Average acceleration problem having trouble

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SUMMARY

The average acceleration of a particle that changes speed from 11 m/s in the positive x direction to -78 m/s in the opposite direction over a time interval of 1.3 seconds is calculated using the formula a = (Vf - Vi) / t. The correct calculation yields an average acceleration of -51.54 m/s². It is crucial to note that the final velocity (Vf) must be negative when the direction reverses, which is essential for accurate results.

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At a certain time a particle had a speed of 11 m/s in the positive x direction, and 1.3 s later its speed was 78 m/s in the opposite direction. What was the average acceleration of the particle during this 1.3 s interval?


Homework Equations


I have Vi=11m/s, Vf=78m/s and t=1.3 m/s


The Attempt at a Solution


My attempt at the solution has been the formula of (78-11)/1.3. This has gotten me the answer of 51.54 and I at am a real loss. I have read my book over and over and I juts need some guidance.
 
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At a certain time a particle had a speed of 11 m/s in the positive x direction, and 1.3 s later its speed was 78 m/s in the opposite direction. What was the average acceleration of the particle during this 1.3 s interval?

a=(vi+vf)/t

a=(11-78)/1.3

a= -51.54 m/s2

Your vf should have been -78 m/s since it was moving in the opposite direction.
 

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