Physics-Work, Energy and Power

Question

The time it takes a car to race from rest through a distance d depends primarily on the engine’s power P.

Assuming the power is constant, derive the time in terms of d and P.

I have followed following steps to solve for this question,

1/2 m v^2 - 0 = P t

t= 1/2 m v^2 / P ---------------> 1

from kinematics (since initial velocity is zero)

v^2= 2 a d

d= 1/2 a t^2

therefore a= 2 d/ t^2

therefore v^2= 2 * (2d/t^2)* d

substitute for v^2 in first equation

I get,

t= (2md^2/P)^(1/3)

However, the solution is supposed to be (3D)^(2/3) * (m/P)^(1/3)

Can any one explain this to me?

Answer 1

Power is defined as Energy generated per unit time or work put into the system or output from the system per unit time. Work is Force vector dotted with distance vector.

Instantaneous power is then F vector dotted with velocity vector.

The premise is that the power is constant for the process. Then Force dotted with velocity must be a constant and independent of time. Several ways that this can be Force is constant and velocity is constant or dot product of the vectors is time independent - constant.

Lets say the Force is a constant....This then requires that the velocity vector is also constant but this leads to an inconsistency as an object experiencing a net force will show a change in motion...the velocity vector is no longer constant with time. So if we have a constant force we can't have a constant velocity and the instantaneous power is not constant with time for the system.

Lets say the velocity vector is constant with time....That leads to the net force being zero on the system and the power involved for the system is zero. A trivial solution.

So this leads to power being constant and non-zero requires that the product of acceleration and velocity be independent of time.

The acceleration is the time derivative of the velocity time dependent function.

lets write the velocity function as v=Kt^p

at t = 0 the velocity of our car is zero and the velocity of our car increases for p greater than zero, which seems reasonable in our model for the speed increase with time for the car under constant power.

We can write acceleration as dv/dt = pKt^(p-1)

It seems that the car needs to experience less and less acceleration as time progresses to still have constant work done upon it in moving along the race track....if p is less than one this is also true.

But note we have vector a dotted with vector v must be independent of time for the power to be constant with time.

This means pK²t^(2p-1) must be equivalent to a constant in time or 2p-1 = 0 Finally p = 1/2 is our solution.

This leads to the solution that you cited.

Your analysis assumes the kinematic equations hold, but these do not hold where the acceleration on the system is not constant with time. This is why your conclusion for the time as related to distance and power does not match.

Using v(t) = Kt^p

we have D= Kt^(p+1)/(p+1) for p=1/2 we have p+1 is 3/2 which is where we see powers of 2/3 in the solution that has been given.

t =(3/2 D/K)^(2/3)

We need to find K

But we have P= Fv= mav= m x p x K x K and K² = P/2m..........seems parallel to v²= P/2m

Then we can write (3/2)^(2/3) x( D²/K²)^1/3 = (3/2)^(2/3) x (D²/(P/2m))^1/3

(3D)^(2/3) x (m/P)^(1/3) x (1/2)^(1/3)

the factor of (1/2) to the power of 1/3 still seems to be at odds to what you have cited as a solution. But I believe the analysis correctly does not assume constant acceleration kinematics.