Finding v(t) for a velocity dependent force

• littlehonda
In summary, the conversation discusses how to find an expression for v(t) given a one-dimensional retarding force. The equation is solved using the method of separable differential equations, and the final expression for v(t) is (ln(\alpha(-At+C3)))/\alpha. The conversation also mentions finding an expression for x(t), which can be done by integrating dx/dt and using initial conditions to solve for a new constant C3.
littlehonda

Homework Statement

Given the one-dimensional retarding force F=-Ae^(-$$\alpha$$v) find an expression for v(t).

Homework Equations

F = m(dV/dt)
A and $$\alpha$$ are constants, v is instantaneous speed.

The Attempt at a Solution

Im not sure how to frame the idea of integrating a velocity dependent force. Can I take simply integrate and use

v= -Ae^(-$$\alpha$$v)t

or should the v be eliminated from the right hand side. I tried doing that by dividing the left by e^(-$$\alpha$$v) and then doing a e^u du type integration but the answer i came up with didn't seem reasonable.

just looking for a little hint or push here, thanks!

Hint: You have a separable differential equation for v(t)

thanks for responding gabba,

I think that I tried the separable differential equations method but couldn't make sense of the answer

what i did was

e^($$\alpha$$v) dv = -Adt

1/$$\alpha$$(e^($$\alpha$$v)-e^($$\alpha$$v0)) = -At

i'm not sure what this means though or where to go from here

littlehonda said:
thanks for responding gabba,

I think that I tried the separable differential equations method but couldn't make sense of the answer

what i did was

e^($$\alpha$$v) dv = -Adt

1/$$\alpha$$(e^($$\alpha$$v)-e^($$\alpha$$v0)) = -At

i'm not sure what this means though or where to go from here

$$\int e^{\alpha v}dv =\frac{e^{\alpha v}}{\alpha}+C_1$$

And

$$\int-Adt=-At+C_2$$

$$\implies \frac{e^{\alpha v}}{\alpha}=-At+C_3$$ where $C_3 \equiv C_1-C_2$

Multiply both sides of the equation by $\alpha$ and then take the natural log. What initial conditions are you given?

ok so that's where I got to pretty much, the C1+C2=C3 makes sense.

Here is all the information given

A= 1/m/s^2
$$\alpha$$=.1 s/m
v0= 20 m/s

I assume $v_0=v(t=0)$? If so, you can solve for $C_3$.

littlehonda said:

Homework Statement

Given the one-dimensional retarding force F=-Ae^(-$$\alpha$$v) find an expression for v(t).

Homework Equations

F = m(dV/dt)
A and $$\alpha$$ are constants, v is instantaneous speed.

The Attempt at a Solution

Im not sure how to frame the idea of integrating a velocity dependent force. Can I take simply integrate and use

v= -Ae^(-$$\alpha$$v)t

or should the v be eliminated from the right hand side. I tried doing that by dividing the left by e^(-$$\alpha$$v) and then doing a e^u du type integration but the answer i came up with didn't seem reasonable.

just looking for a little hint or push here, thanks!

littlehonda said:
thanks for responding gabba,

I think that I tried the separable differential equations method but couldn't make sense of the answer

what i did was

e^($$\alpha$$v) dv = -Adt

1/$$\alpha$$(e^($$\alpha$$v)-e^($$\alpha$$v0)) = -At

i'm not sure what this means though or where to go from here
$$m\frac{dv}{dt}= -Ae^{-\alpha t}[/itex] which separates as [tex]e^{\alpha v}dv= -A/m dt$$
(you dropped the "m")
Integrating,
$$-\frac{1}{\alpha}e^{\alpha v}+ C= -(A/m)t$$
Assuming that your $v_0$ is the velocity when t= 0,
$$-\frac{1}{\alpha}e^{\alpha v_0}+ C= 0$$
so
$$C= \frac{1}{\alpha}e^{\alpha v_0}$$
so
$$-\frac{1}{\alpha}\left(e^{\alpha v}- e^{\alpha v_0}\right)= -(A/m)t$$
which is just what you have (except for the "m").

Are you concerned with solving for v?
$$e^{\alpha v}- e^{\alpha v_0}= (A\alpha/m)t$$
$$e^{\alpha v}= (A\alpha/m)t+ e^{\alpha v_0}$$
Now take the logarithm of both sides.

Last edited by a moderator:
$v_0=v(t=0)$

(e^20$\alpha$)/$\alpha$ = C3

C3 = 73.8

v(t) = (ln(-At$\alpha$)+ ln(73.8$\alpha$))/$$alpha$$

sorry if I am a little slow with this, am I making an error here?

littlehonda said:
$v_0=v(t=0)$

(e^20$\alpha$)/$\alpha$ = C3

C3 = 73.8

v(t) = (ln(-At$\alpha$)+ ln(73.8$\alpha$))/$$alpha$$

sorry if I am a little slow with this, am I making an error here?

Yes, you are making a few errors.

To start with, Halls pointed out that you dropped the mass of the object from the differential equation...when you include it, you should get

$$\frac{me^{\alpha v}}{\alpha}=-At+C_3$$

Multiply both sides of the equation by alpha/m:

$$\implies e^{\alpha v}=\frac{\alpha}{m}(-At+C_3)$$

Then take the natural log of both sides

$$\alpha v =\ln\left(\frac{\alpha}{m}(-At+C_3)\right)$$

You can take it from there...

I was careless putting the force equation up, the right side of the equation also contains a m to cancel the one on the left.

v= (ln($\alpha$(-At+C3)))/$\alpha$

Solving for C3

C3 = (e^v0$\alpha$)/$\alpha$

at V(0) t=0

C3 = 73.89

v= (ln($\alpha$(-At+73.89)))/$\alpha$

Looks good to me

hey thanks a lot both of you guys!

hate to keep pestering but if I want to find x(t) I just integrate dx/dt and find a new constant C3 using the initial condition x(0) t=0?

1. How do I find v(t) for a velocity dependent force?

To find v(t) for a velocity dependent force, you can use the formula: v(t) = v0 + a * t, where v0 is the initial velocity and a is the acceleration.

2. What is a velocity dependent force?

A velocity dependent force is a force that changes in magnitude and/or direction based on the velocity of an object. This means that the force applied to an object is not constant and can vary depending on how fast the object is moving.

3. How does a velocity dependent force affect the motion of an object?

A velocity dependent force can either increase or decrease the velocity of an object depending on its direction and magnitude. For example, if the force is in the same direction as the object's velocity, it will increase the speed of the object. If the force is in the opposite direction, it will decrease the speed of the object.

4. Can a velocity dependent force be negative?

Yes, a velocity dependent force can be negative. This means that the force is acting in the opposite direction of the object's motion, causing it to slow down or change direction.

5. How can I graph v(t) for a velocity dependent force?

To graph v(t) for a velocity dependent force, you can plot the velocity on the y-axis and time on the x-axis. The resulting graph will show how the velocity changes over time due to the velocity dependent force.

Replies
2
Views
2K
Replies
4
Views
2K
Replies
1
Views
2K
Replies
26
Views
3K
Replies
3
Views
26K
Replies
4
Views
4K
Replies
3
Views
1K
Replies
1
Views
1K
Replies
3
Views
2K
Replies
8
Views
3K