How Do You Calculate the Average Acceleration of an Eraser in a Physics Problem?

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Homework Help Overview

The problem involves calculating the average acceleration of an eraser that is thrown and then struck by a metrestick, transitioning from an initial velocity to a final velocity over a specified time interval. The context includes vector components and the effects of direction on acceleration.

Discussion Character

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Approaches and Questions Raised

  • Participants discuss the use of kinematic equations to find average acceleration, with some questioning whether to consider the eraser's momentary stop during the impact. Others suggest focusing on the change in velocity without accounting for the stop.

Discussion Status

There is an ongoing exploration of how to apply kinematic equations, with some participants providing alternative perspectives on handling the initial and final velocities. The discussion includes considerations of vector components and the implications of directionality in the calculations.

Contextual Notes

Participants note confusion regarding the treatment of the eraser's velocity during the impact and the need for vector decomposition due to the differing directions of the velocities involved.

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Homework Statement


Jordyn throws an eraser at her teacher, it moves horizontally at 8m/s when the teacher smacks it with a metrestick. The eraser leaves the metrestick horizontally at 15 m/s [N30W]. It is in contact with the stick for 0.03s. Determine the average acceleration of the eraser while being hit by the metrestick. Draw a vector diagram showing the motion.


Homework Equations


We were not given specific equations for this question, but since we are given two velocities, time, and are asked for average acceleration, i figured that maybe a=(V1-V2)/t might work.


The Attempt at a Solution


So, first i thought that i could use the equation above, and just plug in the given information, but then i figured that the eraser stopped when it hit the stick. So, it would go from 8m/s to 0m/s and then to 15m/s. I then used two equations (both being a=(V1-V2)/t). For the first, i used V1=8m/s and V2=0m/s; for the second, i used V1=0m/s and V2=15m/s. Since we have to find the average acceleration, i thought maybe i could average out both answers from the two equations. I doubt this works though, maybe it does, I'm not sure. It seems too simple. Especially because i don't know if it's right to use the same time for both equations. Also, i have absolutely no idea how this relates to vectors. If i had to add/subtract both vectors then i could have just drawn that, but they're two separate things. Please help and thanks so much!
 
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i didn't read your attempt cause I am really sleepy and the paragraph is really big but

you want to say the eraser went from 8m/s to 15m/s in the other direction in .3 seconds.

so use kinematics equation vfinal=vinitial+at, so vfinal=-vinitial+a(t), a=(vfinal+vinitial)/(time). vinitial is the opposite sign of vfinal as it is in the opposite direction. you might have that written down i dunno. don't worry about the instantaneous velocity of the eraser cause it asks for the average acceleration, i confused myself the other day cause i thought another problem implied infinite acceleration by saying something stopped right away
 
So i completely ignore the fact that it stops and has a velocity of 0m/s while the stick hits it? And for the vector diagram that i draw, do i just just draw 8m/s minus 15m/s[N30W]? Also you said i can just make Vfinal opposite of Vinitial, but they're not in the opposite direction though, one's south and the other is N30W
 
OH i didn't see the degrees part. you want to split it up into vector components then, where -8=15cos30+at i think if the angle looks like i think it does. and also 0=15sin30+at for the x-component of the acceleration and your answer will be a vector with x and y components? I'm pretty sure you don't need to think about the eraser stopping, and that all you neeed is to think ok it goes from 8m/s in this direction and accelerates to 15m/s in this direction in .3 seconds
 

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