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

[MatinSAR]

Homework Statement

We have v-t graph of a moving object. We are asked to guess it's F-x graph.

Relevant Equations

$\vec F=md \vec v/dt$

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?

 

[MatinSAR]

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.

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.

 

[erobz]

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)

 

[Frabjous]

Think about when the acceleration (force) is positive, zero or negative.
Think about what the velocity always being positive means for the position.

 

[MatinSAR]

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.

Think about when the acceleration (force) is positive, zero or negative.
Think about what the velocity always being positive means for the position.

Thanks. Did you see post #2?

 

[Frabjous]

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.

 

[MatinSAR]

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.