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?

Question

Physics

Andres Flores

1 January, 09:22

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?

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Answers (1)

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Trevon Baird · Answer 1

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