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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 −1.29 m/s−1.29 m/s . Then, 2.91 s2.91 s later, it moves with a velocity of 1.77 m/s1.77 m/s . What is the chipmunk's average acceleration during the 2.91 s2.91 s time interval?

Respuesta :

Answer:

1.05 ms⁻²

Explanation:

Acceleration = change in velocity / Time

Change in velocity = Final velocity - initial velocity

= 1.77 - (-1.29)

= 1.77 + 1.29

= 3.06 m/s

Time = 2.91

Acceleration = 3.06 / 2.91

= 1.05 ms⁻² .

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