Question

A busy chipmunk runs back and forth along a straight line of acorns that has been set out between his burrow and a nearby tree. At some instant the chipmunk moves with a velocity of -1.15 m/s. Then, 2.11 s later, it moves at a velocity of 1.63 m/s. What is the chipmunk's average acceleration during the 2.11-s time interval in m/s^2?

Answer 1

Answer: a = 1.32m/s2

Therefore, the average acceleration is 1.32m/s2

Explanation:

Acceleration is the rate of change in the velocity per time

a = change in velocity/time

a = ∆v/t

average acceleration a = (v2 -v1)/t. ....1

Given;

Final velocity v2 = 1.63m/s

Initial velocity v1 = -1.15ms

time taken t = 2.11s

Substituting into eqn 1

a = [1.63 - (-1.15)]/2.11

a = (1.63+1.15)/2.11

a = 2.78/2.11

a = 1.32m/s2

Therefore, the average acceleration is 1.32m/s2