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

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(A) Given velocity $v = 6t - \frac{t^2}{6}$.We know that $v = \frac{ds}{dt}$,therefore $ds = v \, dt$.Integrating both sides: $\int ds = \int (6t - \f

Vedclass · Accepted Answer

$\frac{51}{2} 	ext{ units}$

Vedclass · Answer

(A) Given velocity $v = 6t - \frac{t^2}{6}$.We know that $v = \frac{ds}{dt}$,therefore $ds = v \, dt$.Integrating both sides: $\int ds = \int (6t - \frac{t^2}{6}) dt$.$s = 6 \frac{t^2}{2} - \frac{1}{6} \frac{t^3}{3} + C = 3t^2 - \frac{t^3}{18} + C$.Given that $s = 0$ at $t = 0$,substituting these values gives $0 = 3(0)^2 - \frac{(0)^3}{18} + C$,so $C = 0$.Thus,the displacement equation is $s = 3t^2 - \frac{t^3}{18}$.To find the distance travelled in $3 	ext{ s}$,we calculate $s$ at $t = 3$:$s(3) = 3(3)^2 - \frac{(3)^3}{18} = 3(9) - \frac{27}{18} = 27 - \frac{3}{2} = \frac{54 - 3}{2} = \frac{51}{2} 	ext{ units}$.

The velocity of a particle at time $t$ is given by the relation $v = 6t - \frac{t^2}{6}$. Its displacement $S$ is zero at $t = 0$,then the distance travelled in $3 \text{ s}$ is

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