A funny car accelerates from rest

[wbetting]

Homework Statement

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?

Homework Equations

dT= -T/dP * 3P

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?

 

[Head_Unit]

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.

 

[rcgldr]

The acceleration formula would be d = 05*a*t^2.

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

For constant power you start with

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.

 

[wbetting]

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.

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

 

[Infinitum]

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

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.

 

[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.

 

[Head_Unit]

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

What is the RIGHT answer?

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)

 

[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, [tex]\frac{1}{2}mv^2=Pt[/tex]
that is,
[tex]v=\sqrt{2Pt/m}[/tex].
The displacement is obtained by integrating the velocity with respect to time:[tex]X=\sqrt{\frac{4}{3m}}P^{1/2}t^{3/2}[/tex]. As X is constant,
[tex]Pt^3=const[/tex]

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

[tex]0= t^3 ΔP+ 3 P t^2 Δt \rightarrow \frac{ΔP}{P}=-3\frac{Δt}{t}[/tex].
It is the same equation the OP quoted

dPT^ 3 + 3PT^2 dT = 0

In the numerical example, t=21 s, P=240000 W, ΔW=1 W. Plug in and get Δt.
ehild

 

[rcgldr]

[tex]\frac{1}{2}mv^2=Pt[/tex]

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

[tex]0= t^3 ΔP+ 3 P t^2 Δt \rightarrow \frac{ΔP}{P}=-3\frac{Δt}{t}[/tex]
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.

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

 

[NewtonianAlch]

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