Straight line motion

1. Mar 21, 2006

Jess1986

A particle of mass m moves in a straight line subject only to a resistive force of magnitute a+bv, when the particle's speed is v, where a and b positive constants. If the particle was set moving with initial speed u, obtain formulae for the time taken for it to come to rest and for the distance travelled.

Really stuck on this q. dont really know where to start. Any ideas anyone?

2. Mar 21, 2006

Chi Meson

Hopefully you are using calculus, is this true?

If so, find an expression for the acceleration (Newton's second law).

3. Mar 21, 2006

Jess1986

Yes i am using calculus.
Would a=-(a+bv) be correct?

4. Mar 21, 2006

nrqed

Yes. So ${ dv \over dt} = - (a + b v)$. I would try a solution of the form $v(t) = c_1 + c_2 e^{-c_3 t}$. It seems to me that this should do the trick.

Pat

5. Mar 21, 2006

Hootenanny

Staff Emeritus
Surely the accleration would be

$$a = \frac{a+bv}{m}$$

or am I missing something?

6. Mar 21, 2006

nrqed

Yes, you are right, I forgot the mass. Thanks!
(but it is a minus sign because it is a resistive force)

7. Mar 21, 2006

Jess1986

how do i use $v(t) = c_1 + c_2 e^{-c_3 t}$
to solve this?

8. Mar 21, 2006

Hootenanny

Staff Emeritus
You know that when it is at rest v = 0, this will allow you to find a solution for t. You will then have to integrate v(t) to find x(t). You can then plug your value of t into x(t) to obtain the distance travelled.

9. Mar 21, 2006

Jess1986

i am unsure how to get $v(t) = c_1 + c_2 e^{-c_3 t}$ from the differential eq.?

10. Mar 21, 2006

nrqed

Impose v(o) = u. That gives you one condition. Plug this solution in the differential equation for v(t), and impose that the two sides of the equations must be equal, that will give you two conditions (one condition for the terms proportional to the exponential, one for the terms with no exponentials. altogether, you have three conditions which will allow you to fix the three constants.

Then set v=0 (comes to rest) and solve for t.
Then, integrate v(t) to find x(t) (with the initial condition x(0) = 0) . Plug th et you found above in x(t) and you will find the distance travelled.

Pat

11. Mar 21, 2006

nrqed

I did not get it "from" the diff equation. I just happen to know that this kind of diff equation has a solution of that form. It comes from knowledge of differential equations, not from algebraic manipulations.

12. Mar 21, 2006

Jess1986

ok i see, thanks

13. Mar 21, 2006

Jess1986

i have got u=c1 + c2 at t=0. how do i find the condition with the exponential in?

14. Mar 21, 2006

nrqed

well, that tells you c1= u- c2.

Now, $v(t) = u + c_2 (e^{-c_3 t} - 1)$

Now plug this into the diff equation and that will give you two conditions which will fix c2 and c3.

15. Mar 21, 2006

Jess1986

sorry to keep asking questions but what exactly do you mean when you say ' plug it in the diff. eq' which part? how can i do this? thanks

16. Mar 21, 2006

nrqed

Plug v(t) into $m dv/dt = - (a + b v)$.

Pat

17. Mar 21, 2006

Jess1986

do i need to integrate this equation to sub this in?

18. Mar 21, 2006

Hootenanny

Staff Emeritus
No $v(t)$ is the same as writing $\frac{dv}{dt}$ so, in other words $v(t) = \frac{dv}{dt}$

19. Mar 21, 2006

nrqed

No! You just have to calculate a derivative..
You know v(t). calculate dv/dt. Now, in m dv/dt = - (a +b v), insert your expression fro v(t) and for dv/dt.

20. Mar 21, 2006

Jess1986

so is dv/dt= c2e^-c3t??