# Moving Particle is Brought to Rest by a Resisting Force

1. Apr 4, 2013

### jmlibunao

1. The problem statement, all variables and given/known data
A particle of mass m kg is travelling in a horizontal straight line with a velocity u m/s. It is brought to rest by means of a resisting force of magnitude km(2u - v), where v is the velocity of the particle at any instant and k is a positive constant.

Find the distance travelled by the particle while v decreases from u top zero

2. Relevant equations
F = ma
K = (1/2)(m)(v^2)

I think you're also gonna need the formula for conservation of energy as well
K1 + E1 = K2 + E2

3. The attempt at a solution
I made this equation F = ma = km(2u - v) and then solved for a as a = k(2u - v)
I tried using the kinematic equation vf = vi + at, where vf = 0 and vi = u and solved for time, t. Then I plugged t into xf = xi + vi(t) + (1/2)(a)(t^2) but I just ended up with an ugly equation filled with variables. I think you have to solve for k but I'm not sure how.

Help would be much appreciated!

2. Apr 4, 2013

### Curious3141

Those kinematic equations only apply in constant acceleration scenarios. Here, the force (and hence acceleration) is clearly variable.

You need to use calculus for this. First step: try to express $v$ in terms of $t$ by solving a differential equation. Second step: use that to figure out the time when the particle comes to rest. Final step: use integration to figure out the distance travelled in that time.

3. Apr 4, 2013

### jmlibunao

I forgot that the kinematic equations are only applicable for constant acceleration.

What differential equation are you talking about? dv/dt = k(2u - v) ?? If this is right/wrong, then can you guide me through it? I'm just really having a hard time with this problem.

4. Apr 4, 2013

### Curious3141

Don't forget it's a *retarding* force. Hence the equation should be $\displaystyle \frac{dv}{dt} = -k(2u-v)$. Note the minus sign.

It's a simple ordinary first order differential equation with separable variables - pretty much the most elementary type there is. Have you learnt how to solve them? If not, it'll probably take too long (and be ineffective) to instruct you over this forum, so I suggest you do a little reading around the topic.

5. Apr 4, 2013

### jmlibunao

Don't worry about that. I know how to solve differential equations :)

After solving this I'm gonna be setting the velocity to 0, right? Which one, u or v, or does it not matter?

6. Apr 4, 2013

### HallsofIvy

Staff Emeritus
You say "with a velocity u m/s" and then "v is the velocity of the particle at any instant" so "u" is its initial speed and is constant? In that case v(t) is the variable speed and v should be set to 0.