Finding the instantaneous velocity on a position time graph


by slag1928
Tags: instantaneous, tangent, velocity
slag1928
slag1928 is offline
#1
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.

Thanks
Phys.Org News Partner Science news on Phys.org
NASA's space station Robonaut finally getting legs
Free the seed: OSSI nurtures growing plants without patent barriers
Going nuts? Turkey looks to pistachios to heat new eco-city
Modulated
Modulated is offline
#2
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
slag1928 is offline
#3
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


Register to reply

Related Discussions
Instantaneous Velocity with Position v. Time graph Introductory Physics Homework 1
Finding Instantaneous acceleration from a velocity-time graph Introductory Physics Homework 4
Finding X components for instantaneous velocity with points on a graph Introductory Physics Homework 1
Finding instantaneous velocity at given points on nonlinear graph Introductory Physics Homework 7
velocity time graph &instantaneous acceleration Introductory Physics Homework 4