How to Convert a Velocity-Time Graph to a Position-Time Graph?

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SUMMARY

This discussion focuses on converting a velocity-time graph into a position-time graph using specific calculations. The user converted velocities from km/h to km/min to align with time intervals, calculating displacements for various speeds: 70 km/h (1.16 km/min) for 10 minutes, 80 km/h (1.33 km/min) for 30 minutes, 75 km/h (1.125 km/min) for 5 minutes, 60 km/h (1.0 km/min) for 2.5 minutes, and 55 km/h (0.9 km/min) for 22.5 minutes. The resulting position-time graph raised questions about the appearance of reversing direction, which was clarified by community feedback confirming the calculations were conceptually correct.

PREREQUISITES
  • Understanding of velocity-time graphs
  • Knowledge of unit conversion (km/h to km/min)
  • Familiarity with displacement calculations
  • Basic graphing skills for position-time representation
NEXT STEPS
  • Study the principles of kinematics in physics
  • Learn about graphical analysis of motion
  • Explore the concept of area under the curve in velocity-time graphs
  • Investigate common pitfalls in interpreting position-time graphs
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Students studying physics, particularly those focusing on kinematics, as well as educators and anyone interested in understanding the relationship between velocity and position over time.

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Homework Statement


I have to calculate the displacements from a velocity-time graph to make a position-time graph.
The velocity-time graph from the question is attached.

Homework Equations


d=vt

The Attempt at a Solution


To calculate the values for a position-time graph, I first converted all the velocities from km/h to km/min to match the time.
The truck travels at 70 km/h (1.16 km/min) for 10 mins: 1.16 x 10 = 11.6
It then travels at 80 km/h (1.33 km/min) for 30 mins: 1.33 x 30 = 40
It then travels at 75 km/h (1.125 km/min) for 5 mins: 1.25 x 5 = 6.25
It then travels at 60 km/h (1.0 km/min) for 2.5 mins: 1.0 x 2.5 = 2.5
And then travels at 55 km/h (0.9 km/min) for 22.5 mins: 0.91 x 22.5 = 20.6

I drew a position-time graph which is attached. I just want to know if I did this right because the graph I made looks a little weird to me.
 

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Since the truck keeps traveling in the same direction, how is it that your position graph shows it reversing direction?
 
phinds said:
Since the truck keeps traveling in the same direction, how is it that your position graph shows it reversing direction?

Ah, I think I've got it now. Does this look right?
 

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I didn't check the details but that's certainly the right idea.
 
Ok, thanks!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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