The motion of a particle along a straight line is described by equation $x=8+12 t-t^3$ where, $x$ is in metre and $t$ in second. The retardation of the particle when its velocity becomes zero, is

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Aicharya · Accepted Answer

Given, $x=8+12 t-t^3$
We know $v=\frac{d x}{d t}$
and $a=\frac{d v}{d t}$
So, $v=12-3 t^2$
and $a=-6 t$
At $t=2 \mathrm{~s}$
$y=0$ and $a=-6 t$
So, $a=-12 \mathrm{~m} / \mathrm{s}^2$
So, retardation of the particle $=12 \mathrm{~m} / \mathrm{s}^2$.

The motion of a particle along a straight line is described by equation $x=8+12 t-t^3$ where, $x$ is in…

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