Differentiating Power and Time

[napaul]

Homework Statement

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? (Use any variable or symbol stated above as necessary.)

Homework Equations

I think its:
P = dW/dT
P = Fvcos(Phi) = F dot v
D = 1/2aT^2

The Attempt at a Solution

Sorry this is my first post, please be lenient on me...
Okay, so far I started with

P = dW/dT = F [dot] V

F = (ma)

V = aT

so P = ma(aT)cos(Phi)

P = [mTa^2] cos(Phi)

P = mT[(2d)/T^2]^2 cos(Phi)

P = mT[(4d^2)/T^4] cos(Phi)

P = [4md^2/T^3] cos(Phi)

dP/dT = [8md/3T^2] [-sin(Phi)]

dT = [[3T^2(dP)/8md] [-sin(Phi)]

I know its wrong, but I need some guidance because I'm really confused...

 

[I like Serena]

Welcome to PF, napaul!

Homework Statement

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? (Use any variable or symbol stated above as necessary.)

Homework Equations

I think its:
P = dW/dT
P = Fvcos(Phi) = F dot v
D = 1/2aT^2

Your formulas for P are correct.
However, I'm afraid your formula for D is wrong.

The Attempt at a Solution

Sorry this is my first post, please be lenient on me...
Okay, so far I started with

P = dW/dT = F [dot] V

F = (ma)

V = aT

Similarly your formula for V is wrong.

Your formula for D and V only hold when acceleration a is constant, which in this case it is not.

The proper formulas are:
$$dv = adt \quad \text{or} \quad v=v_0 + \int_0^t a dt$$
$$dx = vdt \quad \text{or} \quad D=\int_0^T v dt$$

You can get "a" as a function of "v" from F=ma and P=Fv=constant.
From there you should solve dv=adt...

Btw, you can leave Phi out of your equations since a car would always accelerate in the direction of its speed.