a 50.0-g Super Ball traveling at 25.0 m/s bounces off a brick wall and rebounds at 22.0 m/s. A high-speed camera records this event. If the ball is in contact with the wall for 3.50 ms, what is the magnitude of the average acceleration of the ball during the time interval?

Question

ash2326 · Answer

Average Acceleration is defined as the change in velocity per unit time,
Let the ball moving towards the ball has  speed vi=> -25 m/s
and when it moves away it has speed vf = 22m/s
$$\Delta a= \frac{vf-vi}{\Delta t}$$
$\Delta t= 3.5 ms$
You can do now, can't you?

lgbasallote · Answer

let me see if i get it right...first i need to change ms into s right? what does ms mean?

anonymous · Answer

miliseconds. 1 second = 1000 ms

lgbasallote · Answer

ohh thanks

lgbasallote · Answer

so it's 22 + 25/ 0.0035? sounds big.

lgbasallote · Answer

wait..if it bounces back...shouldnt the speed decrease..therefore acceleration should decrease too?

anonymous · Answer

Woops.

And yeah, it does sound big. But remember the ball has to turn around almost instantly. So the acceleration has to be big.

anonymous · Answer

You did it right. The velocity does decrease. But aceleration is like, the measurement of a decrease in velocity. So the more it decreases, the bigger the acceleration!