Solving a Displacement-Time Graph with Sharp Turnings

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The discussion focuses on converting a displacement-time graph with sharp turnings into a velocity-time graph, highlighting the challenge posed by non-differentiable points. It notes that real-world graphs cannot have perfectly sharp points due to the implications of infinite force and velocity changes. The sharp points may result from infrequent sampling or insufficient graph scale, suggesting that the professor might be testing students' understanding of these concepts. To represent discontinuities in the velocity graph, one can use mathematical notation, indicating that the velocity graph will not be continuous at sharp corners. Ultimately, the task may involve calculating average velocities for each segment to create a new graph composed of straight lines.
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I got a problem described by a displacement time graph. It has sharp turnings at 2 points ( and sucessives) and the question is to convert the graph to velocity time graph. Since sharp pionts are not differentiable how it can be drawn ?
 
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Listin said:
I got a problem described by a displacement time graph. It has sharp turnings at 2 points ( and sucessives) and the question is to convert the graph to velocity time graph. Since sharp pionts are not differentiable how it can be drawn ?

This is a little tricky to answer because in a class room setting it depends on your professor's perspective.

In the real world there cannot be perfectly sharp points as that implies an infinite rate of change of velocity and therefor infinite force being applied.

Is the graph produced from real data? If so, the sharp point is either a matter of infrequent sampling missing the points that would clearly define the change of velocity there, or it could be that the scale of the graph is not sufficiently fine to show the rapid change.

Could the professor be trying trying teach you this by giving you an apparently impossible graph?

If you want to test the graph as not necessarily bound by natural law (i.e., as a mathematical ideal), could use the math notation of inserting a circle in the velocity time graph to in indicate a discontinuity.
 
If there are sharp corners in the displacement graph, the velocity graph will not be continuous there. For example, if the displacement graph is straight line from (0, 0) to (5, 10) and then changes to the straight line from (5, 10) to (10, 10). The velocity will be the constant (10- 0)/(5- 0)= 2 from 0 to 5, then the constant (10-10)/(10-5)= 0 from 5 to 10. The velocity graph will be the horizontal line from (0, 2) to (5, 2), then the horizontal line from (5, 0) to (10, 0).
 
It is quite common in high school physics to show graphs of the motion (position-time and velocity-time) made from straight line segments connected at various angles.
Usually the questions related to these graphs regard the values at various points on the segments or average values for a segment, in which case there is no problem.
If they ask for instantaneous values at these connection points, then there is a problem.

If the question is just to convert to v-t graph then they may expect you calculate the average velocity on each segment and build another graph made from straight lines.
 
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