5. A 1052 kg truck, starting from rest, reaches a speed of 20.0 m/s in 6.20 s.

a. What is the average acceleration of the truck?

b. What is the magnitude of the average net force on the truck during the 6.20 s?

a. 3,392.7 N

b. 3,392.7 N

Explanation:

We are given the following information;

Mass of the truck as 1052 kg initial speed as 0 m/s Final speed as 20.0 m/s Time taken as 6.20 s

#a. We are required to calculate the acceleration;

We need to know the formula of getting acceleration;

a = (v-u) / t

Where v is the final velocity, u is the initial velocity

Therefore;

a = (20 m/s - 0 m/s) / 6.20s

= 3.225 m/s²

Thus, the average acceleration of the truck is 3.225 m/s²

#b. We are required to calculate the net force on the truck

We need to know that;

According to the second Newton's law of motion, F=ma

Where F is the net force, m is the mass and a is the acceleration.

Therefore;

Net force, F = mass * Acceleration

= 1052 kg * 3.225 m/s²

= 3,392.7 N

Thus, the net force on the truck is 3,392.7 N