What should I do?

Question

A car of mass 700 kg is moving along a straight horizontal road. At the instant when the speed of the car is v m s⁻¹, the resistance to the motion of the car is modelled as a force of magnitude (200 + 10v) N.

The engine of the car is working at a constant rate of 12 kW. At the instant when the speed of the car is w m s⁻¹, the acceleration of the car is 0.6 m s⁻².

Find the value of w.

Answer 1

The first step is to apply Newton's 2nd Law to the car, in the x-direction. We'll assume there are two forces acting on the car in the x-direction;

1) The force from the engine, Fe (positive x-direction)

2) The resistance force, Fr = (200 + 10v) (negative x-direction)

∑F_x = m*a_x

Fe - Fr = m*a

Fe - (200 + 10v) = 700*a

This equation is always true. Per the question, at the instant when the car is at speed w, the acceleration = .6 m/s^2

Plugging into our equation we get:

Fe - (200+10*w) = 700*(.6) = 420

Solving for Fe we get:

Fe = 620 + 10w (Newtons)

Now we also know that at this instant the Power output of the engine is 12 KW = 12,000 W

Power = Force * Velocity

Note that since we are talking about the power output of the engine, the appropriate force to use in this equation is the force of the engine, Fe. Plugging in, we get:

12,000 W = Fe*w = (200+10w)*w = 200w+10w^2

This is a quadratic equation with only one unknown, which we can easily solve.

Solving, we get w = 26 m/s