This question is typical on some driver license exams: A car moving at 43 km/h skids 17 m with locked brakes. How far will the c

Question

2 years ago

12

ar skid with locked brakes at 129 km/h? Assume that energy loss is due only to sliding friction.
Answer in units of m.

1 answer:

3241004551 [841] · Answer 1 · 2022-07-16T09:22:02+03:00

The new stopping distance is 152.6 m

Explanation:

The motion of the car is a uniformly accelerated motion, so we can use the following SUVAT equation:

$v^2-u^2=2as$

where in the initial situation, we have:

v = 0 is the final velocity of the car

u = 43 km/h = 11.9 m/s is the initial velocity

s = 17 m is the stopping distance

a is the acceleration

Solving for a, we find the acceleration:

$a=\frac{v^2-u^2}{2s}=\frac{0-(11.9)^2}{2(17)}=-4.2 m/s^2$

The acceleration depends only on the sliding friction: this means that it is constant also in the second situation, where we have

u = 129 km/h = 35.8 m/s

So we can use the same equation again to find s, the new stopping distance:

$s=\frac{v^2-u^2}{2a}=\frac{0-(35.8)^2}{2(-4.2)}=152.6 m$

Learn more about accelerated motion:

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