Question

PLEASE HELP !! KINEMETICS

A car is traveling at -18 m/s when the driver sees a disabled car in the middle of the road. She takes 0.8 s to react (assume that the car travels at constant speed during this reaction time). She then steps on the brakes and slows at a rate of 9.0 m/s2. How much time does the car need to stop? How far does the car go before it stops? This is a multi step problem.

Please show work !!

Answer 1

Explanation:

Given that,

The initial speed of the car, u = -18 m/s

Reaction time, t = 0.8 s

The acceleration of the car is 9 m/s²

We need to find the time taken by the car to stop. Using first equation of motion to find it as follows :

v = u +at

Put v = 0

$-u=at\\\\t=\dfrac{-u}{a}\\\\t=\dfrac{-(-18)}{9}\\\\t=2\ s$

Let d is the distance covered by the car before it stops. Using third equation of kinematics as follows :

$v^2-u^2=2ad\\\\d=\dfrac{-u^2}{2a}\\\\d=\dfrac{-(-18)^2}{2\times -9}\\\\d=18\ m$

Hence, it will take 2 seconds to stop the car and the distance covered is 18 m.