Question

A car moving along a straight highway with speed of 126 kmh–1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform) and how long does it take for the car to stop ?​

Answer 1

$\underline{ \mathtt{Given}}$

• initial velocity of the car = 126 km / hr
• final velocity of the car = 0 km/hr
• distance = 200 m

$\underline{ \mathtt{Unit \: conversion }}$

in order to convert km/hr to m /s multiply by 5 / 18

initial velocity

$u = 126 \times \frac{5}{18} = 35 \frac{m}{s}$

final velocity = 0

$\underline{ \mathtt{To \: find}}$

• retardation
• time taken by the car to stop

$\underline{ \mathtt{Solution}}$

By using the third kinematic equation ,

$\star\displaystyle\boxed{ \green {\mathtt{ {v}^{2} = {u}^{2} + 2as}}}$

here ,

• v= final velocity
• u = initial velocity
• a = acceleration
• s = distance

Substituting the given values in the above equation

$\displaystyle\implies {\mathtt{ {u}^{2} = - 2as}}$

$\displaystyle\implies{\mathtt{ a = - \frac{1225}{2 \times 200} }}$

$\displaystyle\boxed{ \red{\mathtt{ a = - 3.06 \: \frac{m}{ {s}^{2} } }}}$

Note : here negative sign denotes retardation

By using first kinematic equation

$\star\displaystyle\boxed{ \green{\mathtt{v = u + at}}}$

here ,

• t = time taken

Substituting the given values in the above equation ,

$\displaystyle\implies{\mathtt{ u = - at}}$

$\displaystyle\implies{\mathtt{ t = \frac{35}{ - ( - 3.06)} }}$

$\displaystyle\boxed{ \red{\mathtt{ t = 11.43 \: sec}}}$