Question

The force on a particle of mass $10g$ is $(10\hat{i}+5\hat{j})N$. If it starts from rest, what would be its position at time $t=5s$?

• A. $(12500\hat{i}+6250\hat{j})m$
• B. $(6250\hat{i}+12500\hat{j})m$
• C. $(12500\hat{i}+12500\hat{j})m$
• D. $(6250\hat{i}+6250\hat{j})m$

Answer 1

The correct option is A $(12500\hat{i}+6250\hat{j})m$

Mass of particle $=10g=10\times {10}^{-3}Kg$

$F=10\hat{i}+5\hat{j}$

$Ma=10\hat{i}+5\hat{j}$

$\frac{10}{1000}a=10\hat{i}+5\hat{j}$

$\Rightarrow a=1000\hat{i}+500\hat{j}$

$\Rightarrow \frac{dV}{dt}=1000\hat{i}+500\hat{j}$

$\Rightarrow dV=1000dt\hat{i}+500dt\hat{j}$

$\Rightarrow V=1000t\hat{i}+500t\hat{j}$

$\Rightarrow \frac{dx}{dt}=100t$ and $\frac{dy}{dt}=500t$

$\Rightarrow \int dx={\int }_0^{5}100tdt\int dy={\int }_0^{5}500tdt$

$x=\frac{100{\left(5\right)}^2}{2}\hat{i}y=\frac{500{\left(5\right)}^2}{2}\hat{j}$

$x=12500\hat{i}y=6250\hat{j}$

position $=12500\hat{i}+6250\hat{j}$

So, the correct answer is 'A'.