Kinematics

1. The problem statement, all variables and given/known data
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=

2. Relevant equations

3. 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

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