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Finding Instantaneous acceleration from a velocity-time graph

  • Thread starter MitsuShai
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  • #1
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Graph: http://s324.photobucket.com/albums/k327/ProtoGirlEXE/?action=view&current=Capture.jpg
What is the instantaneous acceleration at t2 = 31 s?

I know that the instantaneous acceleration is the slope of the tangent of that point and yes I do know I have to convert it to (m/s^2). But I am still not getting the right answers. Here are a list of answers I imputed and they are all wrong:
-.1
0
-2
-1
-1.9
1.9
-.278
-.52

on the first few, I forgot to convert. I used 58/31 the last time.....

The question had parts to it, but I don't know if they are necessary to solve this problem...here it is anyways:
A) Compute the average acceleration during the time interval t = 0s to t = 10s. 1.7
B) Compute the average acceleration during the time interval t = 30s to t = 40s. -1.7
C) Compute the average acceleration during the time interval t = 10s to t = 30s. 0
What is the instantaneous acceleration at t1 = 29 s. 0
I already computed this and got them right.
 
Last edited:

Answers and Replies

  • #2
It looks like its either going to be 0 or its going to be at that corner, where the rate of change is not going to be defined.
 
  • #3
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It looks like its either going to be 0 or its going to be at that corner, where the rate of change is not going to be defined.
I put zero and it was wrong and I don't think undefined is an appropriate answer because 30 is exactly at that corner and it has an acceleration, which I used to solve part B with.
 
  • #4
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Graph: http://i324.photobucket.com/albums/k327/ProtoGirlEXE/Capture.jpg [Broken]
 
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  • #5
159
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nevermind I got it now.
 

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