Question

A particle moves in a straight line with an acceleration $a{\text{m/s}}^2$ at time $t$ second where $a=-\frac{1}{t^2}$. At time $t=1\text{s}$ the particle has velocity $3\text{m/s}$ then the velocity at $t=4\text{s}$ is $x\times {10}^{-2}\text{m/s}$ where $x$ is positive integer. The value of $x/5$ is ______.

Answer 1

Given:
$a=-\frac{1}{t^2}$
At $t=1\text{s},u=3\text{m/s}$
At $t=4\text{s},v=x\times {10}^{-2}\text{m/s}$

Now,
$a=\frac{dv}{dt}=-\frac{1}{t^2}$

$\Rightarrow {\int }_3^{v}dv=-{\int }_1^4\frac{dt}{t^2}$

$\Rightarrow {\left[v\right]}_3^v={\left[\frac{1}{t}\right]}_1^4$

$\Rightarrow v-3=\frac{1}{4}-1$

$\Rightarrow v=\frac{9}{4}\text{m/s}=225\times {10}^{-2}\text{m/s}=x\times {10}^{-2}\text{m/s}$

$\therefore x=225$

Hence, $x/5=45$