Question

A 2100 kg truck is moving along a flat, horizontal road at a constant speed of 57 km / h. At one point, the driver lifts his foot off the accelerator pedal. Exactly 22 s after that, the speed of the truck is 35 km / h. Find the intensity of the mean resultant force acting on the truck in the indicated time interval of 22 s.

Answer 1

Let's first find the deceleration of the truck:

$$a=\dfrac{v-v_0}{t}=\dfrac{9.72\ \dfrac{m}{s}-15.83\ \dfrac{m}{s}}{22\ s}=-0.28\ \dfrac{m}{s^2}.$$

The sign minus means that the thruck decelerates.

Then, we can find the mean resultant force acting on the truck from the Newton's Second Law of Motion:

$$F=ma=2100\ kg\cdot(-0.28\ \dfrac{m}{s^2})=-588\ N.$$

The sign minus means that the force acts in the opposite direction to the motion of the truck.