At a certain time a particle had a speed of 87 m/s in the positive x direction, and 6.0 s later its speed was 74 m/s in the opposite direction. What was the average acceleration of the particle during this 6.0 s interval?

Question

Alexandria Benson

Physics

27 October, 12:05

At a certain time a particle had a speed of 87 m/s in the positive x direction, and 6.0 s later its speed was 74 m/s in the opposite direction. What was the average acceleration of the particle during this 6.0 s interval?

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Answers (1)

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Jaydin Prince · Answer 1

The average acceleration during the 6.0 s interval was - 27 m/s².

Explanation:

Hi there!

The average acceleration is defined as the change in velocity over time:

a = Δv/t

Where:

a = acceleration.

Δv = change in velocity = final velocity - initial velocity

t = elapsed time

The change in velocity will be:

Δv = final velocity - initial velocity

Δv = - 74 m/s - 87 m/s = - 161 m/s

(notice the negative sign of the velocity that is in opposite direction to the direction considered positive)

Then the average acceleration will be:

a = Δv/t

a = - 161 m/s / 6.0 s

a = - 27 m/s²

The average acceleration during the 6.0 s interval was - 27 m/s².