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).
To find the magnitude and direction of the average acceleration of the particle during the 2.4s interval, you correctly used the formula: acceleration = (final velocity - initial velocity) / time.
Given that the initial velocity is 18 m/s and the final velocity is -30 m/s (since it is in the opposite direction), and the time is 2.4 seconds, we can substitute these values into the formula:
acceleration = (-30 m/s - 18 m/s) / 2.4 s
acceleration = -48 m/s / 2.4 s
acceleration = -20 m/s^2
So, your calculation of -20 m/s^2 is correct. The negative sign indicates that the acceleration is in the opposite direction of the initial velocity, which aligns with the book's answer.
Therefore, the magnitude of the average acceleration is 20 m/s^2, and the direction is opposite to the initial velocity.