First slide

Integration

Question

The momentum $p$ of a particle change with time t according to the relation $\frac{dp}{dt} = (10 + 2 t) .$ If the momentum is zero at t = 0, then the momentum at $t = 10 s is$

Moderate

A

$200 {kgms}^{- 1}$
B

$300 {kgms}^{- 1}$
C

$300 kgms$
D

$400 {kgms}^{- 1}$

Solution

Given
$\begin{array}{rcl} \frac{dp}{dt} & = & (10 + 2 t) \\ dp & = & (10 + 2 t) dt \\ \Rightarrow & \int_{o}^{p} dp = \int_{0}^{10} (10 + 2 t) dt \\ \Rightarrow & p = {(10 t + t^{2})}_{0}^{10} \\ \Rightarrow & p = [10 (10) + {(10)}^{2}] - [10 (0) + {(0)}^{2}] \\ \Rightarrow & p = 200 {kgms}^{- 1} \end{array}$

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