A motorist driving a 1200-kg car on level ground accelerates from 20.0 m/s to 30.0 m/s in a time of 5.0s. Neglecting friction and air resistance, determine the average mechanical power in watts the engine must supply during this time interval.

Answer 1

a = (Vf - Vo) / t,
a = (30 - 20)m/s / 5s = 2m/s^2.

Fc = mg = 1200kg * 9.8N/kg = 11760N.

(Vf)^2 = Vo^2 + 2ad = (30)^2,
(20)^2 + 2 * 2 * d = 900,
400 + 4d = 900,
4d = 900 - 400 = 500,
d = 125m. = Distance traveled during
acceleration.

P = F*d / t = 11760 * 125 / 5 = 294000W