What average net force is required to stop a 1950 kg car in 10.5 s if it’s initially traveling at 28 m/s?

 -5200 newtons

Your teacher may want you to type in just the positive version of this number.

Explanation:

We have these given values

The acceleration that occurs is

a = (Vf - Vi)/t

a = (0 - 28)/10.5

a = -28/10.5

a = -2.66666666666667

This acceleration is approximate. The units for the acceleration are in m/s^2.

We'll use this acceleration value in Newton's Second Law equation

Force = (mass)*(acceleration)

F = m*a

F = (1950 kg)*(-2.66666666666667 m/s^2)

F = (1950*(-2.66666666666667)) kg m/s^2

F = -5200 newtons

The negative force value is to indicate the force is working against the car's original direction of motion. For instance, if the car is driving to the right, then a force to the left is applied as the braking force to slow the car down. It might be helpful to draw out the free body diagram.

truck y has a greater velocity than truck x.

i already answered it by 34/s