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Solve the following :


The momentum $\rho$ of a particle changes with time $t$ according to the relation $\frac{a \rho}{d t}=(10 \mathrm{~N})+(2 \mathrm{~N} / \mathrm{s}) t$ the momentum is zero at $\mathrm{t}=0$, what will be the momentum be at $\mathrm{t}=10$ s?


It is given that $\frac{\frac{d p}{d t}}{}=(10 \mathrm{~N})+\left(\frac{2 N}{a}\right) t$

Now, $p=0$ at $t=0$. So,

$p=\int_{0}^{10}(10 d t+2 t d t)=10(10-0)+(100-0)=200 \mathrm{~N} / \mathrm{s}=200 \mathrm{~kg} \mathrm{~m} / \mathrm{s}$


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