Monday
March 27, 2017

Post a New Question

Posted by on .

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.

  • Physics - ,

    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

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question