A car is traveling at 26.0 m/s when the driver suddenly applies the brakes,causing the car to slow down with constant acceleration.The car comes to a stop in a distance of 120 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

We can solve this problem using the kinematic equation:

v^2 = u^2 + 2as

Where: 
v = final velocity (0 m/s) 
u = initial velocity 
a = acceleration 
s = displacement (120 m)

We can rearrange this equation to solve for u:

u = sqrt(v^2 - 2as)

u = sqrt((26.0 m/s)^2 - 2(−a)(120 m))

We know that the acceleration is negative because the car is slowing down:

a = - v^2 / 2s
a = - (26.0 m/s)^2 / (2 x 120 m)

a = -11.05 m/s^2

Substituting this value into the previous equation:

u = sqrt((26.0 m/s)^2 - 2(-11.05 m/s^2)(120 m))

u = 18.4 m/s

Therefore, the car was traveling at 18.4 m/s when it was 60.0 m past the point where the brakes were applied. The answer is B.

A Car Is Traveling at 26