Question

a car is moving with a uniform velocity 72 Km/hr . on applying breaks it comes to rest after traveling a distance of 200m. calculate the retardation and tike taken to come to rest.​

Answer 1

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Answer 2

$\bf{\underline{\underline \blue{Solution:-}}}$

$\pink{\bf{Given}\begin{cases}\sf{ \red{Initial\: velocity = }72km/h\: }\\\sf{ \green{Distance \: travelled = 200m}\: }\\ </p><p>\sf{ \red{Final\: velocity = } 0\: }\\</p><p>\sf{ \green{Retardation \:of\: the\: car =\: ?}\: }\\ </p><p>\sf{ \red{Time \:taken \:to\: come \:in \:rest = } \:?\: }\\</p><p>\end{cases}}$

$\bf{\underline{\underline \blue{Explanation:-}}}$

Initial velocity = 72km/h

Initial velocity = 72 × 5/8

Initial velocity = 20m/s

⠀⠀

$\sf\underline{\red{\:\:\: By\:the\:3rd\: equation\:of\: motion:-\:\:\:}}$

$\longrightarrow{\underline{\boxed{\sf\purple{v^2 = u^2 + 2as}}}}$

$\sf\underline{\green{\:\:\: Now,Putting\:the\: values:-\:\:\:}}$

$\dashrightarrow \sf {(0)^2 = (20)^2 + 2 \times a \times 200}$

$\dashrightarrow \sf {0 = 400 + 400a }$

$\dashrightarrow \sf {-400 = 400a}$

$\dashrightarrow \sf {a = \cancel\dfrac{-400}{400}}$

$\dashrightarrow {\boxed{\frak{\pink{ a = -1m/s^2}}}}$

$\sf\underline{\green{\:\:\: ThereFore:-\:\:\:}}$

• The Retardation of the car = 1m/s²

$\rule{200}{2}$

$\bullet\;\;\underline{\textbf{\textsf{Now:-}}}$

$\sf\underline{\red{\:\:\: By\:the\:1st\: equation\:of\: motion:-\:\:\:}}$

$\longrightarrow{\underline{\boxed{\sf\purple{v = u + at}}}}$

$\sf\underline{\green{\:\:\: Now,Putting\:the\: values:-\:\:\:}}$

$\dashrightarrow \sf {0 = 20 + (-1) \times t}$

$\dashrightarrow \sf {0 = 20 - t}$

$\dashrightarrow {\boxed{\frak{\pink{Time = 20\: seconds}}}}$

$\sf\underline{\green{\:\:\: ThereFore:-\:\:\:}}$

• The time taken to come in rest = 20 seconds

$\rule{200}{2}$