A funny car accelerates from rest through a measured track distance in time $\mathit{T}$with the engine operating at a constant power$\mathit{P}$. If the track crew can increase the engine power by a differential amount$\mathbf{d}\mathit{P}$, what is the change in the time required for the run?

It is given that,

1. Time is$T$.
2. Constant power is$P$.
3. Engine power increased by differential amount is$dP$.

The problem deals with the power and the work done. Work, energy, and power are the fundamental concepts of physics. Work is the displacement of an object when force is applied on it. First, evaluate the equation for power$\mathit{P}$, then integrating fromlocalid="1657168415126" $\mathit{t}\mathbf{}\mathbf{=}\mathbf{0}$to$\mathit{t}\mathbf{}\mathbf{=}\mathit{T}$with respect to$\mathit{d}\mathit{t}$, find the solution for$\mathit{T}$. Next, find the solution for total distance$\mathit{L}$travelled by the car. By putting the equation of $\mathit{T}$, further evaluate for$\mathit{P}{\mathit{T}}^{\mathbf{3}}$and differentiating the equations, find the equation forchange in the time$\mathit{d}\mathit{T}$required fortherun.

Formula:

$P=\frac{dW}{dt}$

Where, P is power and $\frac{dW}{dt}$is change in work per unit time.

Fromthegiven information,

$P=\frac{dW}{dt}$

But usingthe work energy theorem, work done is equal to change in kinetic energy, therefore,

$W=\frac{1}{2}m{v}^2$

role="math" localid="1657169407831" $$\begin{gathered} P=\frac{d\left({\displaystyle \frac{1}{2}}m{v}^2\right)}{dt} \\ P=\frac{1}{2}m\frac{d{v}^2}{dt} \\ P=mv\frac{dv}{dt} \\ =cons\tan t \end{gathered}$$

Therefore, the same condition implies that

$dt=mv\frac{dv}{p}$

Therefore, by integration,

$$\begin{gathered} {\int }_o^{T}dt={\int }_o^v\frac{mv}{P}=dv \\ T=\frac{m{v^2}_T}{2P} \end{gathered}$$

Where,$V_T$is the speed of car at$T$.

Now, write the total distance$L$travelled by the car as,

$$\begin{gathered} {\int }_O^{T}vdt={\int }_o^v\frac{m{v}^2}{P}dv \\ L=\frac{m{v^3}_T}{3P} \end{gathered}$$

Taking the square of this equation,

$$\begin{gathered} L^2={\left(\frac{m{v^3}_T}{3P}\right)}^2 \\ {L}^2=\frac{8P}{9m}{\left(\frac{m{v^2}_T}{2P}\right)}^3 \end{gathered}$$

But $T=\frac{m{v^2}_T}{2P}$,

$$\begin{gathered} L^2=\frac{8P}{9m}{\left(T\right)}^{3}P{T}^3=\frac{9}{8}m{L}^2 \\ =\mathrm{constant} \end{gathered}$$

Therefore, differentiating the above equation,

$$\begin{gathered} dP{T}^3+3P{T}^{2}dT=O \\ dT=-\frac{T}{3P}dP \end{gathered}$$

Hence, the change in the time required for the run is$dT=-\frac{T}{3P}dP$.