t = 3 seconds, we need to evaluate x(3) − x(0). Displacement = x(3)−x(0) = (2(3)3 −3(3)2 +6(3)+1)−(2(0)3 −3(0)2 +6(0)+1) = (54 − 27 + 18 + 1) − (0 − 0 + 0 + 1) = 46 m Step 2: To find the distance traveled by the particle during the time interval t = 0 to t = 3 seconds, we need to calculate the total length of the curve x(t) over this interval. We can do this by integrating the absolute value of the velocity function. dx = 6t2 − 6t + 6 v(t) = dt Speed = |v(t)| = |6t2 − 6t + 6| Step 3: To integrate the speed function over the interval t = 0 to t = 3 seconds, we need to break the integral into two parts where the velocity is positive and negative, and then sum them up. Z 3 Distance = |6t2 − 6t + 6| dt 0 Z Distance = 1 (6t2 − 6t + 6) dt + 0 Z 3 −(6t2 − 6t + 6) dt 1 = 9m Therefore, the displacement of the particle during t = 0 to t = 3 seconds is 46 meters, and the distance traveled by the particle during that time interval is 9 meters. Question 5 Question A car traveling at a constant velocity of 25 m/s passes a stopped truck. At the instant the car passes, the truck starts accelerating at a rate of 2 m/s2 . How long will it take for the truck to catch up with the car? 3 Solution Step 1: Define the variables. Let t be the time it takes for the truck to catch up with the car. Step 2: Determine the equations of motion for both the car and the truck. The position of the car as a function of time can be described by: xcar (t) = 25t The position of the truck as a function of time can be described by: xtruck (t) = 1 2 (2t ) 2 Step 3: Set up an equation to find the time at which the truck catches up with the car. This occurs when the positions of the car and the truck are equal: 25t = 2t2 Step 4: Solve for t by rearranging the equation: 2t2 − 25t = 0 2t(t − 12.5) = 0 Step 5: Find the positive value of t by solving for t in the equation t − 12.5 = 0. This gives t = 12.5 seconds. Answer: It will take 12.5 seconds for the truck to catch up with the car. Question 6 Question A car travels along a straight road with a constant velocity of 20 m/s. At time t = 0, the car is 50 m behind a truck that is moving along the same road in the same direction with a constant velocity of 15 m/s. At what time will the car overtake the truck, and how far from the starting point will this occur? Solution Step 1: Determine the relative velocity between the car and the truck. The relative velocity vrel of the car with respect to the truck is the difference between the car’s velocity and the truck’s velocity. vrel = vcar −vtruck = 20 m/s−15 m/s = 5 m/s. Step 2: Determine the time it takes for the car to overtake the truck. Let t be the time it takes for the car to overtake the truck. During this time, the car will have closed the initial distance of 50 meters and some additional distance from the truck. The additional distance closed by the car is equal to the relative 4 velocity multiplied by time: dadditional = vrel · t = 5t. Setting this equal to the initial distance of 50 meters, we have: 50 = 5t =⇒ t = 10 s. Step 3: Determine the distance from the starting point where the overtake occurs. The distance d traveled by the car when it overtakes the truck is given by: d = vcar · t = 20 m/s · 10 s = 200 m. Thus, the car will overtake the truck 200 meters from the starting point. Question 7 Question A car accelerates uniformly from rest at a rate of 3 m/s2 for a distance of 100 m. What is the speed of the car after it has traveled 100 m? Solution Step 1: Determine the final velocity using the equation for uniformly accelerated motion: q vf = vi2 + 2a · d where - vf is the final velocity, - vi is the initial velocity (0 m/s, as the car starts from rest), - a is the acceleration (3 m/s2 ), - d is the distance traveled (100 m). Step 2: Substitute the given values into the equation: p √ √ vf = 0 + 2(3)(100) = 0 + 600 = 600 ≈ 24.49 m/s Therefore, the speed of the car after it has traveled 100 m is approximately 24.49 m/s. Question 8 Question A car traveling at a constant velocity of 25 m/s passes a street lamp. A bicyclist traveling in the same direction has a constant velocity of 10 m/s and passes the same lamp 5 seconds after the car passes it. How far from the lamp will the bicyclist catch up to the car? Solution Step 1: Let’s denote the distance from the lamp as x. We will first find the distance traveled by the car in 5 seconds. Step 2: The distance traveled by the car can be calculated using the formula d = vt, where d is the distance, v is the velocity, and t is the time. 5 Substitute v = 25 m/s and t = 5 s into the formula to find the distance traveled by the car: dcar = (25 m/s) · (5 s) = 125 m Step 3: Now, let’s find the distance between the car and the bicyclist when the bicyclist starts after 5 seconds. This distance is given by the equation: x = (25 m/s − 10 m/s) · 5 s = 15 m/s · 5 s x = 75 m Step 4: At this point, the bicyclist is 75 meters behind the car. Now, both the car and the bicyclist are moving at the same velocity of 15 m/s (25 m/s 10 m/s). To calculate the time it takes for the bicyclist to catch up to the car, we can use the formula: x = vt 75 m = 15 m/s · t Step 5: Solve for t: t= 75 m = 5s 15 m/s Step 6: Finally, we can find how far from the lamp the bicyclist will catch up to the car by using the formula: Distance traveled by the bicyclist = 10 m/s · 5 s = 50 m Adding the 75 meters the bicyclist was behind the car initially, the total distance from the lamp where the bicyclist catches up to the car is: 75 m + 50 m = 125 m Therefore, the bicyclist will catch up to the car 125 meters from the lamp. Question 9 Question 2 A car accelerates from rest at a constant rate of 2.0 m/s for a distance of 100 m. After reaching this distance, the car continues to accelerate but at a lower 2 rate of 1.0 m/s . How long does it take for the car to reach a speed of 25 m/s? 6 Solution Step 1: Determine the time it takes for the car to reach the end of the first acceleration phase. vf = vi + at 2 25 m/s = 0 + 2.0 m/s · t 25 m/s = 2t t= 25 m/s 2 2 m/s t = 12.5 s Step 2: Calculate the distance traveled during the first acceleration phase. 1 d = vi t + at2 2 1 2 d = 0 · 12.5 + · 2.0 m/s · (12.5 s)2 2 d = 0 + 156.25 d = 156.25 m Step 3: Determine how far the car has left to travel after the first acceleration phase. ∆d = 100 m − 156.25 m ∆d = −56.25 m Step 4: Determine the time it takes to reach a speed of 25 m/s after the first acceleration phase. vf = vi + at 2 25 m/s = 0 + 1.0 m/s · t 25 m/s = t t = 25 s The total time for the car to reach a speed of 25 m/s is 12.5 s (first acceleration phase) + 25 s (second acceleration phase) = 37.5 s. Question 10 Question A car initially traveling at a speed of 25 m/s accelerates uniformly at 2 m/s2 for 10 seconds. What is the final velocity of the car? 7 Solution Let’s denote the initial velocity of the car as vi = 25 m/s, the acceleration as a = 2 m/s2 , and the time duration as t = 10 s. We need to find the final velocity vf . Step 1: Use the kinematic equation vf = vi + at to find the final velocity. 2 vf = 25 m/s + 2 m/s × 10 s vf = 25 m/s + 20 m/s vf = 45 m/s Step 2: Therefore, the final velocity of the car after accelerating uniformly at 2 m/s2 for 10 seconds is 45 m/s . Question 11 Question A car travels along a straight road. The car’s velocity as a function of time is given by v(t) = 8t2 − 2t, where v is in m/s and t is in seconds. Find the total distance the car travels between t = 0 and t = 2 seconds. Solution Step 1: To find the total distance traveled by the car, we need to integrate the absolute value of the velocity function over the interval [0, 2]. Z 2 Total distance = |v(t)|dt 0 Step 2: First, we need to find the critical points of v(t) where v(t) = 0. 8t2 − 2t = 0 2t(4t − 1) = 0 1 4 Step 3: Split the integral depending on the sign of v(t). t = 0, t = Z 2 Z 1/4 |v(t)|dt = 0 Z 2 2 −(8t − 2t)dt + 0 (8t2 − 2t)dt 1/4 Step 4: Evaluate the integral.  1/4  2 8 3 8 3 2 2 + t −t = − t +t 3 3 0 1/4 8      8 1 1 8 8 1 3 1 = − ( )3 + ( )2 + (2)3 − (2)2 − ( ) − ( )2 3 4 4 3 3 4 4 1 64 1 65 + −1− = m 3 3 3 3 Therefore, the total distance the car travels between t = 0 and t = 2 seconds is 65 3 meters. = Question 12 Question A car accelerates along a straight road, starting from rest at time t = 0. The 2 acceleration of the car is given by a(t) = 2t, where a is in m/s and t is in seconds. Find the expression for the velocity of the car as a function of time, v(t). Solution Step 1: To find the expression for the velocity of the car as a function of time, we need to integrate the acceleration function with respect to time to get the velocity function. Step 2: The acceleration is given as a(t) = 2t. Integrating with respect to time gives the velocity function: Z Z a(t) dt = 2t dt Z v(t) = 2t dt = t2 + C where C is the constant of integration. Step 3: To determine the value of the constant of integration C, we can use the initial condition that the car starts from rest, meaning v(0) = 0. Substituting t = 0 into the velocity function: v(0) = 02 + C = 0 C=0 Step 4: Therefore, the expression for the velocity of the car as a function of time is: v(t) = t2 9 Question 13 Question An object is thrown vertically upward from the ground. Its initial velocity is 20 m/s. How long does it take for the object to reach its maximum height? 2 (Assume the acceleration due to gravity is −9.81 m/s ) Solution Step 1: We start by finding the time it takes for the object to reach its maximum height using the kinematic equation for vertical motion: vf = v0 + at where: - vf is the final velocity (which is 0 at the maximum height), - v0 is the initial velocity (20 m/s in this case), - a is the acceleration due to gravity (-9.81 m/s2 ), - t is the time. Step 2: Substitute the known values into the equation: 0 = 20 − 9.81t Step 3: Solve for t: 9.81t = 20 20 t= 9.81 t ≈ 2.04 s Therefore, it takes approximately 2.04 seconds for the object to reach its maximum height. Question 14 Question A car initially at rest accelerates at a constant rate for 10 seconds until it reaches a speed of 30 m/s. It then maintains this speed for 20 seconds before coming to a stop with a constant deceleration. If the total distance traveled by the car is 750 meters, what is the magnitude of the deceleration? Solution Step 1: Find the acceleration during the first 10 seconds. The final velocity of the car, vf , is 30 m/s, the initial velocity, vi , is 0 m/s, and the time, t, is 10 seconds. We can use the equation for acceleration in one dimension: a= v f − vi t 10 30 m/s − 0 m/s 2 = 3 m/s 10 s Step 2: Find the distance covered during the acceleration phase. We can use the equation: 1 s = vi t + at2 2 1 s = 0 · 10 + · 3 · 102 = 150 m 2 Step 3: Find the distance covered during the constant velocity phase. During this phase, the car moves at a constant speed of 30 m/s for 20 seconds. Thus, the distance covered is: a= s = v · t = 30 · 20 = 600 m Step 4: Find the deceleration. The total distance traveled by the car is 750 meters, the distance covered during acceleration is 150 meters, and the distance covered during constant speed is 600 meters. The remaining distance covered during deceleration is: 750 − 150 − 600 = 0 Since the car comes to a stop, the final velocity is 0. Therefore, we can use the equation of motion for deceleration: vf2 = vi2 + 2as 0 = 302 + 2 · a · s 900 900 2 =− = −∞ m/s a=− 2s 2·0 Therefore, it appears there may have been a calculation error or assumption made in the problem statement, as the calculated deceleration is not physically possible. Question 15 Question A car is initially traveling at a speed of 25 m/s. It then accelerates uniformly at 2 m/s2 for 10 seconds. What is the final velocity of the car? Solution Step 1: Calculate the acceleration of the car using the formula a = ∆v ∆t where a is the acceleration, ∆v is the change in velocity, and ∆t is the change in time. Given: ∆t = 10 s, a = 2 m/s 11 2 ∆v 2 , we find: ∆v = a × ∆t = 2 m/s × 10 s = 20 m/s ∆t Step 2: Calculate the final velocity of the car using the formula vf = vi + ∆v where vf is the final velocity, vi is the initial velocity, and ∆v is the change in velocity. Given: vi = 25 m/s, ∆v = 20 m/s Using a = vf = vi + ∆v = 25 m/s + 20 m/s = 45 m/s Therefore, the final velocity of the car after accelerating uniformly at 2 m/s2 for 10 seconds is 45 m/s. Question 16 Question A car starts from rest and accelerates uniformly at 2.5 m/s2 for 10 seconds. After this time, the brakes are applied and the car decelerates uniformly at 3.0 m/s2 until it comes to a stop. Determine the total distance the car travels during the entire motion. Solution Step 1: Calculate the distance traveled during the acceleration phase. 2 ˆ Given acceleration, a = 2.5 m/s ˆ Time taken for acceleration, t1 = 10 s ˆ Initial velocity, u = 0 m/s (starting from rest) Using the kinematic equation s = ut + 21 at2 , we can find the distance traveled during acceleration. 1 s1 = 0 × 10 + × 2.5 × (10)2 2 s1 = 0 + 125 s1 = 125 m Step 2: Calculate the distance traveled during the deceleration phase. 2 ˆ Given deceleration, a = −3.0 m/s (negative because it’s deceleration) ˆ Final velocity, v = 0 m/s (the car comes to a stop) Again, using the kinematic equation v 2 = u2 + 2as, we can find the distance traveled during deceleration. 0 = 02 + 2 × (−3.0) × s2 12 0 −6.0 s2 = 0 m s2 = Step 3: Calculate the total distance traveled by summing up the distances from the acceleration and deceleration phases. Total distance = s1 + s2 Total distance = 125 + 0 Total distance = 125 m Therefore, the total distance the car travels during the entire motion is 125 meters. Question 17 Question An object is initially at rest. It undergoes acceleration for 5 seconds, then decelerates for the next 2 seconds until it comes to a stop. The total displacement of the object during this time is 90 meters. If the object moves in a straight line, what is its average velocity during the entire motion? Solution Step 1: Calculate the initial acceleration using the first phase of motion. Given that the object is initially at rest, the final velocity after 5 seconds can be calculated using the equation of motion: vf = u + at Where: - vf = 0 (as the object comes to rest) - u = 0 (initial velocity) - a = acceleration Plugging in the values and solving for a: 0 = 0 + 5a a=0 Therefore, the initial acceleration of the object is 0 m/s2 . Step 2: Calculate the final velocity of the object after the deceleration phase. Given that the final velocity is 0 m/s after the 2-second deceleration, the deceleration b can be calculated: vf = u + bt 0 = 0 + 2b 13 b=0 Therefore, the deceleration is also 0 m/s2 . Step 3: Calculate the total distance traveled during the acceleration phase. During the acceleration phase, the total distance traveled can be calculated using the kinematic equation: 1 s = ut + at2 2 Where: - s = total distance (90 meters) - u = 0 (initial velocity) - a = 0 (acceleration) - t = 5 seconds Plugging in the values: 1 × 0 × 52 2 90 = 0 + 90 = 0 This results in a contradiction, indicating an error in our assumptions or calculations. This suggests that the question may have been set up incorrectly, as the given values do not lead to a consistent solution. Therefore, the average velocity cannot be determined with the provided information due to the inconsistency in the calculations. Question 18 Question A car accelerates from rest at a constant rate of 2 m/s2 for 5 seconds, after which it maintains a constant velocity for 10 seconds, and then decelerates at a rate of 1 m/s2 until it stops. Find the total distance the car travels during this time period. Solution Step 1: Find the distance traveled during the acceleration phase. The distance traveled during the acceleration phase can be calculated using the equation: d= 1 2 at 2 2 where: a = 2 m/s (acceleration) t = 5 s (time) Substitute the given values into the equation: 1 × 2 × (5)2 2 1 d = × 2 × 25 2 d= 14 d = 25 m Therefore, the distance traveled during the acceleration phase is 25 meters. Step 2: Find the distance traveled during the constant velocity phase. During the constant velocity phase, the distance traveled can be calculated using the equation: d = vt where: v = 2 m/s (constant velocity) t = 10 s (time) Substitute the given values into the equation: d = 2 × 10 d = 20 m Therefore, the distance traveled during the constant velocity phase is 20 meters. Step 3: Find the distance traveled during the deceleration phase. The distance traveled during the deceleration phase can be calculated using the equation: 1 d = vi t + at2 2 2 where: vi = 2 m/s (initial velocity) a = −1 m/s (deceleration) t = 15 s (time) Substitute the given values into the equation: 1 × (−1) × (15)2 2 1 d = 30 − × 225 2 d = 30 − 112.5 d = 2 × 15 + d = −82.5 m Therefore, the distance traveled during the deceleration phase is -82.5 meters. Step 4: Find the total distance traveled. To find the total distance traveled, we add up the distances traveled in each phase: Total distance = 25 m (acceleration) + 20 m (constant velocity) + (-82.5) m (deceleration) Total distance = 25 m + 20 m - 82.5 m Total distance = -37.5 m Therefore, the total distance the car travels during this time period is 37.5 meters. 15 Question 19 Question A car is traveling along a straight road with an initial velocity of 25 m/s. The car then accelerates at a constant rate for 10 seconds, reaching a final velocity of 35 m/s. After that, the car decelerates at a constant rate of 2 m/s² until it comes to a stop. Determine the total distance traveled by the car during this journey. Solution Step 1: Calculate the distance traveled during acceleration. vf = vi + at Where: vf = 35 m/s (final velocity), vi = 25 m/s (initial velocity), a is the acceleration, t = 10 s (time taken to accelerate). Solving for the acceleration a: 35 = 25 + a(10)35 = 25 + 10a10 = 10aa = 1 m/s 2 Using the formula for distance traveled during acceleration: 1 d = vi t + at2 2 Substitute vi = 25 m/s, a = 1 m/s², and t = 10 s: 1 d = (25)(10) + (1)(10)2 2 d = 250 + 50 = 300 m Step 2: Calculate the distance traveled during deceleration. The final velocity during deceleration is 0 m/s, and the acceleration a = −2 m/s². Using the formula for distance traveled during deceleration: vf2 = vi2 + 2ad Since vf = 0, the equation simplifies to: 0 = (35)2 + 2(−2)d0 = 1225 − 4d4d = 1225d = 306.25 m Step 3: Calculate the total distance traveled by the car. Total distance = 300 + 306.25 = 606.25 m Therefore, the total distance traveled by the car during the journey is 606.25 meters. 16 Question 20 Question A particle starts from rest at t = 0 and moves along the x-axis with a constant acceleration. If the particle moves a distance of 120 m in the fourth second, what is the acceleration of the particle? Solution Step 1: Let’s denote the acceleration of the particle as a m/s2 . Since the particle starts from rest, its initial velocity v0 = 0 m/s. Step 2: We know that the distance traveled by a particle with constant acceleration is given by the equation: 1 x(t) = v0 t + at2 2 where x(t) is the position of the particle at time t, v0 is the initial velocity, a is the acceleration, and t is the time. Step 3: Given that the particle moves a distance of 120 m in the fourth second, we can write the position equation as: 1 x(4) = 0 · 4 + a · 42 = 120 2 8a = 120 120 2 a= = 15 m/s 8 Step 4: Therefore, the acceleration of the particle is 15 m/s2 . Question 21 Question 2 A car starts from rest and accelerates at a constant rate of 2.5 m/s . How long does it take for the car to reach a speed of 30 m/s? Solution Step 1: Let’s first identify the given values and the unknown in this problem. Initial velocity (u) = 0 m/s (car starts from rest) Acceleration (a) = 2.5 m/s2 Final velocity (v) = 30 m/s Time taken (t) = ? Step 2: We can use the kinematic equation relating final velocity, initial velocity, acceleration, and time: v = u + at 17 Substitute the known values into the equation: 30 = 0 + 2.5t Step 3: Solve for t: 30 = 12 s 2.5 Step 4: Therefore, it takes 12 seconds for the car to reach a speed of 30 m/s. t= Question 22 Question A particle moves along a straight line according to the equation of motion: x(t) = 3t2 − 2t. Determine the velocity and acceleration of the particle at time t = 2 s. Solution Step 1: To find the velocity of the particle, we need to take the derivative of the equation of motion with respect to time. Velocity v(t) = dx dt d(3t2 − 2t) dt v(t) = 6t − 2 v(t) = Step 2: Substitute t = 2 s into the velocity equation to find the velocity at t = 2 s. v(2) = 6(2) − 2 v(2) = 12 − 2 v(2) = 10 m/s Therefore, the velocity of the particle at t = 2 s is 10 m/s. Step 3: To find the acceleration of the particle, we need to take the derivative of the velocity equation with respect to time. Acceleration a(t) = dv dt d(6t − 2) dt a(t) = 6 a(t) = Step 4: Substitute t = 2 s into the acceleration equation to find the acceleration at t = 2 s. 2 a(2) = 6 m/s Therefore, the acceleration of the particle at t = 2 s is 6 m/s2 . 18 Question 23 Question 2 A car accelerates uniformly from rest at a rate of 3 m/s . How long does it take the car to reach a speed of 30 m/s? Solution Step 1: Let’s denote the initial velocity of the car as v0 = 0 m/s, the acceleration 2 as a = 3 m/s , and the final speed reached as v = 30 m/s. We want to find the time t it takes the car to reach this final speed. Step 2: We can use the kinematic equation that relates final velocity, initial velocity, acceleration, and time: v = v0 + at Step 3: Substituting the given values into the equation, we get: 2 30 m/s = 0 m/s + 3 m/s · t Step 4: Solving for t, we find: t= 30 m/s 2 3 m/s = 10 s Step 5: Therefore, it takes the car 10 seconds to reach a speed of 30 m/s. Question 24 Question 2 A car accelerates from rest at a constant rate of 3 m/s for a distance of 100 meters. After reaching a certain velocity, the car decelerates at a constant rate 2 of 2 m/s until it comes to a stop. Determine the total time it takes for the car to come to a stop. Solution Step 1: Let’s first calculate the time it takes for the car to reach a certain velocity before deceleration: The final velocity vf can be calculated using the equation: √ vf = 2a · d 2 where a is the acceleration and d is the distance. Plugging in a = 3 m/s and d = 100 m: √ √ vf = 2 · 3 · 100 = 600 ≈ 24.49 m/s 19 Step 2: Next, we calculate the time taken to reach the final velocity using the equation: v = at Rearranging the equation to solve for time, t: 24.49 v ≈ 8.16 s t= = a 3 2 Step 3: After reaching the final velocity, the car decelerates at 2 m/s until it comes to a stop. The time taken to stop can be calculated using the equation: vf = at 2 where vf = 0, the final velocity after deceleration. Substitute a = −2 m/s for deceleration: 0 = −2t t = 0s Step 4: The total time taken for the car to come to a stop is the sum of the times for acceleration and deceleration: Total time = 8.16 s + 0 s = 8.16 s Therefore, the total time it takes for the car to come to a stop is 8.16 seconds. Question 25 Question A car initially traveling at a speed of 20 m/s accelerates uniformly to a speed of 40 m/s in 5 seconds. What is the distance traveled by the car during this time? Solution vf − v i Step 1: Find the acceleration of the car using the formula a = , where a t is the acceleration, vf is the final velocity, vi is the initial velocity, and t is the time. Given: vi = 20 m/s, vf = 40 m/s, t = 5 s 40 m/s − 20 m/s 20 m/s 2 a= = = 4 m/s 5s 5s Step 2: Use the equation of motion d = vi t + 12 at2 to find the distance traveled by the car, where d is the distance, vi is the initial velocity, a is the acceleration, and t is the time. 2 Given: vi = 20 m/s, a = 4 m/s , t = 5 s 1 2 d = (20 m/s)(5 s) + (4 m/s )(5 s)2 2 d = 100 m + 50 m = 150 m Therefore, the distance traveled by the car during this time is 150 meters. 20 Question 26 Question A car initially at rest accelerates along a straight road. It covers a distance of 400 m in the first 20 s and a distance of 600 m in the next 30 s. Find the acceleration of the car. Solution Step 1: Calculate the average velocity of the car during the first 20 s. Average velocity = displacement 400 m = = 20 m/s time 20 s Step 2: Calculate the acceleration of the car during the first 20 s using the equation: initial velocity + final velocity Average velocity = 2 0 + v1 20 m/s = 2 v1 = 40 m/s Step 3: Calculate the acceleration during the first 20 s using the equation: a= v−u t 40 m/s − 0 2 = 2 m/s 20 s Step 4: Calculate the average velocity of the car during the next 30 s. a= Average velocity = displacement 600 m = = 20 m/s time 30 s Step 5: Calculate the acceleration of the car during the next 30 s using the equation: initial velocity + final velocity Average velocity = 2 v1 + v2 20 m/s = 2 v1 + v2 = 40 m/s Step 6: Calculate the acceleration during the next 30 s using the equation: a= v−u t v2 − 40 m/s 2 = 2 m/s 30 s Step 7: Since the acceleration is constant, the acceleration of the car is 2 2 m/s during both time intervals. a= 21 Question 27 Question A car initially traveling at 20 m/s accelerates uniformly at 2 m/s2 for 10 seconds. After this time, the brakes are applied, causing the car to decelerate at 3 m/s2 . Determine the total distance the car travels during this process. Solution Step 1: Let’s first find the distance traveled during the acceleration phase using the equation: 1 d = vi t + at2 2 where d is the distance traveled, vi is the initial velocity, a is the acceleration, and t is the time. Plugging in the values, we get: 1 2 d = (20 m/s)(10 s) + (2 m/s )(10 s)2 2 d = 200 m + (0.5)(2)(100) m d = 200 m + 100 m = 300 m Step 2: Next, let’s find the distance traveled during the deceleration phase. We can use the same equation, but this time the acceleration is negative: 1 d = vi t + at2 2 Plugging in the values, we get: 1 2 d = (20 m/s)(10 s) + (−3 m/s )(10 s)2 2 d = 200 m + (0.5)(−3)(100) m d = 200 m − 150 m = 50 m Step 3: Finally, to find the total distance traveled, we add the distances traveled during acceleration and deceleration: Total distance = 300 m + 50 m = 350 m Therefore, the car travels a total distance of 350 meters during this process. Question 28 Question A car starts from rest and accelerates at a constant rate of 3.0 m/s2 for 10.0 seconds. After this time, the car maintains a constant velocity for another 20.0 seconds. Finally, the car decelerates at a rate of 2.0 m/s2 until it comes to a stop. Determine the total distance traveled by the car during this entire motion. 22 Solution Step 1: Calculate the distance traveled during the acceleration phase. The distance traveled during the acceleration phase can be calculated using the equation: 1 d = at2 2 2 where a is the acceleration and t is the time. Substitute a = 3.0 m/s and t = 10.0 s into the equation: d= 1 (3.0)(10.0)2 = 150.0 m 2 Step 2: Calculate the distance traveled during the constant velocity phase. The distance traveled during the constant velocity phase can be calculated using the equation: d = vt where v is the constant velocity and t is the time. Since the car maintains a constant velocity, the distance traveled is simply the product of the velocity and time. Substitute v = 3.0 × 10.0 = 30 m Step 3: Calculate the distance traveled during the deceleration phase. The distance traveled during the deceleration phase can be calculated using the equation: 1 d = vi t + at2 2 where vi is the initial velocity (in this case, the constant velocity) and a is the 2 acceleration. Substitute vi = 30.0 m/s and a = −2.0 m/s (negative because it’s decelerating) and t = 10.0 into the equation: 1 d = 30.0 × 10.0 + (−2.0)(10.0)2 = 300.0 − 100.0 = 200.0 m 2 Step 4: Calculate the total distance traveled by the car. The total distance traveled by the car is the sum of the distances traveled during each phase. Total distance = 150.0 + 30.0 + 200.0 = 380.0 m Therefore, the total distance traveled by the car during this entire motion is 380.0 meters. Question 29 Question An object is launched vertically upward from the ground. The initial velocity 2 of the object is 12 m/s and the acceleration due to gravity is −9.81 m/s . How long will it take for the object to reach its maximum height? 23 Solution Step 1: We know that at the maximum height of the object, the final velocity will be 0 m/s since the object momentarily stops before falling back down. We can use the kinematic equation vf = vi + at where vf = 0, vi = 12 m/s, a = 2 −9.81 m/s , and we want to solve for t. 0 = 12 − 9.81t 9.81t = 12 12 t= 9.81 t ≈ 1.22 s Therefore, it will take approximately 1.22 seconds for the object to reach its maximum height. Question 30 Question A particle moves along a straight line according to the equation of motion x(t) = 2 3 (6.0 m/s )t2 − (2.0 m/s )t3 , where x is in meters and t is in seconds. Determine the velocity of the particle at t = 2.0 s. Solution Step 1: Find the expression for the velocity of the particle. The velocity of the particle is given by the derivative of the position function x(t) with respect to time: dx v(t) = dt Step 2: Calculate the derivative of the position function. Differentiating x(t) = 6.0t2 − 2.0t3 with respect to t gives: v(t) = dx = 12.0t − 6.0t2 dt Step 3: Find the velocity at t = 2.0 s. Substitute t = 2.0 s into the expression for velocity: v(2.0) = 12.0(2.0) − 6.0(2.0)2 v(2.0) = 24.0 − 24.0 v(2.0) = 0 m/s Therefore, the velocity of the particle at t = 2.0 s is 0 m/s. 24 Question 31 Question A stone is thrown vertically upward with an initial speed of 15 m/s. How high does it go above its starting point? Take the acceleration due to gravity as 2 9.81 m/s . Solution Step 1: Identify the knowns and unknowns. Let’s denote the initial velocity of the stone as vi = 15 m/s, the acceleration 2 due to gravity as a = −9.81 m/s (where the negative sign indicates that the acceleration is directed downward), and the final velocity when the stone reaches its highest point as vf = 0 m/s. We need to find the height the stone reaches, denoted as h. Step 2: Choose the appropriate kinematic equation. In this particular scenario, the stone is thrown vertically upward and comes to a stop at its highest point before falling back down. Therefore, we can use the kinematic equation: vf2 = vi2 + 2ah Step 3: Substitute the known values into the kinematic equation. 2 Substitute vi = 15 m/s, vf = 0 m/s, and a = −9.81 m/s into the equation to solve for h: (0)2 = (15)2 + 2(−9.81)h Step 4: Solve for the height h. Simplify the equation: 0 = 225 − 19.62h 19.62h = 225 225 h= 19.62 h ≈ 11.47 m Therefore, the stone reaches a height of approximately 11.47 meters above its starting point. Question 32 Question A car initially traveling at 25 m/s accelerates uniformly at 3 m/s2 for 10 seconds. What is the final velocity of the car? 25 Solution Step 1: Identify the given variables and the acceleration equation. The initial velocity of the car, vi , is 25 m/s. The acceleration of the car, a, is 3 m/s2 , and the time interval, t, is 10 seconds. We can use the kinematic equation: vf = vi + a · t where vf is the final velocity. Step 2: Substitute the known values into the equation. 2 vf = 25 m/s + 3 m/s · 10 s Step 3: Calculate the final velocity of the car. vf = 25 m/s + 30 m/s = 55 m/s Therefore, the final velocity of the car is 55 m/s . Question 33 Question A car is initially at rest. It then accelerates at a constant rate of 2 m/s2 for 10 seconds. After this period of acceleration, the car continues to move at a constant velocity for 20 seconds. If the total distance travelled by the car during this time is 300 meters, what is the total displacement of the car? Solution Step 1: Calculate the distance travelled during acceleration phase. The distance travelled during acceleration phase can be calculated using the equation s = ut + 12 at2 , where s is the distance travelled, u is the initial velocity, a is the acceleration, and t is the time. Given that the car starts from rest, u = 0. Therefore, s = 0 + 12 × 2 × (10)2 . s = 0 + 21 × 2 × 100 = 100 meters. Step 2: Calculate the distance travelled during constant velocity phase. During the constant velocity phase, the car moves at a constant speed for 20 seconds. Since speed is constant, we can use the formula s = vt to calculate the distance traveled. The speed v during this phase is the final velocity after acceleration. Using the equation of motion, v = u + at, where u = 0 and a = 2 m/s2 , we have v = 0 + 2 × 10 = 20 m/s. Therefore, s = 20 × 20 = 400 meters. Step 3: Calculate the total displacement of the car. Displacement is a vector quantity that accounts for the change in position from the initial to the final position. Since the car started and ended at the same point, the total displacement is the sum of the displacements during the two phases. The displacement during acceleration is 100 meters in the positive direction, and during constant velocity is 400 meters in the positive direction. Therefore, the total displacement is 100 + 400 = 500 meters. 26 Question 34 Question A car starts from rest and accelerates in a straight line with a constant accel2 eration of 4 m/s for 10 seconds. After this time, the car maintains a constant velocity for another 5 seconds before coming to a stop with a constant deceler2 ation of 2 m/s . Determine the total distance the car travels during this entire motion. Solution Step 1: Find the distance traveled during acceleration. ˆ The initial velocity vinitial is 0 m/s. ˆ The acceleration a is 4 m/s2 . ˆ The time t is 10 s. The distance x traveled during acceleration can be found using the kinematic equation: 1 x = vinitial t + at2 2 1 x = 0 × 10 + × 4 × (10)2 2 x = 0 + 2 × 100 x = 200 m Step 2: Find the distance traveled during constant velocity. ˆ The velocity v is constant during this time. ˆ The time t is 5 s. The distance x traveled during constant velocity is given by: x = vt x=4×5 x = 20 m Step 3: Find the distance traveled during deceleration. ˆ The final velocity vfinal is 0 m/s. ˆ The deceleration a is −2 m/s2 (opposite direction to motion). ˆ The time t is 5 s. 27 The distance x traveled during deceleration can be found using the kinematic equation: 1 x = vfinal t + at2 2 1 x = 0 × 5 + × (−2) × (5)2 2 x = 0 + (−1) × 25 x = −25 m Step 4: Find the total distance traveled. The total distance traveled by the car is the sum of the distances traveled during acceleration, constant velocity, and deceleration. Total distance = 200 m + 20 m + (−25) m Total distance = 195 m Therefore, the total distance the car travels during this entire motion is 195 meters. Question 35 Question A car is traveling along a straight road at a constant speed of 25 m/s. Suddenly, the driver observes an obstruction 120 m ahead. The driver applies the brakes, causing the car to decelerate at a rate of 4 m/s2 . Will the car stop before reaching the obstruction? If not, how far from the obstruction will the car stop? Solution Step 1: Identify the given variables. The initial velocity of the car (vi ) is 25 m/s. The acceleration of the car (a) due to braking is -4 m/s2 (negative because it is deceleration). The distance to the obstruction (d) is 120 m. Step 2: Determine if the car will stop before reaching the obstruction. We will use the equation of motion: vf2 = vi2 + 2ad where vf is the final velocity of the car, and since the car is stopping, vf = 0. Substitute the known values into the equation: 0 = (25)2 + 2(−4)(d) 28 0 = 625 − 8d 8d = 625 625 d= 8 d ≈ 78.125 m Step 3: Conclusion. The car will not stop before reaching the obstruction. It will stop approximately 78.125 meters from the obstruction. 29