How to Draw an Acceleration vs Displacement Graph?

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Homework Help Overview

The discussion revolves around drawing an acceleration versus displacement graph, with references to equations of motion and calculus. Participants are exploring the relationship between velocity, acceleration, and displacement.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to derive the acceleration from a velocity-displacement equation but expresses uncertainty about the correctness of their approach. Some participants suggest avoiding differentiation to simplify the problem, questioning the introduction of additional variables.

Discussion Status

Some participants have provided hints and guidance regarding the relationships between the variables involved. There is an acknowledgment of different approaches being considered, but no explicit consensus has been reached.

Contextual Notes

Participants are discussing the implications of differentiating equations and the potential introduction of extra variables, particularly time, in their analysis.

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Homework Statement


The graph is as folllows-(SEE ATTACHMENT)
Q.draw acceleration vs disp. graph.

Homework Equations


Basic equations of motion, basic calculus

The Attempt at a Solution


I have tried to write a linear equation in v and s and then differentiating it w.r.t. to time. but I am not sure if this is the correct way to do it. It gives me a positive slope and -ve y intercept.
Also if there is/are any other ways please do tell
 

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first write the v-s equation, but do not diffrentiate it with time(as it would add one more variable which is not needed)

Hint: What is the relation between v,a and s?
 
thanks for your reply.
i have solved the question.
1 doubt though which extra variable is generated? on differentiation
v becomes a
s becomes v
constant is eliminated
 
hav0c said:
thanks for your reply.
i have solved the question.
1 doubt though which extra variable is generated? on differentiation
v becomes a
s becomes v
constant is eliminated

I thought 't" would create some some problems but thanks it is eliminated. You can also do it using the relation a = v*dv/ds. This was what i was referring to!
 

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