Motion along a straight line – problem 5
Tennis ball acceleration
A tennis ball had velocity of 100 km/h when it touched the wall and 0.2 s later its velocity was 80 km/h in opposite direction. What was the magnitude of the average acceleration of the ball? Compare it to gravitational acceleration g = 9.81m/s2.
We denote:
v1 = 100 km/h - initial velocity
v2 = -80 km/h - final velocity, with a minus sign because of opposite direction
Dt = 0.2 s – time interval during which velocity was changing (acceleration was different from 0)
a = ? - acceleration to be calculated.
The average acceleration in straight line motion is the total change of velocity divided by time during which this change occurred.
We now insert the numerical values into this equation.
We now change kilometers to meters and hours to seconds, to get result in SI units.
After a little algebra we have
The minus sign indicates that acceleration had a direction opposite to the initial velocity, to which we assigned a positive direction.
We now divide this acceleration by gravitational acceleration g. We will omit signs of acceleration as we are comparing only magnitudes.
The average acceleration of the tennis ball was about 25 larger in magnitude than gravitational acceleration g.