Finding the instantaneous velocity on a position time graph

by slag1928
Tags: instantaneous, tangent, velocity
slag1928 is offline
Sep26-11, 05:06 PM
P: 13
I need help with a very general question. I was asked to find the instantaneous velocity of a position time graph at .5 seconds. i know to do this i need to create a line that is the tangent to that point. Here lie the problem... how on earth do i make that line, and how do i measure the slope?

i think (y2-y1/x2-x1)for finding the slope of the tangent line? but i have no idea where that line should begin or end.

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Modulated is offline
Sep26-11, 05:52 PM
P: 44
It doesn't matter where the line begins or ends, it still has the same slope... So I'd draw a reasonably long line so that both the "rise" ([itex]y_2-y_1[/itex]) and "run" ([itex]x_2-x_1[/itex]) are biggish numbers, and then calculate their ratio as you suggest. (If they're both big numbers, they're easier to measure precisely so your value for the gradient will be more accurate.)
slag1928 is offline
Sep26-11, 05:55 PM
P: 13
thank you. turns out i was over complicating things. the velocity was constant for the points i was finding so it was as simple as finding the slope. T.T

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