A 2.3 × 103 kg car accelerates from rest under the action of two forces. One is a forward force of 1141 N provided by traction between the wheels and the road. The other is a 947 N resistive force due to various frictional forces.

Question

A 2.3 × 103 kg car accelerates from rest under the action of two forces. One is a forward force of 1141 N provided by traction between the wheels and the road. The other is a 947 N resistive force due to various frictional forces.
How far must the car travel for its speed to reach 3.0 m/s?

Henry · Answer

a = (F1-F2)/m F1 = 1141 N. F2 = 947 N. M = 2300 kg Solve for a. V^2 = Vo^2 + 2a*d V = 3 m/s. Vo = 0 a: Calculated in step #1. Solve for d in meters. .