Ask Question
28 November, 23:21

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?

+5
Answers (1)
  1. 29 November, 02:54
    0
    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
Know the Answer?
Not Sure About the Answer?
Get an answer to your 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 ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers