Question

A shopper is pushing a cart down a grocery store aisle. Starting from rest, the shopper applies a constant force to the cart for 4.0 s. From t = 4s until t = 4 s until t = 8 s, the shopper applies just enough force to balance the friction on the cart. Finally, the shopper applies a constant force to slow the cart until it comes to rest att = 12 s. The resulting position-time graph is shown below. What is the average acceleration of the shopping cart during its 12s motion?​

Answer 1

Answer:0 m/s2

Explanation: Cus i said so.

Answer 2

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:

$a = \frac{v_{12}-v__o}{t_{12}-t_0} = \frac{0m/m-0m/s}{12s-0s} = 0 m/s^{2}$

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.

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