At a certain time a particle had a speed of 16 m/s in the positive x direction, and 2.2 s later its speed was 30 m/s in the opposite direction. What is the magnitude of the average acceleration of the particle during this 2.2 s interval?
change in velocity = end - begin = -30 - 16 = -46 m/s
a = delts v/ delta t = -46/2.2 = -20.9 m/s^2
Thank you so much!
To find the magnitude of the average acceleration of the particle during the 2.2 s interval, we need to use the formula for average acceleration:
Average Acceleration = (Change in Velocity) / (Change in Time)
First, let's find the change in velocity. The initial velocity is 16 m/s in the positive x direction, and the final velocity is 30 m/s in the opposite direction. Since the velocities have opposite directions, we can consider them as negative and find the change in velocity as follows:
Change in Velocity = Final Velocity - Initial Velocity
= -30 m/s - 16 m/s
= -46 m/s
Next, we need to find the change in time. The time interval given in the question is 2.2 s.
Change in Time = 2.2 s
Finally, we can calculate the magnitude of the average acceleration using the formula:
Average Acceleration = (Change in Velocity) / (Change in Time)
= (-46 m/s) / (2.2 s)
≈ -20.91 m/s² (rounding to two decimal places)
The magnitude of the average acceleration of the particle during this 2.2 s interval is approximately 20.91 m/s².