Posted by **Sarah** on Sunday, September 10, 2006 at 4:25pm.

A particle had a speed of 18m/s at a certain time and 2.4 s laer its speed was 30 m/s in the opposite direction. What were the magnitude and direction of the average acceleration of the particle during the 2.4s interval?

I got -5m/s^2 but the book said it was 20m/s^2 in the opposite direction

acceleration= change velocity/time

= ( finalvelocity- initial)/time

= ( 18-(-30)/2.4 m/s^2

THe book is right. If you let the direction of the 18m/s as positive, then the direction of the 30 m/s is negative, thus the negative sign in the (-30).

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