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Straight line motion

  • Thread starter Jess1986
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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?
 

Answers and Replies

  • #2
Chi Meson
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Hopefully you are using calculus, is this true?

If so, find an expression for the acceleration (Newton's second law).
 
  • #3
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Yes i am using calculus.
Would a=-(a+bv) be correct?
 
  • #4
nrqed
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Jess1986 said:
Yes i am using calculus.
Would a=-(a+bv) be correct?
Yes. So [itex] { dv \over dt} = - (a + b v) [/itex]. I would try a solution of the form [itex] v(t) = c_1 + c_2 e^{-c_3 t} [/itex]. It seems to me that this should do the trick.


Pat
 
  • #5
Hootenanny
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Surely the accleration would be

[tex] a = \frac{a+bv}{m}[/tex]

or am I missing something?
 
  • #6
nrqed
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Hootenanny said:
Surely the accleration would be

[tex] a = \frac{a+bv}{m}[/tex]

or am I missing something?
Yes, you are right, I forgot the mass. Thanks!
(but it is a minus sign because it is a resistive force)
 
  • #7
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how do i use [itex] v(t) = c_1 + c_2 e^{-c_3 t} [/itex]
to solve this?
 
  • #8
Hootenanny
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Jess1986 said:
how do i use [itex] v(t) = c_1 + c_2 e^{-c_3 t} [/itex]
to solve this?
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. :smile:
 
  • #9
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i am unsure how to get [itex] v(t) = c_1 + c_2 e^{-c_3 t} [/itex] from the differential eq.?
 
  • #10
nrqed
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Jess1986 said:
how do i use [itex] v(t) = c_1 + c_2 e^{-c_3 t} [/itex]
to solve this?
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
nrqed
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Jess1986 said:
i am unsure how to get [itex] v(t) = c_1 + c_2 e^{-c_3 t} [/itex] from the differential eq.?
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
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ok i see, thanks
 
  • #13
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i have got u=c1 + c2 at t=0. how do i find the condition with the exponential in?
 
  • #14
nrqed
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Jess1986 said:
i have got u=c1 + c2 at t=0. how do i find the condition with the exponential in?
well, that tells you c1= u- c2.

Now, [itex] v(t) = u + c_2 (e^{-c_3 t} - 1) [/itex]

Now plug this into the diff equation and that will give you two conditions which will fix c2 and c3.
 
  • #15
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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
nrqed
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Jess1986 said:
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
Plug v(t) into [itex] m dv/dt = - (a + b v) [/itex].

Pat
 
  • #17
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do i need to integrate this equation to sub this in?
 
  • #18
Hootenanny
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Jess1986 said:
do i need to integrate this equation to sub this in?
No [itex]v(t)[/itex] is the same as writing [itex]\frac{dv}{dt}[/itex] so, in other words [itex]v(t) = \frac{dv}{dt}[/itex] :smile:
 
  • #19
nrqed
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Jess1986 said:
do i need to integrate this equation to sub this in?
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
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so is dv/dt= c2e^-c3t??
 
  • #21
nrqed
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Hootenanny said:
No [itex]v(t)[/itex] is the same as writing [itex]\frac{dv}{dt}[/itex] so, in other words [itex]v(t) = \frac{dv}{dt}[/itex] :smile:
? I lost you on that one!:wink:
 
  • #22
nrqed
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Jess1986 said:
so is dv/dt= c2e^-c3t??
No... [itex] {d \over dt} ( u + c_2 (e^{-c_3 t} - 1 ) = -c_2 c_3 e^{-c_3 t} [/itex]
 
  • #23
Hootenanny
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nrqed said:
? I lost you on that one!:wink:
Damn! I thought she said v'(t). What I meant to say was [itex]v'(t) = \frac{dv}{dt}[/itex] Thank's for pulling me up.
 
  • #24
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oh yes oh course i forgot to bring the -c3 down, thanks
 
  • #25
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ive ended up with what looks a complicated expression with many c1,c2,c3...is there an easy way which i can solve for t? thanks
 

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