Answer to Question #128704 in Classical Mechanics for Tony

Question

A driver starts from rest at t=0 and has traveled along a circular track of radius 400m the tangential component of the total force on the vehicle from t=0 to t=30 is given as F(t) = 200-6t. the combined mass of the vehicle and driver is 150kg.
what is the average of the tangential component of the total force from t=0 to t=30s
what is the average of the normal component of the total force from t=0s to t=30s
what is the magnitude of the velocity at t=30s

Answer 1

The average of the tangential component of the total force from t=0 to t=30 s can be calculated as a middle point of force graph because the function of the force is linear. At the beginning, t=0 and F(0)=200. In the end, t=30 and F(30)=20.

Thus, the average force (middle point of the force graph between t=0 and t=30) is

$$F_\text{avg}^\text{tan}=\frac{F_\text{max}^\text{tan}-F_\text{min}^\text{tan}}{2}=\frac{200-20}{2}=90\text{ N}.$$

Calculate the average of the normal component of the total force from t=0s to t=30s. The normal force does not change because we assume the track to be horizontal. Thus, the average normal force is

$$N_\text{avg}=N=mg=150\cdot9.8=1470\text{ N}.$$

The magnitude of the velocity at t=30s can be found if we compute the acceleration. Acceleration is force over mass according to Newton's second law. Since the force changes with time, the acceleration changes as well and we deal with a non-uniformly accelerated motion. The velocity, however, is still acceleration times time:

$$v=at,\\ a=\frac{F}{m}=\frac{200-6t}{m},\\\space\\ v=\int_0^{30}\frac{F}{m}dt=\frac{4t}{3}-\frac{t^2}{50}\bigg|^{30}_0=22\text{ m/s}.$$