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Finding instantaneous velocity at given points on nonlinear graph

  • Thread starter exi
  • Start date
  • #1
exi
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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?
 

Answers and Replies

  • #2
58
1
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.
 
  • #3
exi
85
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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.
 
  • #4
1,707
5
well at 2s it's zero
 
  • #5
exi
85
0
well at 2s it's zero
And how is that?
 
  • #6
1,707
5
And how is that?
im sorry i meant 4s
 
  • #7
exi
85
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im sorry i meant 4s
Alright... but how is that?
 
  • #8
1,707
5
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.
 

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