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.
First, let's calculate the change in velocity. The initial velocity is 17 m/s in the positive x direction, and 2.0 seconds later, the velocity is 32 m/s in the opposite direction.
The change in velocity is given by the final velocity minus the initial velocity:
Change in velocity = 32 m/s - (-17 m/s) = 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
Therefore, the magnitude of the average acceleration of the particle during this 2.0 s interval is 24.5 m/s^2.