Question

A car starts out traveling at 35 m/s. The car hits the brakes and decelerates at a rate of 3 m/s^2 for 5 seconds. What Distance does the car travel during this period of time?

Answer 1

Answer:

Time needed: 2.5 s
Distance covered: 31.3 m

Explanation:

I'll start with the distance covered while decelerating. Since you know that the initial speed of the car is 15.0 m/s, and that its final speed must by 10.0 m/s, you can use the known acceleration to determine the distance covered by

v2f=v2i−2⋅a⋅d

Isolate d on one side of the equation and solve by plugging your values

d=v2i−v2f2a

d=(15.02−10.02)m2s−22⋅2.0ms−2

d=31.3 m

To get the time needed to reach this speed, i.e. 10.0 m/s, you can use the following equation

vf=vi−a⋅t, which will get you

t=vi−vfa

t=(15.0−10.0)ms2.0ms2=2.5 s