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Physics magnitude of avg acceleration

  • Thread starter sycho2
  • Start date
  • #1
4
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A baseball player hits a line drive. Just before the ball is struck, it is moving east at a speed of 37.6 m/s (84 mi/h). Just after contact with the bat, 1.05x10^-3 s later, the ball is moving west at a speed of 51 m/s (114 mi/h). Find the ball's average acceleration.

So i know that to find avg acc is the change in velocity over the change in time. But I've tried doing 51-37.6 over .00105 but that gives me a ridiculously large number and i know its wrong. I realized that the time interval isn't the overall time, I don't understand how to get that time or maybe is there another way to solve this problem?
 

Answers and Replies

  • #2
50
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You solved the problem correctly!

It was good of you to second guess your answer when you found a number larger than any you have seen before, however, when the time for acceleration is that small, and the change in velocity is not, the acceleration MUST be very large.

As a general rule of thumb, very small things accelerate very quickly, so expect larger numbers from things like baseballs, bugs, or atoms; large things accelerate slowly, so expect small numbers from things like cars, trains, or planets.
 
  • #3
gneill
Mentor
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Sure, expect large accelerations when the action takes place over tiny time intervals. Just make sure that you calculate the change in velocity properly. Watch out for signs when directions change.
 
  • #4
4
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so then I would just do change in velocity which is 51-37.6 and divide that by 1.05x10^-3? that give my answer?
 
  • #5
gneill
Mentor
20,781
2,759
so then I would just do change in velocity which is 51-37.6 and divide that by 1.05x10^-3? that give my answer?
Suppose the ball were initially moving at 50 m/s to the east, and afterward at 50 m/s to the west. What would be the magnitude of the change in velocity?
 
  • #6
4
0
it would be 0. Oh yeah i forgot about that...so it would be (51+37.6)/1.05x10^-3
 

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