Define average power.

Hint: We know that pressure is force applied per unit area. But when there are many forces acting, work is done by those forces for a certain amount of time. Hence we cannot check what power is given out by individual forces. Hence we use average power for the same.

Complete step-by-step solution:
Let us start with what power is : power is force: per unit area. It is expressed as :
\[P = \dfrac{F}{A}\].its SI unit is ‘watt’
 But it can also be defined as the amount of work done per unit time:
\[P = \dfrac{W}{t}\]
When there are many forces acting some amount of work is done by each force and to calculate power at such times we use the term average power. Let’s look at its definition:
Average power is defined as the ratio of total work done to the total time taken by the body. Let us look down at its expression:
Average power:\[{P_{avg}} = \dfrac{{\Delta W}}{{\Delta t}}\]
Where, \[\Delta W\]- work done in time \[\Delta t\]and,
\[{P_{avg}}\]-average power.

ADDITIONAL INFORMATION:
instantaneous power: it is defined as power delivered by a body at an instant (when time approaches to 0) . expression:
 \[P = \mathop {\lim }\limits_{\Delta t \to 0} {P_{avg}}\]
\[P = \mathop {\lim }\limits_{\Delta t \to 0} \dfrac{{\Delta W}}{{\Delta t}}\]
\[P = \dfrac{{\delta W}}{{\delta t}}\]
power can also be written in terms of force and velocity:
\[W = \int {F.dr} \]
\[W = \int\limits_{\Delta t} {F.\dfrac{{dr}}{{dt}}dt = \int\limits_{\Delta t} {F.vdt} } \]
And \[P = \dfrac{{\delta W}}{{\delta t}} = \dfrac{\delta }{{\delta t}}\int\limits_{\Delta t} {F.vdt = F.v} \].
Where F=force and v=velocity.
(Here we have written work done In terms of force and distance and then changed distance in terms of velocity and then integrated to get power in terms of F and v respectively.)

Note: 1) Power is a scalar quantity.
2) SI unit of average power is kilowatt hour.
3) do not confuse instantaneous power and average power.
4) in many terms average power is just written as power, it should be understood based on context