Question

The momentum p of a particle changes with time according to the relation, $\frac{dp}{dt}=10+2tN$. If the momentum is zero at t = 0 s. What will the momentum $\left(in\frac{N}{s}\right)$ at t=10 s.

Answer 1

$\frac{dp}{dt}=10+2t$
$\int dp=\int (10+2t)dt$
$p=10t+\frac{2t^2}{2}+c$
$p=10t+t^2+c$
At $t=0;p=0\Rightarrow c=0$
$\Rightarrow p=10t+t^2$
$\text{p at t}=10\text{s}$
$p=10\times 10+10\times 10=200\frac{N}{s}$