Question

when a driver hits the brakes, his car declelerates at a uniform rate of 2.0m/s^2. his car stops in 9.0m how long does it take him to stop​

Answer 1

Time taken to stop: 3.0 s

Explanation:

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

$s=vt-\frac{1}{2}at^2$

where

s is the displacement

v is the final velocity

t is the time

a is the acceleration

For the car in this problem, we have:

s = 9.0 m (displacement)

v = 0 (the car comes to a stop)

$a=-2.0 m/s^2$ (deceleration)

t is time

And solving for t, we find the time the car rakes to stop:

$t=\sqrt{-\frac{2s}{a}}=\sqrt{\frac{2(9.0)}{-2.0}}=3.0 s$

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