Question

A car of mass 100kg accelerates from rest to 100km/h in 5s. Find the average power of car.
A. $7.71 \times {10^5}W$
B. $7.71 \times {10^4}W$
C. $15.42 \times {10^4}W$
D. $15.42 \times {10^4}W$

Answer 1

Hint: We can assume a constant acceleration for the car and find the power at every instant. Then the average is obtained through integration. Don’t forget to convert speed into m/s.
Formula used :
${P_{av}} = \dfrac{1}{T}\int_0^T {Power(t)dt}$
$P(t) = F(t) \times v(t)$
$F = ma$

Complete step-by-step answer:
Before answering, let's get an Idea of how to take averages of continuous quantities.
As in this question, say we have to find the average of power. We first integrate Power to find its total value over the range and then divide by the range to get average.
That means- ${P_{av}} = \dfrac{1}{T}\int_0^T {Power(t)dt}$

Now let’s start answering the question by first finding the force on the car
Since the car accelerated from an initial velocity of $0m/s$ to a final velocity of $100km/h$ in $5s$, the acceleration of the car is $a = \dfrac{{v - u}}{t} = \dfrac{{100\dfrac{{km}}{h} - 0}}{{5s}} = \dfrac{{100\dfrac{{1000m}}{{3600s}}}}{{5s}} = \dfrac{{100\dfrac{5}{{18}}m/s}}{{5s}} = \dfrac{{100}}{{18}}m/{s^2}$.
Since no more information is given regarding what happened in between this 5s , we can take this acceleration as a constant for this time period.
Now the force is given by $F = ma$. Thus
$F = 1000 \times \dfrac{{100}}{{18}} = \dfrac{1}{{18}} \times {10^5}N$
Now we know the power delivered at any instant is:
Force at that instant $\times$ Velocity at that instant
Since the speed of the car is not the same in between these 5s, we have to take the average though integration. So average of power for $T$ seconds is:
${P_{av}} = \dfrac{1}{T}\int_0^T {F(t) \times v(t)dt}$
Since $F$ is assumed to be constant,
${P_{av}} = F\dfrac{1}{T}\int_0^T {v(t)dt}$
${P_{av}} = F\dfrac{1}{T}\int_0^T {atdt}$
and,
${P_{av}} = \dfrac{1}{5}F \cdot a\left[ {\dfrac{{{t^2}}}{2}} \right]_0^5$
${P_{av}} = \dfrac{1}{5}F \cdot a\dfrac{{{5^2}}}{2}$
${P_{av}} = F \cdot a\dfrac{5}{2}$
${P_{av}} = \dfrac{{{{10}^5}}}{{18}} \cdot \dfrac{{100}}{{18}}\dfrac{5}{2} = 7.71 \times {10^4}W$

Note : A common possible mistake is to simply multiply the force obtained $\dfrac{1}{{18}} \times {10^5}$ with final velocity $100\dfrac{5}{{18}}$. This gives a wrong answer because the velocity has not been $100\dfrac{5}{{18}}$ always. It increased from 0 slowly.

In competitive exams, such long and tedious methods are unnecessary and also time consuming. So It is worth to remember that if a quantity is proportional to time - Just like the power in this case - then it's time average would be half of the final value.
ie : Power at any instant : $P(t)= F\cdot a \cdot t$. So power is proportional to time and hence the average would be ${P_{av}} = \dfrac{{F \cdot {v_{final}}}}{2}$
So a quick solution would be : $F = m \times a = m \times \dfrac{{v - u}}{t} = 1000 \times \dfrac{{100\dfrac{5}{{18}} - 0}}{5} = \dfrac{1}{{18}} \times {10^5}N$.
$P = F \cdot v(t) = F.at$
${P_{av}} = \dfrac{{F \cdot {v_{final}}}}{2}$ since P$$$ \propto $$ t$${P_{av}} = \dfrac{{{{10}^5}}}{{18}} \cdot \dfrac{{100}}{{18}}\dfrac{5}{2} = 7.71 \times {10^4}W$