Question

A car moving along a straight road with a
speed of 120 km/hr, is brought to rest by
applying brakes. The car covers a distance
of 100 m before it stops. Calculate (i) the
average retardation of the car (ii) time
taken by the car to come to rest.
[Ans: 50/9 m/sec, 6 sec)

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Answer 1

$\huge\mathcal{\fcolorbox{cyan}{black}{\pink{ANSWER}}}$

Answer 2

Given : Car has a velocity of 120km/h and came to rest. [v = 0] & It covers a distance of 100m before it stops.

To Find : Find The average retardation of the car & Time taken by the car to come to rest ?

________________________

Solution : Initial velocity (u) = 120km/h

$~~~~~~~~~~~~~~~~~$Final velocity (v) = 0

$~~~~~~~~~~~~~~~~~$Distance covered (s) = 100m

• $\leadsto{\sf{u~=~120~×~\dfrac{5}{18}}}$

• $\leadsto{\sf{u~=~\dfrac{100}{3}m/s}}$

$~$

$\pmb{\sf{According ~to~ third~equation~of~motion~:}}$

$~$

${\sf\qquad\qquad:\implies{v^2~=~u^2~+~2as}}$

${\sf\qquad\qquad:\implies{(0)^2~=~\bigg(\dfrac{100}{3}\bigg)^2~+~2~×~a~×~100}}$

${\sf\qquad\qquad:\implies{0~=~\dfrac{10000}{9}~+~200a}}$

${\sf\qquad\qquad:\implies{-~\dfrac{10000}{9}~=~200a}}$

${\sf\qquad\qquad:\implies{a~=~\dfrac{- 10000}{9}~×~200}}$

${\sf\qquad\qquad:\implies{a~=~\dfrac{- 100}{18}}}$

$\qquad\qquad:\implies{\underline{\boxed{\frak{\purple{a~=~\dfrac{- 50}{9}}}}}}$★

$~$

Therefore,

• $\therefore\underline{\sf{Retardation~ of ~a ~car ~is ~\bf{\underline{\dfrac{50}{9}}}\underline{m/s}}}$

$~$

$\pmb{\sf{First~ equation ~of~ motion~:}}$

$~$

${\sf\qquad\qquad:\implies{v~=~u~+~at}}$

${\sf\qquad\qquad:\implies{0~=~\dfrac{100}{3}~+~\dfrac{(- 50)}{9}~×~t}}$

${\sf\qquad\qquad:\implies{0~=~\dfrac{100}{3}~-~\dfrac{- 50t}{9}}}$

${\sf\qquad\qquad:\implies{\dfrac{- 50t}{9}~=~\dfrac{100}{3}}}$

${\sf\qquad\qquad:\implies{t~=~9~×~\dfrac{100}{50}~×~3}}$

${\sf\qquad\qquad:\implies{t~=~3~×~2}}$

$\qquad\qquad:\implies{\underline{\boxed{\frak{\pink{t~=~6}}}}}$★

$~$

Hence,

$\therefore\underline{\sf{Time~Taken~=~\bf{\underline{6~second}}}}$

$~$

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V E R I F I C A T I O N :

${\rm\qquad\qquad\dashrightarrow{v~=~u~+~at}}$

${\rm\qquad\qquad\dashrightarrow{0~=~\dfrac{100}{3}~+~\bigg(\dfrac{- 50}{9}~×~6\bigg)}}$

${\rm\qquad\qquad\dashrightarrow{0~=~\dfrac{100~×~3}{3~×~3}~-~\dfrac{300}{9}}}$

${\rm\qquad\qquad\dashrightarrow{0~=~\dfrac{300}{9}~-~\dfrac{300}{9}}}$

$~~~~~~~~~~~~~~~~\dashrightarrow\large\pmb{\frak\red{0~=~0}}$