Question

A car is traveling with a speed of 22 m/s. The brakes can produce a maximum deceleration of 8.0 m/s. What is the minimum stopping distance for the car?​

Answer 1

Explanation:

In the given question,

Initial speed = 22m/s

Deceleration = 8m/s

Final speed = 0m/s [because the car is supposed to stop, i.e. come to rest]

We shall express deceleration as an acceleration of -8m/s.

Hence, by the equation $\mathsf{2as=v^2-u^2}$, where s is the distance covered,

$\mathsf{2\times(-8)\times{s}=0^2-22^2}$

$\mathsf{==>0-484=-16s}$

$\mathsf{==>s=\frac{-484}{-16}}$

∴  $\mathsf{s=30.25\:m}$

Hence, 30.25 m is the minimum stopping distance for the car.

Answer: 30.25 metres