It is given that $\mathrm{t}=\mathrm{p} x^{2}+\mathrm{q} x$, where $x$ is displacement and $\mathrm{t}$ is time. The acceleration of particle at origin is

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Aicharya · Accepted Answer

$t=px^2+qx$
Differentiate with respect to time,
$1=2 px \frac{dx}{dt}+q \frac{dx}{dt}$
or $2 p x v+q v=1$
At $x=0, q v=1$ or $v=\frac{1}{q}$
Again differentiate with respect to time,
Hence, $2 p x a+2 p v^2+q a=0$
At $x=0$, put $v=\frac{1}{q}$
Thus, we get $a=\frac{-2 p}{q^3}$

It is given that $\mathrm{t}=\mathrm{p} x^{2}+\mathrm{q} x$, where $x$ is displacement and $\mathrm{t}$ is…

Solution