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

In summary, to find force as a function of position \( F = F(x) \) using velocity \( v = v(t) \), one must first express acceleration \( a \) in terms of position. This can be achieved by using the relationship \( a = \frac{dv}{dt} \) and applying the chain rule to rewrite it as \( a = \frac{dv}{dx} \cdot \frac{dx}{dt} \), where \( v = \frac{dx}{dt} \). By substituting \( a \) into Newton's second law \( F = ma \), the force can then be expressed as a function of position \( x \), allowing for the analysis of motion in terms of position
  • #1
MatinSAR
614
188
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?
 
Physics news on Phys.org
  • #2
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.
 
  • Like
Likes Frabjous
  • #3
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)
 
Last edited:
  • Like
Likes MatinSAR
  • #4
Think about when the accleration (force) is positive, zero or negative.
Think about what the velocity always being positive means for the position.
 
  • Like
Likes MatinSAR and erobz
  • #5
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?
 
  • #6
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.
 
  • Like
Likes MatinSAR
  • #7
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.
 

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

What is the relationship between velocity and force in this context?

Velocity \( v(t) \) is the rate of change of position with respect to time, and force \( F(x) \) is related to the acceleration, which is the rate of change of velocity with respect to time. Using Newton's second law, \( F = ma \), where \( a = \frac{dv}{dt} \), we can derive the force as a function of position if we know the velocity as a function of time.

How can we express acceleration in terms of position and velocity?

Acceleration \( a \) can be expressed as \( a = \frac{dv}{dt} \). Using the chain rule, this can be rewritten as \( a = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx} \). Thus, acceleration can be expressed in terms of position \( x \) and velocity \( v \).

How do we derive force as a function of position from velocity as a function of time?

First, express acceleration in terms of velocity and position: \( a = v \frac{dv}{dx} \). Then, apply Newton's second law \( F = ma \). Substituting for \( a \), we get \( F(x) = m v \frac{dv}{dx} \). If \( v \) is known as a function of time \( t \), we need to find \( x(t) \) by integrating \( v(t) \) and then use it to express \( v \) as a function of \( x \).

What if the velocity is not directly given as a function of position?

If velocity \( v \) is given as a function of time \( t \), we need to find the position \( x \) as a function of time by integrating \( v(t) \). Once we have \( x(t) \), we can invert this relation to express \( t \) as a function of \( x \), and substitute back into \( v(t) \) to get \( v(x) \). Then we can use the relationship \( F(x) = m v \frac{dv}{dx} \) to find the force as a function of position.

Can you provide an example of finding force as a function of position?

Suppose \( v(t) = kt \), where \( k \) is a constant. Integrating \( v(t) \) gives \( x(t) = \frac{kt^2}{2} \). Solving for \( t \), we get \( t = \sqrt{\frac{2x}{k}} \). Substituting this back into \( v

Similar threads

Calculating Energy Dissipation for a Moving Object Using Integral Calculus
Replies
14
Views
2K
Why is a 1d conservative force approximately linearly related to position?
Replies
3
Views
455
Understanding the generic nature of linearity in Physics
Replies
2
Views
631
Vector is a function of its position or not?
Replies
16
Views
1K
Find the distance a particle travels
Replies
15
Views
736
How Does the Positive Spring Constant k = F/x Apply in Bow Mechanics?
Replies
4
Views
1K
Find dissipative function for the non-linear force f=-bv^n
Replies
3
Views
1K
What is meant by small oscillations for this potential energy function?
Replies
2
Views
462
Finding shortest distance between an observer and a moving object
Replies
13
Views
1K
Finding the y-component of a velocity vector
Replies
2
Views
1K

Hot Threads

Recent Insights

Back
Top