A car traveling 56 km/h is 21.0 m from a barrier when the driver slams on the brakes. The car hits the barrier 2.17 s later.

Question

a)What was the car's constant deceleration before impact?

b)How fast was the car traveling at impact?

Answer 1

First you must change 56 km/h to m/s. In m/s the Initial Velocity is 15.56 m/s. Next use the equation:
S = ViT + (1/2)at^2 and solve for a

21 = (15.56)(2.17) + (1/2)(a)(2.17)^2

a = -5.42 m/s^2

Now that you have the acceleration you can solve for the final velocity using either:

Vf = Vi + at ---or--- Vf^2 = Vi^2 + 2as

Vf = 15.56 + (-5.42)(2.17) = 3.8 m/s
or
Vf^2 = 15.56^2 + 2(-5.42)(21) = 3.8 m/s

As you can see each equation gave the same answer. Overall your answers would be:

a.) -5.42 m/s^2
b.) 3.8 m/s

Answer 2

thanks