Simple physics(acceleration) problem

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Homework Help Overview

The problem involves a pitcher throwing a baseball and calculating its average speed and acceleration as it comes to rest in the catcher's mitt. The context is within the subject area of kinematics, specifically focusing on speed and acceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the initial speed of the ball, the time taken to decelerate, and the resulting acceleration values. There are questions about the feasibility of the calculated acceleration and the implications of a very short time interval on deceleration rates.

Discussion Status

The discussion is actively exploring the relationship between speed, time, and acceleration. Some participants are questioning the large values obtained for acceleration and discussing the physical implications of such values. There is an acknowledgment of the concept of deceleration in relation to high initial speeds and short stopping times.

Contextual Notes

Participants are grappling with the implications of the calculated acceleration values and the assumptions made about the time interval for deceleration. There is a focus on understanding how these factors interact in the context of the problem.

ucdawg12
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here is the problem:

A pitcher throws a baseball from the mound to homeplate, a distance of 18.5m, in .411s. What was the average speed of that ball in miles/h? If the catcher allowed his mitt to recoil backward 7.5cm while stopping the ball, what was the acceleration of the ball while it was being brought to rest?

Okay, the first part of the question I got, it was 101 mi/h, however the 2nd part has an answer of -1.4x10^4 m/s^2 for acceleration and I just can't seem to figure out how the number got that big...
 
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hmm
seems like the ball would be slowing down not speeding up

if you assume that it was
[tex]Vo=45 m/s (101mi/hr) Vf=0 x=.075m[/tex]
[tex]x=.5(Vo + Vf)[/tex]
[tex].075=.5(45)t[/tex]
then t = .003 seconds
[tex]V=Vo + at[/tex]
[tex]0=45 + a*.003[/tex]
[tex]a= aprox -13500 m/s^2[/tex]
 
one question i don't c how the ball can decelarate at 13500 miles per hour, onless the pitcher is an alien i don't c that becoming posible.
 
The ball has a large initial speed and it has been slown down in a very small time interval. Thats why the ratio of the initial speed to this time interval is quite a large quantity (mathematically). Think of it like car moving at high speed brought to rest by a decelerating force. The magnitude of the deceleration produced is proportional to the initial velocity of the car and inversely proportional to the time taken for the car to come to rest. The smaller the time taken to do so and the larger the initial velocity of the car, the greater is the deceleration.
 
i see what you are saying but, if someone throws a ball at let's say 100 miles per hour calculate that in seconds will be like nothing. So how can it decelarate at 13500 miles per second.
 
if you are moving at a constant speed
your deceleration is dependent upon how fast you slow from this speed
since the time is so slow
the deceleration must be very fast... or the catcher is going to be forced to catch it over a longer time span
 
I see now, thank you. Never did really understand that concept, well I do now. :smile:
 

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