Question

Akaal moving along a long straight road with a speed of 10 metre per second is brought to rest within 10 seconds after applying the break what is the magnitude of the retardation of car

Answer 1

$\color{red}{\large\underline{\underline\mathtt{Question:}}}$

$\textit{Akaal moving along a long straight}$ $\textit{road with a speed of 10 metre per second}$ $\textit{is brought to rest within 10 seconds after applying}$ $\textit{the break what is the magnitude of the retardation of car.}$

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$\color{purple}{\large\underline{\underline\mathtt{To\:Find:}}}$

$\textsf{The retardation of the body}$

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$\color{blue}{\large\underline{\underline\mathtt{Concept:}}}$

• $\textit{In this case , the final velocity will be 0}$

• $\textit{In this case , the acceleration will be negative.}$

• $\textit{In this case , the retardation will be positive.}$

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$\color{yellow}{\large\underline{\underline\mathtt{Given:}}}$

• $\mathtt{v = 0}$
• $\mathtt{u = 10ms^{-1}}$
• $\mathtt{t = 10sec}$

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$\color{cyan}{\large\underline{\underline\mathtt{We\:know:}}}$

$\color{lime}{\mathtt{\underline{\boxed{v = u - at}}}}$

$where,$

• $\mathrm{v \rightarrow final\:velocity}$

• $\mathrm{u \rightarrow initial\:velocity}$

• $\mathrm{a \rightarrow acceleration}$

• $\mathrm{t \rightarrow Time\:taken}$

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$\color{magenta}{\large\underline{\underline\mathtt{Solution:}}}$

$\textsf{Using the equation}$

$\mathtt{v = u - at}$

$\textit{we get ,}$

$\Rightarrow 0 = 10ms^{-1} - a \times 10sec$

$\Rightarrow -10ms^{-1} = -10 a$

$\Rightarrow \dfrac{-10ms^{-1}}{10s}= a$

$\Rightarrow \dfrac{\cancel{-10ms^{-1}}}{\cancel{10s}}= a$

$\Rightarrow 1 ms^{-2} = a$

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$\color{orange}{\mathtt{\underline{\boxed{ a = 1 ms^{-2} }}}}$

Answer 2

Answer:

$\large\tt\color{orange}\underline{\underline{Question:}}$

$\textsf{Akaal moving along a long straight road with a}$

$\textsf{speed of 10 meter per second is brought to}$

$\textsf{rest within 10 seconds after applying the}$

$\textsf{break what is the magnitude of the}$

$\textsf{retardation if car}$

$\large\tt\color{red}\underline{\underline{Given:}}$

$\textsf{▪︎Speed of the car = 10m/s-1}$

$\textsf{▪︎Time taken to rest the car = 10 sec}$

$\large\tt\color{green}\underline{\underline{To\:find:}}$

$\textsf{The Retardation of the car}$

$\large\tt\color{pink}\underline{\underline{Concept:}}$

▪︎Final velocity will be 0.

▪︎Acceleration will be negative because it means retardation.

▪︎And the retradation will be positive.

$\large\tt\color{blue}\underline{\underline{We\:know:}}$

$\textsf{v=u+at}$

$\textsf{V=Final velocity}$

$\textsf{U=Initial velocity}$

$\textsf{A=Acceleration}$

$\textsf{T=Time taken}$

$\large\tt\color{purple}\underline{\underline{Solution:}$

$\textsf{v=u+at}$

$\mapsto{0=10m\:s-1+-a×10\:s}$

$\mapsto{0+10m\:s-1=-10a}$

$\mapsto{\frac{10m\:s-1}{10\:s}=a}$

$\mapsto{a=1\:m\:s-2}$

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Retradation = 1 m s -² .

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