# Finding Instantaneous acceleration from a velocity-time graph

by MitsuShai
Tags: acceleration, graph, instantaneous, velocitytime
 P: 161 Graph: http://s324.photobucket.com/albums/k...nt=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.
 P: 83 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.
P: 161
 Quote by Whitishcube 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.

P: 161

## Finding Instantaneous acceleration from a velocity-time graph

 P: 161 nevermind I got it now.

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