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 little creature moves with a velocity of -1.03 m/s. Then, 2.47 s later, it moves at the velocity 1.51 m/s. What is the chipmunk\'s average acceleration during the 2.47-s time interval?

Answer 1

Answer:

a =  1.02834008 m/s2

Explanation:

given  data:

initial velocity u = -1.03 m/s

time t = 2.47 s later

final velocity v = 1.51 m/s

average acceleration is given as a

$a = \frac{(v - u)}{t}$

putting all value to get required value of acceleration:

   $= \frac{(1.51 -(- 1.03))}{2.47}$

   $= \frac{1.51+1.03}{2.47}$  

   = 1.02834008 m/s2