Help with instantaneous velocity

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Homework Help Overview

The discussion revolves around finding the instantaneous velocity from a given distance vs. time graph. The original poster is specifically interested in determining the instantaneous velocity at T=1.4 seconds, while some participants express confusion regarding the definitions and formulas related to velocity and acceleration.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants discuss the formula for instantaneous velocity and its relationship to the graph. There are attempts to clarify whether the focus is on velocity or acceleration, with some suggesting the use of the slope of the graph to find the instantaneous velocity.

Discussion Status

The conversation is ongoing, with various interpretations of the problem being explored. Some participants have provided guidance on how to approach finding the instantaneous velocity, while others have questioned the assumptions made regarding the formulas used.

Contextual Notes

There is a mention of the graph being a straight line, which implies constant velocity, and the need to clarify the difference between instantaneous and average values. The original poster is working within the constraints of a homework assignment, which may limit the information available for discussion.

ramin86
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I was given the following graph:

http://www.webassign.net/pse/p2-03.gif

I must find the instantaneous velocity at T=1.4

The formula for instantaneous velocity is delta x/delta t as t approaches zero. However, I'm still not sure how to work this problem. Please help
 
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Find the equation for x(t) in the interval t = [0,2]. You should know the velocity by then.
 
The formula you gave is for instantaneous acceleration, not velocity. So I will assume you are looking for instantaneous acceleration.
You will note the graph is a straight line from 0 - 2 seconds which means the acceleration is constant. The instantaneous acceleration is therefore the same as the average acceleration over that interval.
 
No, its instantaneous velocity that I'm looking for. I thought instantaneous acceleration is the change in velocity over the change in time.
 
Chronos said:
The formula you gave is for instantaneous acceleration, not velocity. So I will assume you are looking for instantaneous acceleration.
You will note the graph is a straight line from 0 - 2 seconds which means the acceleration is constant. The instantaneous acceleration is therefore the same as the average acceleration over that interval.

HOWEVER though, the graph does not show change in velocity over change in time. It's simply distance vs time. Besides, Ramin gave the right formula, distance over change in time. i think you should ignore chrono's comments Ramin...

Since it is a straight line though, the velocity is not changing, so you can just find the instantaneous velocity by finding the slope of the 0-2 second interval(since t = 1.4 has the same velocity as that interval)
 
Last edited:
My bad, you are correct, dx/dt. Make a table x1, t1, x2, t2. x1 and t1 will always be 0,0. x2 and t2 will be per the graph. solve for dx/dt at t2 = 1.4 seconds. Hint dx is always x2 - x1 and dt is always t2 - t1
 

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