The distance ' $s$ ' in meters covered by a body in $t$ seconds is given by $s=3 t^2-8 t+5$. The body will stop after

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Aicharya · Accepted Answer

$\begin{aligned} & s=3 t^2-8 t+5 \ 	herefore \quad & \frac{d s}{d t}=6 t-8 \end{aligned}$ when body stops, $\frac{\mathrm{ds}}{\mathrm{dt}}=0$ $	herefore \quad 6 \mathrm{t}-8=0 \Rightarrow t=\frac{4}{3}$

The distance ' $s$ ' in meters covered by a body in $t$ seconds is given by $s=3 t^2-8 t+5$. The body will…

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