After $t$ seconds, the acceleration of a particle, which starts from rest and moves in a straight line is $\left(8-\frac{\mathrm{t}}{5}\right) \mathrm{cm} / \mathrm{s}^2$, then velocity of the particle at the instant, when the acceleration is zero, is

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Aicharya · Accepted Answer

Acceleration $=\left(8-\frac{\mathrm{t}}{5}\right) \mathrm{cm} / \mathrm{s}^2$ $\Rightarrow \frac{\mathrm{dv}}{\mathrm{dt}}=8-\frac{\mathrm{t}}{5}$
Integrating on both sides, we get $v=8 t-\frac{t^2}{10}+c...(i)$
At $t=0, v=0$ $\begin{array}{ll} \therefore & 0=8(0)-0+c \Rightarrow c=0 \\ \therefore & v=8 t-\frac{t^2}{10}...ii[From(i)] \end{array}$
Acceleration $=0$ $\begin{aligned} & \Rightarrow \frac{\mathrm{dv}}{\mathrm{dt}}=0 \\ & \Rightarrow 8-\frac{\mathrm{t}}{5}=0 \\ & \Rightarrow \mathrm{t}=40 \end{aligned}$
Substituting $\mathrm{t}=40$ in (ii), we get Velocity $(v)=8(40)-\frac{(40)^2}{10}=160 \mathrm{~cm} / \mathrm{s}$

After $t$ seconds, the acceleration of a particle, which starts from rest and moves in a straight line is…

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