Acceleration Kinematics

Question

1. A truck is travelling at 22 m/s when the driver notices a speed limit sign for the town ahead. He slows down to a speed of 14 m/s. He travels a distance of 125 m while he is slowing down.
2. How long did it take the truck driver to change his speed?
3. Calculate the acceleration of the truck.
4. An arrow is accelerated for a distance of 75 cm [fwd] while it is on the bow. If the arrow leaves the bow at a velocity of 75 m/s [fwd], what is the average acceleration while on the bow?
5. In a test run, a rocket-powered car is driving down a test track at a constant speed when the rockets are fired. It then accelerates at 4 m/s² for 8.0 s covering a distance of 224 m. At what speed was the car travelling when the rockets were fired?

Answer 1

1. Truck
2. We know that distance is equal to average velocity times time. If the truck slowed from 22 m/s to 14 m/s, the average velocity was (22+14)/2 = 18 m/s (using the mean formula). So, using Δx = vΔt, we find 125 = 18 * t ⇒ t = 125/18 = 6.94 s
3. We know that a = Δv/Δt and we just solved for Δt = 6.94s. We also know that Δv = v_f - v_i = 14-22 = -8 m/s. So a = Δv/Δt = -8/6.94 = -1.15 m/s²
4. We use the equation v_f² = v_i² + 2aΔx. From the problem we are given v_i = 0 m/s (because the arrow starts stationary), v_f = 75 m/s, and Δx = 0.75 m. Plugging in we find 75² = 0 + 2*a*0.75 ⇒ 75² / 2*0.75 = a ⇒ a = 3750 m/s²
5. We use the equation x_f = x_i + v_i*t + 1/2*a*t². We can assume the object starts at some initial position x_i = 0 m, making x_f = 224 m. Plugging in the given variables we get: 224 = 0 + v_i*8 + 1/2*4*8² ⇒224 = 8*v_i + 128 ⇒ 224 - 128 = 8*v_i ⇒ 96/8 = v_i ⇒ v_i = 12 m/s