Question

An object moving at 40 m/s decelerates steadily at 2.8 m/s². Calculate:

a) The distance traveled by the object before it comes to rest.
b) The time it takes to stop.
c) The displacement it traveled during the last 2 seconds.

A. a) 200 m, b) 14.29 s, c) 28.57 m
B. a) 400 m, b) 14.29 s, c) 28.57 m
C. a) 200 m, b) 14.29 s, c) 57.14 m
D. a) 400 m, b) 14.29 s, c) 57.14 m

Answer 1

The distance traveled before the object comes to rest is incorrect in the provided options, and it should be 285.71 m. The time it takes to stop is 14.29 s. The displacement in the last 2 seconds is 0 m, since the object stops right at the 2-second mark.

Step-by-step explanation:

To solve for the distance, time, and displacement, we can use the kinematic equations for uniformly accelerated motion.

Part a) Distance traveled before coming to rest

We'll use the equation v^2 = u^2 + 2as, where v is the final velocity (0 m/s), u is the initial velocity (40 m/s), a is the acceleration (-2.8 m/s²), and s is the distance. Solving for s, we get s = (v^2 - u^2) / (2a) = (0 - 1600) / (-5.6) = 285.71 m.

Part b) Time to stop

Using the equation v = u + at, and solving for t, we find t = (v - u) / a = (0 - 40) / (-2.8) = 14.29 s.

Part c) Displacement in the last 2 seconds

We calculate the initial speed for the last 2 seconds using v = u + at again, but t here will be the time until 2 seconds before the object stops, so t = 14.29 - 2 = 12.29 s.

The initial velocity for the last 2 seconds is u = 40 + (-2.8)(12.29) = 5.6 m/s.

Then we use s = ut + (1/2)at^2 for the last 2 seconds to get s = (5.6)(2) + 0.5(-2.8)(2^2) = 11.2 + (-11.2) = 0 m as the displacement during the last 2 seconds since the object comes to rest exactly at 2 seconds before stopping.