Faith L.

asked • 10/01/21# ...A 1169 kg car accelerates from

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

## 1 Expert Answer

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_{f}^{2} - (1/2)mv_{i}^{2}

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.

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Marcos O.

10/03/21