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.

 

[HalfLostAndSearching]

Therefore, the car travels 292 meters before coming to a stop. This solution assumes that the car stops at 0 m/s and that there are no other external forces acting on the car. It also assumes that the acceleration remains constant throughout the braking process. If these assumptions are not met, the solution may vary.