Physics, Homework 1, Week 2

Physics Homework, Week 2 Week 2, Homework 1 Prob. 2.1.1 A race car can be slowed with a deceleration of -11. m/s2. (a) If it is going +55. m/s, how many meters will it take to stop? (b) Repeat part (a) for a car going 110 m/s. a) vf2 = vi2 + 2ad v final = 0m/s v initial = 55.m/s a = -11m/s^2 d=? (o m/s)^2 = (55. m/s)^2 + 2(-11 m/s^2)d 0 = (3025 m^2/s^2) – 22 m/s^2 * d (-3025 m^2/s^2) = - 22 m/s^2 * d (-3025 m^2/s^2) = d
137.5 m = d
(-22 m/s^2) b) vf2 = vi2 + 2ad v final = 0m/s v initial = 110m/s a = -11m/s^2 d=? (o m/s)^2 = (110 m/s)^2 + 2(-11 m/s^2)d 0 = (12100 m^2/s^2) – 22 m/s^2 * d d= 1.4 x 10^2 m (-12100m^2/s^2) = – 22 m/s^2 * d (-12100m^2/s^2) = d
550 m = d
– 22 m/s^2 d= 5.5 x10^2 m Prob. 2.1.2 An astronaut dropped a feather from 1.2 m above the surface of the moon. If the acceleration of gravity on the moon is 1.6 m/s2, how long did it take to hit the surface? d= vi*t + 1/2at2 d = 1.2 m a=1.6 m/s^2 v initial = 0 m/s t=? 1.2 m = (0 m/s)*t + ½(1.6 m/s^2) * t^2 1.2 m = ½(1.6 m/s^2) * t^2 (1.2 m)*2 = (1.6 m/s^2) * t^2 (2.4 m) = (1.6 m/s^2) * t^2 2.4 m (1.6 m/s^2) = t^2 1.5 s^2 = t^2 1.2247 s = t
t = 1.2 x 10^0 s Prob. 2.1.3 Find the uniform acceleration that will cause an object’s speed to change from 32. m/s to 96. m/s in an 8.0 s period. vf = vi + at v final = 96 m/s v initial = 32 m/s t = 8.0 s a= ? 96 m/s = 32 m/s + a(8.0 s) 96 m/s – 32 m/s = a(8.0 s) 64 m/s = a(8.0 s) 64 m/s = a
8.0 s a = 8.0 x 10^0 m/s^2 Prob. 2.1.4 A rocket traveling at +155. m/s is decelerated at a rate of −31. m/s2. (a) How long will it take before the instantaneous speed is 0 m/s? (b) How far will it travel during this time? (c) What will be its velocity after 8.0 s? a) vf = vi + at v initial = 155 m/s v final = 0 m/s a = -31 m/s^2 0 m/s = 155 m/s + (-31 m/s^2) * t -155 m/s = (-31 m/s^2) * t 155 m/s =t (-31 m/s^2) 5s=t
t = 5.0 x 10^1 s b) d = vi + vf * t 2 v initial = 55 m/s v final = 0m/s t = 5.0 s d = (155m/s) + (0 m/s) * 5.0 s 2 d= (155 m/s) * 5.0 s 2 d = 77.5 m/s * 5.0 s d = 387.5 m
d = 3.9 x 10^2 m c) vf = vi + at v initial = 155 m/s a= -31 m/s^2 t = 8.0 s v final = 155 m/s + (-31 m/s^2)(8.0 s) v final = 155 m/s – 248 m/s v final = -9.3 x 10^1m/s prob. 2.1.5 A car with a velocity of 22. m/s is accelerated uniformly at the rate of 1.6 m/s2 for 6.8 s. What is its final velocity? vf = vi + at v initial = 22 m/s a = 1.6 m/s^2 t = 6.8 s v final = 22 m/s + (1.6 m/s^2)(6.8 s) v final = 22 m/s + 10.88 m/s v final = 32.88 m/s
v final = 3.3 x 10^1m/s Prob. 2.1.6 Determine the displacement of a plane that is uniformly accelerated from 66. m/s to 88. m/s in 12.0 s. First solve for the acceleration with the formula vf = vi + at v final = 88. m/s v initial = 66.m/s t= 12.0 s 88 m/s = 66 m/s + a(12.0 s) 22m/s = a(12.0 s) 1.8 m/s^2 Use the acceleration that was solved for in the previous section in the following formula: d = vi * t + 1/2 at2 v initial = 66.m/s t= 12.0 s a = 1.8 m/s^2 d = (66. m/s)(12.0 s) + ½ (1.8m/s^2)(12.0 s)^2 d = 792 m + ½(1.8 m/s^2)(144 s^2) d = 792 m + 129.6 m d = 921.6 m
d = 9.2 x 10^2 m Prob. 2.1.7 A car comes to rest after uniform deceleration at the rate of 9.0 m/s2 for 8.0 s. What distance does it travel during this time? In order to solve for distance, v initial needs to be solved for first by using the formula: vf = vi + at v final = 0 m/s a = 9.0 m/s^2 t = 8.0 s 0 m/s = v initial + (-9.0 m/s^2)(8.0 s) v initial = - (-9.0 m/s^2)(8.0 s) v initial = 72 m/s Now that I have v initial; I will use it in the formula: d = vi * t + 1/2 at2 To solve for distance d = (72 m/s)(8.0 s) + 1/2(-9.0 m/s^2)(8.0 s)^2 d = 576 m + (-4.5 m/s^2)(64 s^2) d = 576m - 288 m d = 288 m
d = 2.9 x 10^2 m Prob. 2.1.8 A plane travels a distance of 5.0 × 102 m while being accelerated uniformly from rest at the rate of +5.0 m/s2. What final speed does it attain? vf2 = vi2 + 2ad v initial = 0 m/s a = +5.0 m/s^2 d = 5.0 x 10^2 m vf2 = (0 m/s)^2 + 2(5.0 m/s^2)(5.0 x 10^2 m) vf2 = (10.0 m/s^2)(5.0 x 10^2 m) vf2 = 50.0 x 10^2 m^2/s^2 v final = 70.7107 m/s
V final = 7.1 x 10^1m/s Prob. 2.1.9 A stone falls freely from rest for 8.0 s. (a) Calculate its final velocity. (b) What distance does the stone fall during this time? a) vf = vi + at v initial = 0 m/s a = 9.8 m/s^2 t = 8.0 s v final = 0 m/s + (-9.8 m/s^2)(8.0 s) v final = -78.4 m/s
V final = -7.8 x 10^1 m/s b) distance d= (v final – v initial)/2 * time (780 m/s – 0 m/s)/2 * 8.0 s = 310 m. distance = 3.1 x 10^2 m Prob. 2.1.10 A weather balloon is floating at a constant height above the earth when it releases a pack of instruments. (a) If the pack hits the ground with a speed of 73.5 m/s, how far does the pack fall? (b) How long does the pack fall? a) vf2 = vi2 + 2ad v initial = 0 m/s v final = 73.5 m/s a = 9.8 m/s^2 (0 m/s)^2 = -(73.5 m/s)^2 + 2(-9.8 m/s^2)d (0 m/s)^2 = -5402.25 m^2/s^2 + (-19.6 m/s^2)d 5402.25 m^2/s^2 = (-19.6 m/s^2)d 5402.25 m^2/s^2 = d (-19.6 m/s^2) -275.625 m = d
c) vf = vi + at d = -2.76 x 10^2 m -73.5 m/s = 0 m/s + (-9.8m/s^2)t -73.5 m/s = (-9.8m/s^2)t -73.5 m/s = t -9.8m/s^2 7.5 s = t
t = 7.50 x 10^0 s