Position-Time Graph: Average & Instantaneous Velocity

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Homework Help Overview

The discussion revolves around understanding position-time graphs, specifically focusing on calculating average and instantaneous velocity. The original poster describes a graph resembling an upside-down parabola and seeks clarification on how to determine the instantaneous velocity at a specific point in time.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss methods for finding instantaneous velocity by measuring the slope of the tangent line at a given point. There are questions about how to find this slope without a time interval and how to interpret the graph's features, such as when the velocity is zero.

Discussion Status

Several participants have offered insights into the process of drawing a tangent line and calculating its slope. There is an ongoing exploration of the relationship between the position function and velocity, with some participants expressing uncertainty about specific steps in the process.

Contextual Notes

The original poster mentions difficulty in drawing the graph and understanding the concept of instantaneous velocity at a single point, indicating potential gaps in foundational knowledge or graphical interpretation skills.

mooneh
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heyy, i have question abt position-time graph, i can't draw it so i'll try to explain it, it looks like an upside prabola, and the question asks first for the average velocity in certain time interval; i knew how to do that by subtracting xf-xi/tf-ti.
but then another branch of te question is to determine the instantaneous velocity at t=2 by measuring the slope of the tangent ?
how do u find the slope at a certain second ??
and another brach is at what value of t is the velocity zero ??
 
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Well the instantaneous velocity is dealing with calculus. To approximate the instantaneous velocity we need to find the slope of the line tangent to the particular point on the curve.

You need to realize the the change in position is essentially velocity (This is the slope of the curve at any particular point). So if the position is increasing (concave up) then the velocity is positive (note not necessarily increasing, just positive). If the position is decreasing then velocity is negative. To go from positive to negative or negative to positive the velocity has to be 0 at some point. Can you go on from there?
 
To find the slope at a particular time "t," you need to draw a line that is tangent to the curve at time "t." This should be a straight line thaextends from the top of the curve to the x-axis. Then you simply take that tangent line to be the hypotenuse of a right triangle. You want to find how long the triagngle is at its base ( this will be your "run"), and the height of the triangle (this will be your "rise"). To find the slope, you divide the rise by the run.

We know that the slope of the position function gives us the velocity, so think about when the slope is zero.

Edit: Well Feldoh beat me to it :)
 
i get how to find the velocity is zero but i still don't understand how to find the slope when i don't have a time interval, i only have one point; at t=2
i drew the tangent but then wut ?
i sound extremely stupid but i'd appreciate the help :P
 
mooneh said:
i get how to find the velocity is zero but i still don't understand how to find the slope when i don't have a time interval, i only have one point; at t=2
i drew the tangent but then wut ?
i sound extremely stupid but i'd appreciate the help :P

You don't sound stupid, you're just trying to learn like everyone else :)


Your tangent line should extend from the x-axis to the top of the curve. make sure that your line just touches the point "t." Make this line the hypotenuse of a right triangle, and find its x distance, and its y distance. Divide the y distance by the x distance, and you'll have your slope, does that help?
 
The tangent line is well, a line. Why do we like lines compared to curves? It's a lot easier to find the slope of a line.
 

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