# Calculating power without distance?

So here's my hw problem:

Calculate KE of a 1500 kg vehicle moving at 30 m/2. Which I did, and got 675000 joules, using the standard KE formula.

But then it asks, if the vehicle accelerated to this speed in 20 seconds, what average power was developed?

Any formula for power involves a distance, which is not given here at all. Can someone help me out? Is it a trick question?

berkeman
Mentor
So here's my hw problem:

Calculate KE of a 1500 kg vehicle moving at 30 m/2. Which I did, and got 675000 joules, using the standard KE formula.

But then it asks, if the vehicle accelerated to this speed in 20 seconds, what average power was developed?

Any formula for power involves a distance, which is not given here at all. Can someone help me out? Is it a trick question?

If it acquires its energy in some amount of time, that implies an average power level input during that time interval...

HallsofIvy
Homework Helper
As long as the acceleration is constant, the average velocity is the average of the starting and ending velocities. You can use that to find the average kinetic energy.

And I am surprised to hear that "Any formula for power involves a distance". I was under the impression that "power" was "rate of change of energy over time".

(You could, of course, calculate the distance moved at a constant acceleration in a given time. For constant acceleration, a, and time, t, the velocity is at. You can use that to find a from the information given. The distance moved is $(1/2)at^2$.)

Well the reason I thought I needed a distance was because the formula for power is the change in work over the change in time. SOOOO the formula for work is force times DISTANCE. So I don't understand how can I calculate power if I can't find work.

berkeman
Mentor
Well the reason I thought I needed a distance was because the formula for power is the change in work over the change in time. SOOOO the formula for work is force times DISTANCE. So I don't understand how can I calculate power if I can't find work.

It's not change in work, it's amount of work (or energy) in some amount of time.

It's not change in work, it's amount of work (or energy) in some amount of time.

So I just divide the numerical answer for kinetic energy by 20?

berkeman
Mentor
So I just divide the numerical answer for kinetic energy by 20?

I believe that is correct.