A testing cart is 18 meters from a marker at $t=2$ seconds and 51 meters from it at $t=8$ seconds. What is the cart's average rate of change in distance?

Answer and derivation

Correct answer: A

The average rate is $33/6=11/2$ meters per second.

1. The distance changes by $51-18=33$ meters.
2. The elapsed time is $8-2=6$ seconds.
3. Divide to obtain $33/6=11/2$ m/s.

Why the other options fail

A

Correct difference quotient.

B

This uses an incorrect elapsed time.

C

This divides the final distance by 6.

D

This is only the change in distance.

Common misconception: An average rate needs changes in both quantities and a unit ratio.

China/AP bridge: This is the slope formula already studied in China; retain the contextual unit in an AP response.