A baseball player hits a line drive. Just before the ball is struck, it is moving east at a speed of 43.2 m/s (97 mi/h). Just after contact with the bat, 9.30 10-4 s later, the ball is moving west at a speed of 52 m/s (116 mi/h). Find the ball's average acceleration.
Magnitude___________m/s^2
Calulate the VECTOR velocity change and divide it by the time interval. Since the direction changes by 180 degrees, the velocity change is the sum of the before and after speeds.
To find the ball's average acceleration, we need to calculate the change in velocity and divide it by the time taken.
Step 1: Convert the speeds into m/s, as the units should be consistent for the calculation.
Given:
Initial velocity (before striking the ball), v1 = 43.2 m/s
Final velocity (after striking the ball), v2 = -52 m/s (since it is moving west)
Step 2: Calculate the change in velocity, Δv = v2 - v1
Δv = -52 m/s - 43.2 m/s
Δv = -95.2 m/s
Step 3: Calculate the time taken, t = 9.30 * 10^-4 s
Step 4: Calculate the average acceleration, a = Δv / t
a = (-95.2 m/s) / (9.30 * 10^-4 s)
Now, we divide the numerator by the denominator:
a = -102376.3 m/s^2
Since acceleration has both magnitude and direction, we can disregard the negative sign and take the magnitude of the acceleration.
Therefore, the ball's average acceleration has a magnitude of approximately 102376.3 m/s^2.