Integral of (Force * Velocity).

[Skatch]

Homework Statement

A particle moves in the x-y plane having the components of its velocity to be:

x = 64\sqrt{3}t and y = 64t - 16t^2,

and a force acting on this particle is proportional to its velocity. Find:

\int(F \cdot V)dt

from t = 0 to t = 4. Give a physical meaning to your result.

Homework Equations

Not sure.

The Attempt at a Solution

I'm having a hard time getting started here, because I don't know what F is. I've got:

V = (64\sqrt{3}t)i + (64t - 16t^2)j,

right? But I don't know what to dot it with inside the integral. I'm not looking for a total solution here, I'm just wondering if someone can quickly tell me what exactly F is. I should be ok from there.

If F is proportional to V, do I just set

F = (64a\sqrt{3}t)i + (64at - 16at^2)j

for some unknown constant a?

Thanks.

 

[Skatch]

So, if I set F equal to what I stated above, I get an answer of:

\int( F \cdot V ) dt = \int(16384 a t^2 - 2048 a t^3 + 256 a t^4)dt = (4063232 a)/15

I don't even know what to make of that...

 

[nasu]

Why would the force be proportional to the velocity?
You know (I hope) that F=m*a.
And acceleration is the derivative of velocity, isn't it?

The significance of the integral... v*dt is ds (distance). Force time distance = ?

 

[ify7007]

Why would the force be proportional to the velocity?
You know (I hope) that F=m*a.
And acceleration is the derivative of velocity, isn't it?

The significance of the integral... v*dt is ds (distance). Force times distance = ?

Force times Distance equals Work

 

[vela]

Why would the force be proportional to the velocity?

Because it's a given in the problem.

You know (I hope) that F=m*a.
And acceleration is the derivative of velocity, isn't it?

The significance of the integral... v*dt is ds (distance). Force time distance = ?