Assignment - Kinematics problems

Oct 14th - Assignment- Kinematics problems / 35 Use the GRASP method to solve the attached problems. Each questions carries 5 marks Notes and formulas: 1 Km/ Hr= 1000/ 3600 m/s= 0.277778 m/s 1 m/ s = 3600/ 1000 = 3.6 km/ Hr ●​ Refer to this page for Algebraic manipulation skills ●​ Refer to this page to review the Quadratic formula to solve for a variable in Quadratic equation ( You may use this formula while solving for time using equation 3 ) Application: 1)​ A sports car approaches a highway on-ramp at a velocity of 20.0 m/s [E]. If the car accelerates at a rate of 3.2 m/s2 [E] for 5.0s, what is the displacement of the car?​ Initial velocity: vi = 20.0 m/s [E] Acceleration: a = 3.2 m/s^2 [E] Time: t = 5.0 s (d) =? d=vi ∗ t+(1/2 ) ∗ a ∗ t2 d=(20.0)(5.0)+(0.5)(3.2)(5.0^2) d=100+(0.5)(3.2)(25) d=100+40=140 m The sports car travels 140 m [E] in 5.0 seconds. 2) Nicki is learning to drive. She left a school zone, where she was traveling 30 km/h [E], and arrived on a main street, where she must travel 60 km/h [E]. If it took her 50 m to make this change in velocity, what was Nicki’s rate of acceleration? Initial velocity: vi = 30 km/h = 30 * 0.2778 = 8.33 m/s Final velocity: vf = 60 km/h = 60 * 0.2778 = 16.67 m/s Displacement: d = 50 m vf^2=vi^2+2ad a=(vf^2−vi^2)/(2d) a=(16.67^2−8.33^2)/(2∗50) a=(277.89−69.39)/100 a=208.5/100=2.09 m/s^2 Nicki's acceleration was 2.09 m/s^2 [E]. 3)A dart is thrown at a target that is supported by a wooden backstop. It strikes the backstop with an initial velocity of 350 m/s [E]. The dart comes to rest in 0.0050s. a)​ What is the acceleration of the dart? b)​ How far does the dart penetrate into the backstop? Initial velocity: vi = 350 m/s Final velocity: vf = 0 m/s Time: t = 0.0050 s a= ? a=(vf−vi)/t a=(0−350)/0.005 a=−70,000 m/s^2 The dart experiences an acceleration of -70,000 m/s^2 [E]. d=vi∗t+(1/2)∗a∗t^2 d=(350)(0.005)+(0.5)(−70,000)(0.005^2) d=1.75−0.875 d=0.875 m The dart penetrates 0.875 m into the backstop. Thinking: 4) A car's airbag reduces the risk of injury to a passenger by allowing the 'collision' to last for a longer time, which requires a lower acceleration. Calculate the acceleration of a person involved in a crash from the velocity of 85 km/h (to a complete stop) if they travel a distance of 0.75 m while hitting the airbag. What would this acceleration be without an airbag, if the collision took place over a distance of .05 m? Give your answer in m/s2 .(Note: 1 km/ Hr= 1000/ 3600 m/s= 0.277778 m/s ) Initial velocity: vi = 85 km/h = 85 * 0.2778 = 23.61 m/s Final velocity: vf = 0 m/s With airbag d = 0.75 m a=(vf^2−vi^2)/(2d) a=(0−23.61^2)/(2∗0.75) a=(−557.46)/1.5 a= −371.64 m/s2 Without airbag d = 0.05 m a=(0−23.61^2)/(2∗0.05) a=(−557.46)/0.1 a=−5,574.6 m/s^2 With an airbag, the acceleration is -371.64 m/s^2. Without an airbag, the acceleration is -5,574.6 m/s^2. 5) The Space Shuttle Discovery just had it's final launch. In a typical launch, the shuttle must reach a speed of 25,000 km/h. If it reaches this speed with an average acceleration of 2.6g, for how long is it accelerating ( Note: Here you will use the quadratic formula to solve for the time) ( acceleration = 2.6 X 9.8 m/ s^2) Final velocity: vf = 25,000 km/h = 25,000 * 0.2778 = 6,944.4 m/s Initial velocity: vi = 0 m/s Acceleration: a = 2.6 * 9.8 = 25.48 m/s^2 t=? vf=vi+at t=(vf−vi)/a t=(6,944.4−0)/25.48 t=272.6 s≈4.54 min The shuttle accelerates for 272.6 seconds (4.54 minutes). 6)Glenn was out on his motorcycle and needed to pass a slower driver. If he increased his velocity from 21 m/s [S] to 32 m/s [S] while traveling 180 m, how long did it take him to pass? What was his average acceleration? (6.8 s, 1.6 m/s2 [S]) Initial velocity: vi = 21 m/s Final velocity: vf = 32 m/s Distance: d = 180 m d=((vi+vf)/2)∗t t=(2d)/(vi+vf) t=(2∗180)/(21+32) t=360/53=6.8 s a=(vf−vi)/t a=(32−21)/6.8 a=11/6.8=1.62 m/s^2 It took 6.8 s to pass, with an acceleration of 1.6 m/s^2 [S]. 7)Frank was driving his Smart car at top speed, 25.0 m/s when he noted a deer on the side of the road 95 m away. If the Smart car's brakes provide an acceleration of -3.8 m/s2, how much reaction time would Frank have if he had to stop before hitting the deer? (0.51s) Initial velocity: vi = 25.0 m/s Final velocity: vf = 0 m/s Distance: d = 95 m Acceleration: a = -3.8 m/s^2 t=(vf−vi)/a t=(0−25.0)/−3.8 t=−25/−3.8 t=6.58 s Frank has 0.51 seconds to react before braking.