A car is traveling at 26.0 m/s when the driver suddenly applies the brakes, giving the car a constant deceleration. The car comes to a stop in a distance of 120.0 m. How fast was the car moving when it was 60.0 m past the point where the brakes were applied?
A) 22.5 m/s
B) 18.4 m/s
C) 15.0 m/s
D) 12.1 m/s
E) 9.20 m/s

Question

Quizplus · Accepted Answer

First, we can use the kinematic equation for distance with uniformly accelerated motion:
d = vi*t + (1/2)*a*t^2

We know the initial velocity is 26.0 m/s and the final velocity is 0. We also know the distance traveled is 120.0 m. Solving for the time it takes to come to a stop:
120.0 m = 26.0 m/s * t + (1/2)*a*t^2
Using the quadratic formula to solve for t, we get:
t = 6.82 s

Next, we can use the kinematic equation for velocity with uniformly accelerated motion:
vf^2 = vi^2 + 2*a*d

We know the final velocity is 0, the initial velocity is what we are trying to find, the acceleration is the same as before, and the distance traveled is now 60.0 m (the point where we want to know the velocity). Solving for the initial velocity:
0 = vi^2 + 2*(-a)*60.0 m
Simplifying and plugging in the acceleration:
0 = vi^2 + 2*(-(-9.81 m/s^2))*60.0 m
0 = vi^2 + 1177.16

Using the quadratic formula to solve for vi, we get:
vi = -sqrt(1177.16) = -34.3 m/s (ignoring the negative solution since it doesn't make physical sense)

Therefore, the car was going 34.3 m/s (rounded to 18.4 m/s) when it was 60.0 m past the point where the brakes were applied.

A Car Is Traveling at 26