The displacement of a particle at the time $\mathrm{t}$ is given by $\quad s=\sqrt{1+t}$, then its…
The displacement of a particle at the time $\mathrm{t}$ is given by $\quad s=\sqrt{1+t}$, then its
acceleration 'a' is proportional to
square of the velocity
$\sqrt[3]{S}$
$\sqrt{S}$
cube of the velocity
Solution
Given $s=\sqrt{1+t}$
Differentiating with respect to $\mathrm{t}$, we get
$\begin{aligned}
\mathrm{v} &=\frac{\mathrm{ds}}{\mathrm{dt}}=\frac{1}{2 \sqrt{1+\mathrm{t}}} \\
\mathrm{a} &=\frac{\mathrm{d}^{2} \mathrm{~s}}{\mathrm{dt}^{2}}=\frac{1}{2} \frac{\mathrm{d}}{\mathrm{dt}}(1+\mathrm{t})^{\frac{-1}{2}}=\frac{1}{2}\left(-\frac{1}{2}\right)(1+\mathrm{t})^{\frac{-3}{2}} \\
&=\left(\frac{-1}{4}\right)\left(\frac{1}{(1+\mathrm{t})^{\frac{3}{2}}}\right)=-2\left(\frac{1}{\left[(1+\mathrm{t})^{\frac{1}{2}}\right]^{3}}\right)=-2\left(\frac{1}{2 \sqrt{1+\mathrm{t}^{2}}}\right)^{3}=-2 \mathrm{v}^{3}
\end{aligned}$