Question

A 1500-kg car accelerates from 0 to 30 m/s in 6.0 s. What is the minimum average power delivered by the engine?

Answer 1

Power, P = 112500 watts

Step-by-step explanation:

It is given that,

Mass of the car, m = 1500 kg

Initial velocity of the car, u = 0

Final velocity of the car, v = 30 m/s

Time taken, t = 6 s

The minimum average power delivered by the engine is given by :

P = W/t

Where

W = work done

t = time taken

Work done = change in kinetic energy

So,
`P=(\Delta K)/(t)=((1)/(2)m(v^2-u^2))/(t)`

`P=((1)/(2)* 1500\ kg* (30\ m/s)^2)/(6\ s)`

P = 112500 watts

So, the minimum average power delivered by the engine is 112500 watts. Hence, this is the required solution.