Question

the displacement of a partial starting from rest (at t=0)is given by S=6t²-t³ the time in seconds at which the particle will attain zero velocity again is ​

Answer 1

$\sf{\underline{\boxed{\large{\blue{\mathsf{Solution}}}}}}$

$\sf{\bold{\green{\underline{\underline{Given}}}}}$

• $\sf{\bold{ S = 6t^2 - t^3 }}$

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$\sf{\bold{\green{\underline{\underline{To\:Find}}}}}$

• Time in which the particle will attain zero velocity .

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$\sf{\bold{\green{\underline{\underline{Solution}}}}}$

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★$\sf{\red{\boxed{\bold{V = \dfrac{d_s}{d_t}}}}}$

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: $\sf\implies \: {\bold { V = \dfrac{d ( 6t^2 - t^3 )} { dt } }}$

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: $\sf\implies \: {\bold { V = \dfrac{ 6dt^2 - dt^3 )} { dt } }}$

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: $\sf\implies \: {\bold { V = \dfrac{6dt^2} { dt } - \dfrac{ dt^3 }{ dt }}}$

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: $\sf\implies \: {\bold { V = 6d^{1-1}t^{2-1} - d^{1-1}t^{3-1}}}$

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: $\sf\implies \: {\bold { V = 6t - t^2 }}$

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• Here in question it is asked that when velocity will be zero , therefore V = 0

: $\sf\implies \: {\bold { 0 = 6t - t^2 }}$

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: $\sf\implies \: {\bold { 0 = t ( 6 - t ) }}$

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: $\sf\implies \: {\bold { ( t ) ( 6 - t ) = 0 }}$

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$\begin{tabular}{|c|c|}\cline{1-2}\sf t = 0 &\sf 6 - t = 0 \\\cline{1-2}\sf t = 0 &\sf t = 6 - 0 \\\cline{1-2}\sf t = 0 &\sf t = 6 \\\cline{1-2}\end{tabular}$

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: $\sf\implies \: {\bold { t = 6sec \: or \: 1sec }}$

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$\sf{\bold{\green{\underline{\underline{Answer}}}}}$

• Time in which the particle will attain zero velocity is 0 or 6 seconds .
• In the question 0 velocity is at 0 seconds ( given )
• Therefore the answer is 6 seconds .