Calculate the speed of a point based on a graph

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Homework Help Overview

The discussion revolves around a problem involving the motion of a point moving rectilinearly, with a focus on analyzing a graph that depicts the distance traveled over time. Participants are tasked with calculating average speed, maximum speed, and identifying the moment when instantaneous speed equals average speed over a specified interval.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of average speed using the formula v=Δs/Δt and question the validity of their results based on the graph. They explore how to determine maximum speed by identifying segments of the graph with the greatest change in distance over the least change in time. Questions arise regarding the interpretation of slopes and the relationship between average and instantaneous speeds.

Discussion Status

The conversation is active, with participants providing insights and questioning each other's reasoning. Some participants have offered guidance on identifying segments of the graph for maximum speed and clarifying the relationship between average and instantaneous speeds. There is an ongoing exploration of how to accurately interpret the graph and apply the relevant equations.

Contextual Notes

Participants note the limitations of the information provided, indicating that they are working solely from the graph without additional data. There is a focus on ensuring the correct interpretation of the graph's scale and values, which has led to some confusion regarding the calculations.

  • #31
haruspex said:
The line from the origin to the curve at t=10s is the same as the tangent at t=10s? I don't think so.
But I think there isn't any line from the origin that would be equal to the tangent at t=10s. At leaks it looks like that.
 
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  • #32
Davidllerenav said:
But I think there isn't any line from the origin that would be equal to the tangent at t=10s. At leaks it looks like that.
Right, so t=10s is not the point we're after - that was just an example.
Find a point where the two lines to that point, as we have discussed, would be the same.
 
  • #33
haruspex said:
Right, so t=10s is not the point we're after - that was just an example.
Find a point where the two lines to that point, as we have discussed, would be the same.
It seems that t=16 would do.
 
  • #34
Davidllerenav said:
It seems that t=16 would do.
Bingo.
 
  • #35
haruspex said:
Bingo.
Ok! So for the second question, I'll have to calculate the maximum speed (slope) at the interval from t=10 sto t=14s, where the distance goes from s=1.0m to s-1.4m.
For the third, I must say that the instantaneous speed at t=16 is equal to the average speed from t=0s to t=16s.
That's all, right?
 
  • #36
Davidllerenav said:
Ok! So for the second question, I'll have to calculate the maximum speed (slope) at the interval from t=10 sto t=14s, where the distance goes from s=1.0m to s-1.4m.
For the third, I must say that the instantaneous speed at t=16 is equal to the average speed from t=0s to t=16s.
That's all, right?
Yes.
 
  • #37
haruspex said:
Yes.
Thanks for your help!
 
  • #38
Davidllerenav said:
Thanks for your help!
You are welcome. You understood the method, yes?
 
  • #39
haruspex said:
You are welcome. You understood the method, yes?
Yes. When I'm asked to find the maximum velocity on a graph of distance vs. time, I must look for the interval where the point travels more distance in the less time. When I'm asked to find the point where the instantaneous speed is equal to the average speed, I must find the point where its tangent line (slope/derivative) is equal to the line from the origin to that point, which is the average speed.
 
  • #40
Davidllerenav said:
When I'm asked to find the point where the instantaneous speed is equal to the average speed, I must find the point where its tangent line (slope/derivative) is equal to the line from the origin to that point, which is the average speed.
Yes, but I meant more in terms of the principles behind that... that the tangent slope gives the instantaneous speed and the slope from the start point to current point gives current average speed.
 
  • #41
haruspex said:
Yes, but I meant more in terms of the principles behind that... that the tangent slope gives the instantaneous speed and the slope from the start point to current point gives current average speed.
Yes. I understood that.
 

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