What Is the Power Output of a Car Accelerating from Rest in First Gear?

[TheRedDevil18]

Homework Statement

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

Homework Equations

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

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

Could Someone Please Check?

 

[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.

 

[haruspex]

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.

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.

 

[TheRedDevil18]

So what's the difference between instantaneous and average power?

 

[PhanthomJay]

So what's the difference between instantaneous and average power?

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 ___?___?