A particle moves along a straight line according to the law $s=\frac{1}{3} t^3-3 t^2+9 t+17$, where $s$ is in metre and $t$ is in second. Its velocity decreases in

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Aicharya · Accepted Answer

We have, $ \begin{aligned} & S=\frac{1}{3} t^3-3 t^2+9 t+17 \ & V=\frac{d s}{d t}=t^2-6 t+9 \ & V=t^2-6 t+9 \ & \frac{d V}{d t}=2 t-6 \end{aligned} $ Velocity is decreasing, if $\frac{d V}{d t} 0] \end{array} $

A particle moves along a straight line according to the law $s=\frac{1}{3} t^3-3 t^2+9 t+17$, where $s$ is…

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