- #1
ellieee
- 78
- 6
- Homework Statement
- qn (di)-> the graph is a v-t graph, so shouldn't we say "constant velocity" instead of "speed"?
- Relevant Equations
- nil
Can you use speed and velocity interchangeably?
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Speed and velocity are closely related concepts, particularly in one-dimensional motion, where they can often be used interchangeably. Speed refers only to magnitude, while velocity includes direction, which is crucial for understanding motion. In a straight line, if speed is x, velocity can be either x or -x, but this relationship does not apply to other types of motion, such as circular motion. The discussion highlights the importance of accurately interpreting graphs, particularly velocity-time graphs, to determine the nature of the motion. Accurate labeling of such graphs is essential for confirming that the motion is indeed linear.
- #2
Doc Al
Mentor
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Yes. It's better to describe it as constant velocity, as speed just describes the magnitude and not the sign. (I presume this is one-dimensional motion.)
- #3
Delta2
Homework Helper
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In motion in a straight line such as in your example, the concepts of velocity and speed are almost identical: If speed is x, then velocity will be x or -x. I repeat, this holds only in motion in a straight line, it does not hold for other motions e.g. circular motion.
- #4
ellieee
- 78
- 6
how can you tell it's in a straight line ?Delta2 said:
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- #5
Delta2
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I can not infer it from the given data (the screenshot of the graph isn't complete, doesn't completely show the text that describes the graph), somehow I thought it was given to you.ellieee said:
- #6
Doc Al
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Since they label it a velocity-time graph, where the velocity is constant it must be moving in a straight line. (Assuming it's labeled accurately.)ellieee said:
My attempt:
Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE.
PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##.
PE of part in the tube = ##\frac{m}{l}(l - h)gh##.
Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##.
Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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