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;

#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

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