Find force as a function of position: F=F(x) using v=v(t)

AI Thread Summary
The discussion revolves around finding the force as a function of position using the velocity equation v = -3t^2. The user initially calculates position and acceleration but concludes that their derived force F = Ax^(1/3) contradicts the given options, which include F = -Ax^(1/3). The main error identified is the assumption that velocity can be negative while still representing a positive motion, as this leads to incorrect conclusions about acceleration and force. The user acknowledges this mistake and expresses gratitude for the guidance received. Understanding the relationship between velocity, acceleration, and position is crucial for accurate force calculations.
MatinSAR
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Homework Statement
We have v-t graph of a moving object. We are asked to guess it's F-x graph.
Relevant Equations
##\vec F=md \vec v/dt ##
1708036321695.png

If we consider ##v=-3t^2## then: $$x=-t^3$$$$a=-6t$$
Using ##t=-x^{1/3}## we have : ##a=-6(-x^{1/3})=6x^{1/3}##. My answer suggust that ##F=Ax^{1/3}## but in options we have ##F=-Ax^{1/3}##.

Can someone guide me where my mistake is?
 
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My 2nd idea:
Forget about the formulas and math ...

This object has positive velocity so when "t" is increasing "x" is also increasing. We know that before ##t=t_0## the acceleration is posotive but it's magnitude is decreasing with time so it is decreasing with position also.
1708038376724.png

After ##t=t_0## the acceleration becomes negative and it's magnitude is increasing by time and by position.
So "b" is the correct option.
 
MatinSAR said:
Homework Statement: We have v-t graph of a moving object. We are asked to guess it's F-x graph.
Relevant Equations: ##\vec F=md \vec v/dt ##

View attachment 340389
If we consider ##v=-3t^2## then: $$x=-t^3$$$$a=-6t$$
Using ##t=-x^{1/3}## we have : ##a=-6(-x^{1/3})=6x^{1/3}##. My answer suggust that ##F=Ax^{1/3}## but in options we have ##F=-Ax^{1/3}##.

Can someone guide me where my mistake is?
I think the mistake was to consider ## v = -3t^2##. If you are saying ##v## is an inverted parabola, then its graph would be like: ##v = -pt ( t-q)##. Two distinct roots (one zero), not a single repeated root (both zero)
 
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Think about when the accleration (force) is positive, zero or negative.
Think about what the velocity always being positive means for the position.
 
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erobz said:
I think the mistake was to consider ## v = -3t^2##. If you are saying ##v## is an inverted parabola, then its graph would be like: ##v = -pt ( t-q)##. Two distinct roots (one zero), not a single repeated root (both zero)
Thanks.
Frabjous said:
Think about when the accleration (force) is positive, zero or negative.
Think about what the velocity always being positive means for the position.
Thanks. Did you see post #2?
 
MatinSAR said:
Did you see post #2?
Missed it. Sorry. You did what I suggested before I suggested it.

BTW, the mistake with -3t2 is that it is always negative.
 
Frabjous said:
Missed it. Sorry. You did what I suggested before I suggested it.

BTW, the mistake with -3t2 is that it is always negative.
Thank you for your time. I see that big mistake in my first approach now.
 
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