At a certain time a particle had a speed of 17 m/s in the positive x direction, and 2.0 s later its speed was 32 m/s in the opposite direction. What is the magnitude of the average acceleration of the particle during this 2.0 s interval?
To find the magnitude of the average acceleration, we need to calculate the change in velocity and divide it by the time interval.
The change in velocity can be found by subtracting the final velocity from the initial velocity:
Change in velocity = final velocity - initial velocity
In this case, the initial velocity is 17 m/s in the positive x direction, and the final velocity is 32 m/s in the opposite direction. Since the final velocity is in the opposite direction, we can consider it as -32 m/s:
Change in velocity = -32 m/s - 17 m/s
= -49 m/s
Next, we divide the change in velocity by the time interval of 2.0 s:
Average acceleration = change in velocity / time interval
= -49 m/s / 2.0 s
= -24.5 m/s^2
The magnitude of the average acceleration is the absolute value of the average acceleration:
Magnitude of average acceleration = |-24.5 m/s^2|
= 24.5 m/s^2
Therefore, the magnitude of the average acceleration of the particle during this 2.0 s interval is 24.5 m/s^2.