# Finding instantaneous velocity at given points on nonlinear graph

## Homework Statement

Problem is presented as a nonlinear graph. Y-axis is in meters and ranges from 0.0 to 40.0m in increments of 10m, and the X-axis is in seconds, 0.0-5.0s (increments of 1). Points on graph include (0,10), (1,~19), (2, ~23), (3, ~26), (4, ~26), and (5, ~20).

Question: "Estimate the instantaneous velocity of the train at 2.0s." (and again at 4.0s).

## Homework Equations

V(av) = delta(x) / delta(t)
V(as delta(t) approaches 0) = delta(x) / delta(t).

## The Attempt at a Solution

Unsure.

I understand the concept behind finding instantaneous velocity in that it's an instant measure of motion and includes a directional component, but damn if I can remember how to do it. The notes from class are a bit too complicated to be of use, and guesstimating the slope of a line tangent to x=2 isn't much help, given this graph.

Any thoughts?

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guesstimating the slope of a line tangent to x=2 isn't much help, given this graph.

Any thoughts?
Do you have a copy of the graph online? You should be able to obtain a reasonable tangent by using a ruler and a pencil, and drawing your tangent onto your graph.

Do you have a copy of the graph online? You should be able to obtain a reasonable tangent by using a ruler and a pencil, and drawing your tangent onto your graph.
I'm afraid I don't. Besides drawing in more precise increments on the axes, sketching the tangent line, and taking the slope of that where it intersects the point in question, how should I be going about this? All of the questions before that one on these notes are in reference to distance, displacement, average speed and velocity - all the usual stuff - but I'm still a little unsure as to the most efficient way to solve it.

well at 2s it's zero

well at 2s it's zero
And how is that?

And how is that?
im sorry i meant 4s

im sorry i meant 4s
Alright... but how is that?

Alright... but how is that?
(3, ~26), (4, ~26) because dx/dt is zero there? as in it doesn't move as in it's position doesn't change, again if i could see the graph it would be much easier to tell.

i guess it could have a steep negative slope around there somewhere.