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

Answer

To solve the problem, we need to derive the relationship between the change in time (dt) and the change in power (dP) for a funny car accelerating from rest over a distance L with a constant power P. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the relationship between power, work, and time
The power (P) is defined as the work done (W) over time (T):
\[ P = \frac{W}{T} \]
Since the car is accelerating from rest, the work done can be expressed in terms of kinetic energy:
\[ W = \frac{1}{2} mv^2 \]
where m is the mass of the car and v is its final velocity.

### Step 2: Relate velocity to time
Using the definition of power, we can express the work done in terms of velocity:
\[ P = \frac{dW}{dT} = \frac{d\left(\frac{1}{2} mv^2\right)}{dT} \]
This implies:
\[ P = m v \frac{dv}{dt} \]
Rearranging gives us:
\[ dt = \frac{m v}{P} dv \]

### Step 3: Integrate to find the total time T
Since the car starts from rest (v = 0) and accelerates to a final velocity (v = v_T) over a time T:
\[ T = \int_0^{v_T} \frac{m v}{P} dv \]
Integrating this expression:
\[ T = \frac{m}{P} \left[\frac{v^2}{2}\right]_0^{v_T} = \frac{m v_T^2}{2P} \]

### Step 4: Relate distance L to time T
The distance covered by the car can also be expressed in terms of the final velocity and time:
\[ L = v_T T \]
Substituting for T from the previous step:
\[ L = v_T \left(\frac{m v_T^2}{2P}\right) = \frac{m v_T^3}{2P} \]

### Step 5: Square the distance equation
From the equation \( L = \frac{m v_T^3}{2P} \), we can square both sides:
\[ L^2 = \left(\frac{m v_T^3}{2P}\right)^2 \]
This gives us:
\[ L^2 = \frac{m^2 v_T^6}{4P^2} \]

### Step 6: Differentiate the relationship
Now, we differentiate the squared distance equation with respect to time:
\[ 2L dL = \frac{m^2}{4} \left(6v_T^5 dv_T\right) \frac{1}{P^2} - \frac{m^2 v_T^6}{2P^3} dP \]
This leads to:
\[ 2L dL = \frac{3m^2 v_T^5}{4P^2} dv_T - \frac{m^2 v_T^6}{2P^3} dP \]

### Step 7: Solve for dt
From the earlier derived relationships, we can isolate dt:
\[ dt = -\frac{T}{3P} dP \]

### Final Result
Thus, the change in time required for the run when the engine power is increased by a differential amount dP is given by:
\[ dt = -\frac{T}{3P} dP \]