Calculate the speed of a point based on a graph

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SUMMARY

The discussion focuses on calculating the average and maximum speed of a point moving rectilinearly based on a distance-time graph. The average speed is determined using the formula v=Δs/Δt, yielding a result of 0.1 m/s for a total distance of 2.0 m over 20 seconds. The maximum speed is identified by analyzing the slope of the graph, specifically between time intervals t=10s and t=16s, where the distance changes from s=4.0 m to s=18.0 m. The point where the instantaneous speed equals the average speed is found at t=16s, where the tangent line's slope matches the average speed line from the origin.

PREREQUISITES
  • Understanding of basic kinematics, specifically speed and velocity.
  • Familiarity with graph interpretation, particularly distance vs. time graphs.
  • Knowledge of calculus concepts, including derivatives and slopes.
  • Proficiency in applying the formula v=Δs/Δt for speed calculations.
NEXT STEPS
  • Study the concept of derivatives to understand instantaneous speed calculations.
  • Learn how to analyze slopes on distance-time graphs for maximum speed determination.
  • Explore the relationship between average speed and instantaneous speed in motion analysis.
  • Practice problems involving distance, time, and speed to reinforce these concepts.
USEFUL FOR

Students studying physics, particularly those focusing on kinematics, as well as educators looking for examples of graph analysis in motion. This discussion is also beneficial for anyone interested in understanding the principles of speed and velocity in a practical context.

  • #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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