Question

A busy chipmunk runs back and forth along a straight line of acorns that has been set out between its burrow and a nearby tree. At some instant, it moves with a velocity of â’1.47 m/s. Then, 2.07 s later, it moves with a velocity of 1.73 m/s. What is the chipmunk's average acceleration during the 2.07 s time interval?

Answer 1

Answer:

Chipmunk's average acceleration during the 2.07 s time interval = 0.1256 $m/s^2$

Explanation:

  We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

  In this case we need to find acceleration value, we have initial velocity = 1.47 m/s, final velocity = 1.73 m/s and time taken = 2.07 seconds.

 Substituting

       1.73 = 1.47 + a * 2.07

       a = 0.1256 $m/s^2$

  Chipmunk's average acceleration during the 2.07 s time interval = 0.1256 $m/s^2$