Instantaneous Velocity from the graph

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Homework Help Overview

The problem involves determining instantaneous velocity from a distance-time graph, specifically at given time intervals such as 1.1s, 7.4s, and 2.7s. The original poster references a limit formula for instantaneous velocity and expresses difficulty in applying it to the graph provided.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the nature of the graph, noting that it consists of straight lines rather than curves. There is an exploration of how to calculate slopes to find instantaneous velocity and a questioning of the relationship between average and instantaneous velocity in this context.

Discussion Status

The discussion is ongoing, with participants providing insights about the graph's characteristics and the implications for calculating velocity. Some guidance has been offered regarding the interpretation of slopes, but there remains uncertainty about the distinction between average and instantaneous velocity.

Contextual Notes

Participants note discrepancies in the graph's data points and the implications for calculations. There is an acknowledgment of the limitations in drawing tangents to straight lines, which complicates the original poster's approach.

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Homework Statement



The problem is finding the instantaneous velocity during certain time intervals from the graph.
p2-03.gif


Examples of time intervals: 1.1s, 7.4s, 2.7s

Homework Equations



I know this is the formula for finding the instantaneous velocity

lim Δv=(Δx/Δt)
t->0

The Attempt at a Solution



I tried locating the points on the curve, by taking two obvious points like 1 on the x and 4 on the y and then making a cross multiplication to find the y in terms of 1.1 and then dividing by the T, since I do not think drawing a tangent to the point is possible here. So, should I use the equation somehow? I made several attempts but they all turned out to be incorrect.
 
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According to your graph, there are no "curves", but only straight lines. Recall that on a distance-time graph, the slope is the velocity. The lines' slopes are easy to calculate because they intersect on reasonable numbers.

So, the slope at 1.1 sec would be equal to the slope at 1 sec or any time within 0-2 seconds.

Also, at x=1, y does not equal 4, but actually equals 5.
 
Shootertrex said:
According to your graph, there are no "curves", but only straight lines. Recall that on a distance-time graph, the slope is the velocity. The lines' slopes are easy to calculate because they intersect on reasonable numbers.

So, the slope at 1.1 sec would be equal to the slope at 1 sec or any time within 0-2 seconds.

Also, at x=1, y does not equal 4, but actually equals 5.

But isn't the average velocity different than the instantaneous one? Like, I know the slope can represent the average velocity, but is that true on this graph for instantaneous velocity too because it's not a curve?
 
Instantaneous velocity is the slope of the line tangent to a point on a curve. Is there really a way to make a line that will be tangent to another line? Not really. Let's say that you can. This 'tangent' line have the same slope as the line it is tangent to, and is actually the same line.

Lets say that f(x)=4x will be the equation that illustrates the distance of an object. Velocity is the change in distance over the change in time. The change in distance for this function is the slope of the line. Therefore the slope of this line will equal the velocity, both average and instantaneous.
 
Yes, that is true. Thank you VERY much for that ^^
 

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