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A car braked with a constant deceleration of 16 ft per sec squared producing skid marks measuring 200ft before coming to a stop How fast was the car traveling when the brakes were first applied?
Wiki User
∙ 15y ago
If you have taken physics, you should know the basic kinematic formula
vf²=vi²+2ad
0=vi²-2*16*200
vi=80m/s
Alternatively, if you haven't taken physics yet, you will have to derive an equivalent formula on your own.
We know that the integral of acceleration is equal to velocity so integrate -16 with respect to time to get
v=-16t + vi
for some constant vi.
We also know that the integral of velocity with respect to time is distance so
d=-8t²+vi*t+c
for a constant value c.
Since the distance traveled by the car was 0 at t=0, the value of c is 0.
We also know that there was a constant acceleration so
a=(vf-vi)/t
-16=-vi/t
vi=16t
Combining this with the preceding equation for distance gives us
d=-8t²+vi*t
200=-8t²+16t²
t=5s (reject the negative root)
vi=16t
vi=80m/s
Wiki User
∙ 15y ago
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