Velocity and position from forces with mixed dependences

In summary, the problem involves a particle of mass m moving under the influence of a force ##\vec{F}(t,v)=-ktv \hat{v}## and the goal is to calculate the velocity of the particle, the time it takes for the particle to stop, and the maximum x-position reached by the particle. The solution involves using Newton's Laws and integrating to obtain the equations for velocity and time, with the recommendation to use Wolfram Alpha for the integral involving position.
  • #1
QuantumCurt
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Homework Statement



This is for a junior level classical mechanics course. We're currently working with Newton's Laws.

A particle of mass m moves under the influence of a force ##\vec{F}(t,v)=-ktv \hat{v}##, where k is a positive constant. At time t=0, ##v=v_0 \hat{x}## at the origin, where ##v_0## is a positive constant.

a. Calculate the velocity of the particle ##\vec{v}(t)## for times t>0.

b. How much time does it take for the particle to stop?

c. What is the maximum x-position reached by the particle? Hint: you will need to look up ##\int_{0}^{\infty} (something) dt## on Wolfram Alpha, which is a tool that you should take the time to learn how to use!

Homework Equations



$$\vec{F}=m \frac{dv}{dt}$$

The Attempt at a Solution



For part a I figured that since F=m dv/dt, I could simply rewrite this as
$$-ktv \hat{v}=m \frac{dv}{dt}$$
And then solve for v

$$\frac{dv}{dt}=\frac{-ktv \hat{v}}{m}$$

$$\int{dv}=\int{\frac{-ktv \hat{v}}{m}}dt$$

$$\int{dv}=\frac{-kv \hat{v}}{m} \int{t}~dt~\hat{v}$$

So,

$$v=-\frac{kv~\hat{v}}{2m}t^2+v_0 \hat{x}$$

Where ##v_0 \hat{x}## is the constant of integration evaluated at time t=0.

Then for part b, I used this equation and set v=0, and arrived at

$$t=\sqrt{\frac{2v_0m}{kv}}$$

This seemed okay to me, but for part c it specifies that we're going to want to look up an integral. Obviously I can integrate velocity to obtain position, but the simplicity of the integral, along with the recommendation to use Wolfram Alpha, makes me think I'm doing something wrong. If anyone has any input, it would be greatly appreciated.
 
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  • #2
QuantumCurt said:
$$\frac{dv}{dt}=\frac{-ktv \hat{v}}{m}$$

$$\int{dv}=\int{\frac{-ktv \hat{v}}{m}}dt$$

$$\int{dv}=\frac{-kv \hat{v}}{m} \int{t}~dt~\hat{v}$$
.

First off, you should be clear that you have vectors on both sides of the equation, though maybe you just had problems making this clear in the LaTeX codes. Second, and more important for this problem, here ##v=v(t)##, so we can't just pull it out of the integral. Instead you should have separated variables before the integration by writing

$$\frac{1}{v} \frac{dv}{dt}=\frac{-kt}{m}.$$

You can then integrate both sides of this. For part (c), it looks like you will get an incomplete Gaussian integral, which is why they suggest using wolframalpha to get a numerical result for it.
 
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  • #3
I actually realized my mistake shortly after posting this. It was indeed what you're referring to. I managed to get it figured out though. Thank you!
 

Related to Velocity and position from forces with mixed dependences

1. What is the relationship between velocity and position in a system with mixed dependences?

The relationship between velocity and position in a system with mixed dependences is complex and depends on the specific forces acting on the system. In general, velocity is the rate of change of position over time, and forces can either accelerate or decelerate the system's motion. The resulting velocity and position will depend on the direction and magnitude of the forces.

2. How do forces with mixed dependences affect an object's motion?

Forces with mixed dependences can have a significant impact on an object's motion. They can cause changes in both the speed and direction of the object's movement, as well as its position in space. The specific effects will depend on the specific forces and their interactions within the system.

3. Can forces with mixed dependences cancel each other out?

Yes, forces with mixed dependences can cancel each other out. This can happen when two forces act in opposite directions and have equal magnitudes, effectively canceling each other's effects on the object's motion. This is known as a balanced force and can result in the object maintaining a constant velocity or coming to a complete stop.

4. How can the velocity and position of a system with mixed dependences be calculated?

The velocity and position of a system with mixed dependences can be calculated using mathematical equations and principles such as Newton's laws of motion and the equations of motion. These equations take into account the forces acting on the system and the initial conditions of the object's motion to determine its velocity and position at any given time.

5. How can the effects of forces with mixed dependences be manipulated or controlled?

The effects of forces with mixed dependences can be manipulated or controlled by adjusting the magnitude and direction of the individual forces. This can be done through various means, such as changing the angle or strength of a force, or adding or removing forces from the system. By understanding the relationships between forces, velocity, and position, scientists can manipulate and control the motion of objects in a variety of systems.

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