Question On Displacement - Velocity-Time Graph

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SUMMARY

The discussion clarifies the calculation of displacement using a velocity-time graph, specifically when the graph starts with a non-zero velocity. In the example provided, with an initial velocity of 20 m/s and a final velocity of 30 m/s over 2 seconds, the total displacement is calculated by combining the area of the triangle (10 m) and the rectangle (40 m), resulting in a total displacement of 50 m. The key takeaway is that displacement must account for the entire area under the velocity curve, regardless of the starting velocity, confirming that displacement is determined solely by initial and final positions.

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  • Knowledge of basic kinematic equations, specifically v^2 = u^2 + 2as
  • Ability to calculate areas of geometric shapes (triangles and rectangles)
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Linday12
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[SOLVED] Question On Displacement -- Velocity-Time Graph

When calculating displacement, if you have a graph that starts with a non zero velocity, do you calculate the area from zero still, or do you calculate only from where it starts?

For instance, a linear problem with a velocity starting at 20m/s 0s and goes to 30m/s 2s.

Do I go Area=1/2bh=1/2(30m/s-20m/s x 2s-0s)=10m and then.. this is the part, do I calculate the left over rectangle under it even though the velocity started at a non zero number? 20m/s x 2s=40m

40m+10m=50m, or does it just stay with where the linear velocity started (10m)?
 
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Displacement is independent of the path taken.
It depends only on the initial and the final positions.

Using the equation.:

v^2=u^+2as

we get s= 50m

It does not matter if the particle started with a non zero velocity.You must include that part of the rectangle also for calculating the displacement.The negative part of the graph does not mean you have to reduce that from the final answer.It must be added to get the final Result.
 
Thank you very much!
 
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