...A 1169 kg car accelerates from

Question

A 1169 kg car accelerates from 67.26 m/s to 93.84 m/s in 12 s. How much power does that require?

Answer 1

Given m = 1169 kg, v_i = 67.26 m/s, v_f = 93.84 m/s, t = 12 s

Find P = ?

Start with definition of power, P = W/t or P = ΔE/Δt or P = ΔK/Δt

(all are equivalent since work W done on the car by its engine equals gain in car's energy, which will be kinetic energy K assuming horizontal road and neglecting air resistance & friction)

Note: when figuring ΔK you must set up to calculate each energy separately:

ΔK = K_f - K_i = (1/2)mv_f² - (1/2)mv_i²

You can factor out the (1/2)m, but you cannot simplify to (v_f - v_i)². No no no. Check and see for yourself.

You'll get an answer in the hundred of thousands of joules per second. I suggest you round to proper number of sig figs, and it would be nice to express in kilowatts (yes, divide by 1000) rather than as hundreds of thousands of watts.