Kinematics - How fast does the car stop?

[kenji1992]

Kinematics -- How fast does the car stop?

Homework Statement

A car is moving with a speed of 32.0 m/s. The driver sees an accident ahead and slams on the brakes, giving the car an acceleration of -3.50 m/s2. How far does the car travel after the driver put on the brakes before it comes to a stop?

vi=32 m/s

vf=

t=

a=-3.5 m/s2

delta-x=

Homework Equations

The Attempt at a Solution

How would this problem be solved? Does it require two steps?

Is Vf=0m/s and Vi= 32 m/s?

Because if that's the case, wouldn't I use the formula a=vf-vi/t

So:

-3.5 m/s2=(0 m/s - 32 m/s)/t

-3.5 m/s2 = (-32 m/s)/t

t= -32 m/s/-3.5 m/s2

t= 9.14 s

Then to find distance, use s=d/t, rearrange as d=s*t

d=32 m/s*9.14 s

d=292 m

 

[lewando]

Your "t" is correct, but your "d" is not. d=s*t is valid only when you have a constant s (rate), which is not the case here-- s is decreasing. Pick a relevant kinematic equation and you will be home free.

 

[kenji1992]

d=0.5(vi+vf)*t
d=0.5(0 m/s + 32 m/s)*9.14 s
d=146.24 m

 

[lewando]

Looks good. Watch your significant digits.