The velocity of a particle at time t is given | vedclass

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(C) Given,$v = \frac{ds}{dt} = 6t - \frac{t^2}{6}$.Integrating both sides with respect to $t$,we get:$s = \int (6t - \frac{t^2}{6}) dt = 3t^2 - \frac{

Vedclass · Accepted Answer

$\frac{51}{2}$

Vedclass · Answer

(C) Given,$v = \frac{ds}{dt} = 6t - \frac{t^2}{6}$.Integrating both sides with respect to $t$,we get:$s = \int (6t - \frac{t^2}{6}) dt = 3t^2 - \frac{t^3}{18} + C$.Given that $s = 0$ at $t = 0$,we substitute these values to find the constant $C$:$0 = 3(0)^2 - \frac{(0)^3}{18} + C \implies C = 0$.Thus,the displacement function is $s(t) = 3t^2 - \frac{t^3}{18}$.To find the distance traveled in $3 \ s$,we calculate $s(3)$:$s(3) = 3(3)^2 - \frac{(3)^3}{18} = 3(9) - \frac{27}{18} = 27 - \frac{3}{2} = \frac{54 - 3}{2} = \frac{51}{2}$.

The velocity of a particle at time $t$ is given by the relation $v = 6t - \frac{t^2}{6}$. The distance traveled in $3 \ s$ is,if $s = 0$ at $t = 0$.

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