Finding Instantaneous acceleration from a velocitytime graphby MitsuShai Tags: acceleration, graph, instantaneous, velocitytime 

#1
Aug3110, 03:23 PM

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. 



#2
Aug3110, 03:35 PM

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.




#3
Aug3110, 03:50 PM

P: 161





#4
Sep210, 04:58 PM

P: 161

Finding Instantaneous acceleration from a velocitytime graph 



#5
Sep210, 07:42 PM

P: 161

nevermind I got it now.



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