# Work, Energy And Power Problem

1. Feb 12, 2013

### TheRedDevil18

1. The problem statement, all variables and given/known data
The operation manual of a car, mass 760kg, claims that the car, starting from rest, can reach a speed of 15m.s in first gear in 3 seconds. The acceleration of the car is constant and friction can be ignored. If the car operates under these conditions, calculate:

9.1) The power output in first gear
9.2) The maximum distance covered in 3 seconds

2. Relevant equations

ek=1/2mv^2
p=w/t
delta y=vf+vi/2 *t

3. The attempt at a solution

9.1) ek = 1/2mv^2
= 1/2*760*15^2
= 85500 J

p = w/t
= 85500/3
= 28500 W

9.2) delta y = vf+vi/2 *t
= 15/2 *3
= 22.5m

2. Feb 12, 2013

### PhanthomJay

Distance looks good, calc for power is the average power delivered by the engine over that time and distance, not the instantaneous power delivered at the 3 second mark, which is higher, since instantaneous power varies from zero at the start to a max at the given time. I don't know if the problem is asking for average power or instantaneous power.

3. Feb 12, 2013

### haruspex

Since it asks for 'power output', peak power seems the more likely interpretation to me. I.e., what power is it capable of. But I agree the question is poorly worded.

4. Feb 13, 2013

### TheRedDevil18

So whats the difference between instantaneous and average power?

5. Feb 13, 2013

### PhanthomJay

The average power is the equal to the rate that work is done over a given time period, that is, $P_{avg} = \Delta W/\Delta t$.

The instantaneous power is limiting value of the average power as the time interval Δt approaches zero, that is, $P_{instant} = dW/dt$, which, in this example for constant force, is equal to $Fds/dt$, or $Fv_{instant}$.

The average power is 28.5kW, the instantaneous power at t=0 is 0, and at t=3, the instantaneous power is ___?___?