Finding this displacement on a v/t graph

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SUMMARY

The discussion centers on calculating displacement from a velocity-time (v/t) graph over the first 18 seconds. The correct method involves determining the area between the velocity line and the horizontal axis, with areas below the axis counted as negative. The user initially calculated the total area as 171 or -171 meters, which was incorrect. The accurate approach requires careful consideration of the areas above and below the axis to derive the correct displacement value.

PREREQUISITES
  • Understanding of velocity-time graphs
  • Knowledge of calculating areas of triangles and rectangles
  • Familiarity with the concept of displacement in physics
  • Ability to interpret graphical data
NEXT STEPS
  • Study the principles of kinematics in physics
  • Learn how to calculate areas under curves in physics
  • Explore the relationship between velocity and displacement
  • Practice with various v/t graphs to reinforce understanding
USEFUL FOR

Students studying physics, particularly those focusing on kinematics and graph interpretation, as well as educators looking for examples of displacement calculations from v/t graphs.

jeffc93
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Homework Statement



2j3p8x2.gif


there's a picture of the graph and i need to determine the displacement of the object within the first 18seconds of it. (in meters)

Homework Equations



area below line is displacement (i thought?)

The Attempt at a Solution



i just did some basic things:

i found the height which is 18.

and then i found the areas of all the triangles and rectangles under the line and added them up and got 171 or -171 but its not right

any help?

any help would be greatly appreciated, therefore i could do the rest on my own :)
 
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Welcome to PF!

Hi jeffc93! Welcome to PF! :smile:
jeffc93 said:
area below line is displacement (i thought?)

Nope :redface: …

area between line and horizontal axis (with areas below the axis counting as negative) is displacement. :wink:
 

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