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A shopping cart that starts from rest, is accelerated for 4 s, moves at constant velocity for 4 s, and is decelerated for 4s until returning to rest, has an average acceleration of 0 m/s².
A shopper is pushing a cart down a grocery store aisle. The movement of the cart is:
- It starts from rest.
- From t = 0 s to t = 4.0 s it is accelerated with a constant force.
- From t = 4 s to t = 8.0 s it receives just enough force to balance the friction on the cart.
- From t = 8 s to t = 12 s it is decelerated until it comes to rest.
All in all, at the initial time (t = 0 s), the velocity is 0 m/s (rest) and at the final time (t = 12 s) the velocity is 0 m/s as well (rest). The average acceleration in that period is:

A shopping cart that starts from rest, is accelerated for 4 s, moves at constant velocity for 4 s, and is decelerated for 4s until returning to rest, has an average acceleration of 0 m/s².
Learn more: brainly.com/question/16274121
To solve this problem it is necessary to apply the concepts related to Normal Force, frictional force, kinematic equations of motion and Newton's second law.
From the kinematic equations of motion we know that the relationship of acceleration, velocity and distance is given by

Where,
Final velocity
Initial Velocity
a = Acceleration
x = Displacement
Acceleration can be expressed in terms of the drag coefficient by means of
Frictional Force
Force by Newton's second Law
Where,
m = mass
a= acceleration
Kinetic frictional coefficient
g = Gravity
Equating both equation we have that



Therefore,


Re-arrange to find x,

The distance traveled by the car depends on the coefficient of kinetic friction, acceleration due to gravity and initial velocity, therefore the three cars will stop at the same distance.