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Instantaneous Acceleration on a Velocity-Time Graph

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Linday12
#1
Mar16-08, 04:37 PM
P: 54
I need to find the acceleration at a specific time (for example, 6s). I know how to do the acceleration between time intervals, slope=rise/run, a=vf-vi/t2-t1, but what do I do when I need the acceleration at a specific time?

I've tried to do a=v/t, but it seems like there's something I'm missing. Like for instance, what do you do for a straight line, obviously the acceleration is 0, but this method gives a different acceleration.
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Snazzy
#2
Mar16-08, 04:47 PM
P: 459
The tangent of a velocity-time graph represents instantaneous acceleration.
Linday12
#3
Mar16-08, 06:38 PM
P: 54
Ok, The velocity-time graph in this case is simpler, and is a series of straight lines (not curved). So, I wouldn't have to draw the tangent, would I? Do I have to pick another point on the line and use it? How would this give the acceleration for the given point, and not for a time interval? (Say I need the acceler. for 6s, wouldn't taking the data for 6s-5s be giving a time interval instead of the acceleration for 6s?)

Sorry, not entirely sure I'm making sense. Any help is greatly appreciated!

Snazzy
#4
Mar16-08, 06:43 PM
P: 459
Instantaneous Acceleration on a Velocity-Time Graph

For a straight curve on a V-t graph, it doesn't matter which point you pick or if you pick an interval because the acceleration is constant for all points on that interval. Remember, the acceleration is the derivative of velocity with respect to time, so if you have a linear relationship between velocity and time, you'll have a constant acceleration.
Linday12
#5
Mar16-08, 07:08 PM
P: 54
Thank you. That was exactly what I was lost on!
elemis
#6
Oct18-10, 08:52 AM
P: 161
http://www.xtremepapers.net/CIE/Inte..._s10_qp_11.pdf

So if you have a curved velocity time graph like in question 8 in the above link, all you have to do is draw a tangent to the required time ?


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