Finding the greatest distance from the origin given a velocity vs. time graph.

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Homework Help Overview

The discussion revolves around determining the greatest distance from the origin based on a velocity vs. time graph, where an object moves with a maximum constant velocity of 2 m/s for the first 2 seconds, then reverses direction with a maximum negative constant velocity of -2 m/s for the next 2 seconds.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the implications of constant velocity on distance calculation and the effect of direction change on overall distance. Questions arise about the specific calculations needed to find the greatest distance and the interpretation of the graph's area.

Discussion Status

Some participants have offered insights into calculating distance using the area under the graph and the formula for distance with constant velocity. There is ongoing exploration of how to interpret the graph between the defined time intervals, and multiple interpretations of the problem are being considered.

Contextual Notes

There is a lack of clarity regarding the behavior of the object between 2 and 4 seconds, which is affecting participants' understanding of the problem. The original poster has requested hints rather than direct solutions.

1irishman
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Homework Statement


If given a velocity vs. time graph where max. constant velocity is 2m/s for 2s and the object turns around at 4s with max negative constant velocity is -2m/s for 2s. How do i find the greatest distance from the origin?



Homework Equations


d= vi+vf/2 *t
d=vit + 1/2at^2


The Attempt at a Solution


i'm really not sure what to do here...please hints?
 
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Okay, so you have constant speed for the first two seconds and the distance increases steadily at d = vt. Then the direction reverses and the distance diminishes to zero over the next 2 seconds. The greatest distance is pretty clear . . .
 
is it 5m?
 
is there a formula to help figure this out from just looking at a v-t graph coordinates?
 
Terribly sorry, I didn't see the bit about "turning around at 4s". Now I don't understand it because there is no indication of what happens between 2 and 4 seconds.

If you are looking at the graph, I suggest you find the distance traveled up to the turnaround time of 4 seconds. The area under the graph is the distance. Or use d = vt for the first 2 seconds and whatever is appropriate for the next two.
 

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