# Finding Instantaneous Acceleration/Velocity

Hi, you guys seem really helpful so I didnt think the template was necessary as my question dosent really fit those guidelines.

Im currently working on a review sheet for my Physics test tomorrow, and am having trouble finding the instantaneous velocity from looking at a acceleration vs. time graph and the instantaneous velocity from a position vs. time graph.

the problems i need help with are 1c and 2d on this page:http://www.jburroughs.org/science/mschober/consta/sframe.htm [Broken]

Im just looking for a general method on how to complete these problems.

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djeitnstine
Gold Member
using an acceleration vs. time graph you simply need to find the tangent at that point. remember $$Lim _{\Delta x \rightarrow 0}$$ $$\frac{f(x+\Delta x)-f(x)}{\Delta x}$$

And for velocity from a position vs. time graph you just need to find the area under the graph. $$\sum f(x) \Delta x$$

using an acceleration vs. time graph you simply need to find the tangent at that point. remember $$Lim _{\Delta x \rightarrow 0}$$ $$\frac{f(x+\Delta x)-f(x)}{\Delta x}$$

And for velocity from a position vs. time graph you just need to find the area under the graph. $$\sum f(x) \Delta x$$
thanks for the reply, but Im still kind of confused. When looking at a position vs. time graph, wouldnt the area under the curve be the velocity over that time interval, not the specific time?

djeitnstine
Gold Member
Oops I have all of that backwards.

let me rewrite that:

using a position vs. time graph you simply need to find the tangent at that point. remember $$Lim _{\Delta x \rightarrow 0}$$ $$\frac{f(x+\Delta x)-f(x)}{\Delta x}$$

And for velocity from a acceleration vs. time graph you just need to find the area under the graph. $$\sum f(x) \Delta x$$

I'm really sorry if I confused you.

ok gotcha. Just curious, is there any other simpler way to find the inst. velocity that finding the tangent? It takes a long time and seems to be pretty inaccurate.

djeitnstine
Gold Member
Just take the derivative of the function at hand for the position/time graph and take the integral of the acceleration/time graph (derived using the formulas I gave you above.)