# A funny car accelerates from rest

1. Jun 7, 2012

### wbetting

1. The problem statement, all variables and given/known data
A funny car accelerates from rest through a measured track distance in time 21 s with the engine operating at a constant power 240 kW. If the track crew can increase the engine power by a differential amount 1.0 W, what is the change in the time required for the run?

I learned how to do the problem from the original problem: "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?" but i do not know how to change it to a number answer. explained "variable" answer below (i only put last line that has final equation to use!

￼Differentiating the above equation gives dPT^ 3 + 3PT^2 dT = 0, or
dT = − T/dP * 3P how do i fill numbers into that^^? like how do i find the derivative of power if i know the actual number for power?

2. Relevant equations
dT= -T/dP * 3P

3. The attempt at a solution
We found how to solve equation without numbers but how to i solve to get one single number?like i know power and time but how do i take a derivative of a whole number?

2. Jun 7, 2012

I kinda follow that, but I think of it quite differently. kW are like horsepower and horsepower is torque*rpm/5250. (So actually constant power acceleration would mean continuously decreasing acceleration).

The torque translates through the gears to force at the tires. That accelerates the car via F=ma.

If you have 241 kW instead of 240 kW, the car will accelerate 241/240 as fast.

The acceleration formula would be d = 05*a*t^2. To keep d the same, if a becomes 241/240 as much, then t^2 must become sqrt(240/241) as much.

3. Jun 7, 2012

### rcgldr

That's the formula for constant acceleration, not constant power.

f = force
p = power
m = mass
v = velocity
a = accelertaion

a = f / m
p = f v

a = dv/dt = p / (m v)

v dv = (p/m) dt

and continue from there. The orignal post doesn't include equations for velocity or position versus time.

Last edited: Jun 7, 2012
4. Jun 8, 2012

### wbetting

i did this and got .502 which was the wrong answer!

5. Jun 8, 2012

### Infinitum

As rcgldr noted above, the formula Head_Unit used is for constant acceleration and not constant power, so you can't use it. Try using his hint related to constant power.

6. Jun 8, 2012

### rcgldr

It would help to see more of the equations the orgiinal poster was given, or to know if the problem includes doing the integration of the equations by the students.

7. Jun 11, 2012

And the above are correct, I got too sidetracked with torque. Constant torque would be increasing power over the RPM range; constant power would mean decreasing torque with RPM. That means decreasing acceleration, so maybe you have to integrate after all? (i.e. cannot use a typical plug-in formula)

8. Jun 11, 2012

### ehild

I do not see the formula in the OP, but constant power means work W=Pt, and the work done is equal to the change of KE. Starting with zero velocity, $$\frac{1}{2}mv^2=Pt$$
that is,
$$v=\sqrt{2Pt/m}$$.
The displacement is obtained by integrating the velocity with respect to time:$$X=\sqrt{\frac{4}{3m}}P^{1/2}t^{3/2}$$. As X is constant,
$$Pt^3=const$$

Changing the power with a small amount Δp, while the displacement is the same, t will change by Δt:

$$0= t^3 ΔP+ 3 P t^2 Δt \rightarrow \frac{ΔP}{P}=-3\frac{Δt}{t}$$.
It is the same equation the OP quoted
In the numerical example, t=21 s, P=240000 W, ΔW=1 W. Plug in and get Δt.

ehild

Last edited: Jun 11, 2012
9. Jun 11, 2012

### rcgldr

You also get this if you integrate m v dv = p dt.

That should be ΔP=1 W. I was waiting for the OP to explain how he got his answer.

10. Jun 11, 2012

### NewtonianAlch

A car with constant 240kW would definitely be a funny car, especially when you go to fill up.