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What is the average power output by the car?

A 1500kg car driving at 20 m/s, in 4 seconds it accelerates to 40m/s.
1. Assume that there is no wind friction, calculate the average power put out by the car.

1.)We know that power is the change in energy divided by the change in time. There are many ways to power is written, but let’s use this one because it is the easiest (to me) for this particular part.
Power = ΔE/Δt
Take the kinetic energy equation, which is (1/2mv^2). The change is energy is final KE – initial KE. Also, change is time is the final time – initial time.
= (1/2 * m * v(f)^2) – (1/2 * m * v(i)^2) / t(f) – t(i)
You have all the variables, so plug it in.
= [(.5 * 1500 kg * (40m/s)^2) - (.5 * 1500 kg * (40m/s)^2)]/(4s-0s)
= 225,000 W = 225 kW
2.) This one is a little tricky. Because you have friction, it would be a difficult to use the ΔE/Δt. So another way to write power is W/t, which is work divided by time.
Since, W = Force * distance
Force = m*a
We need to find acceleration
v = v(i) + at
40m/s = 20m/s + (a)4s
a = 5 m/s^2
We need to find distance.
Δx = v(i)t + 1/2at^2
= (20m/s)(4s) + 1/2(5m/s^2)(4s^2)
= 120 m
Now, we plug in variables for work.
W of friction = 3000N * 120m
= 360,000 J
W of normal = 7500 N * 120m
= 900,000 J
Now add the work together to get the work done by the car.
W = (900,000 J + 360,000 J)
= 1260000 J
Finally, we can use the W/t to get the average power.
Power = 1260000 J / 4s
= 315,000 W = 315 kW

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