Question

A time varying $P=2t$ is applied on a particle of mass m. Find average power over a time interval from t=0 to t=t:
A.${ P }_{ av }=t$
B. ${ P }_{ av }=2t$
C. ${ P }_{ av }=4t$
D. ${ P }_{ av }=8t$

Answer 1

Use the formula for average power over a time interval. Substitute the values given in the question. And then integrate it. The answer obtained is average power over a time interval.

Complete answer:
Given: Power P= 2t
Average power over a time interval from t=0 to t=t is given by,
${ P }_{ avg }=\dfrac { \int _{ 0 }^{ t }{ P } dt }{ \int _{ 0 }^{ t }{ dt } }$
Substituting the value above we get,
${ P }_{ avg }=\dfrac { \int _{ 0 }^{ t }{ 2t } dt }{ \int _{ 0 }^{ t }{ t } }$
Integrating the above expression, we get,
${ P }_{ avg }=\dfrac { { { \left[ \dfrac { 2{ t }^{ 2 } }{ 2 } \right] }_{ 0 }^{ t } } }{ t }$
$\therefore{ P }_{ avg }=\dfrac { { { t }^{ 2 } } }{ t }$
$\therefore{ P }_{ avg }=t$
Therefore, the average power over a time interval from t=0 to t=t is t.

Hence, the correct answer is option A i.e. ${ P }_{ av }=t$.

Note:
The average power in a short interval of time at a particular instant is called Instantaneous Power. Formula for average power is also given by,
${ P }_{ avg }=\dfrac{\delta W}{\delta t}$
Where, $\delta W$ is the amount of work done during a time period of $\delta t$.