Finding x(t) from a time dependant force

  • Thread starter Keshroom
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  • #1
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Homework Statement


A particle of mass m is subject to a force
F(t) = mae-bt

The initial position and speed are zero. Find x(t)

Homework Equations





The Attempt at a Solution



how would i begin to start this??
 
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Answers and Replies

  • #2
tiny-tim
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  • #3
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yeah...

this is what i tried

F(t) = m(dv/dt) e-bt

F(t)ebtdt = mdv

∫F(t)ebtdt = ∫mdv (with bounds t=0 to t=t, and v=0 to v=v)

∫F(t)ebtdt = mv (only integrating right side yields)

∫F(t)ebtdt = m(dx/dt)

[∫F(t)ebtdt]dt = dx (integrating both sides again)

∫ [∫F(t)ebtdt]dt = x(t)

is this correct? can it be simplified further?
 
  • #4
tiny-tim
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A particle of mass m is subject to a force
F(t) = mae-bt
F(t) = m(dv/dt) e-bt

You mean "a" is the acceleration?

No, that makes no sense.

If F(t) is the force, and m is the mass, then F(t) = ma,

so F(t) cannot be mae-bt. :redface:

What exactly was the question? :confused:
 
  • #5
I think I get it:

F(t)=ma(t) in which a(t)=a0e-bt

If you integrate you get:

v(t)=v0-(a0/b)e-bt

And again:

x(t)=x0+v0t+(a0/b2)e-bt

Since x0 and v0 are both zero.

x(t)=(a0/b2)e-bt
 
  • #6
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You mean "a" is the acceleration?

No, that makes no sense.

If F(t) is the force, and m is the mass, then F(t) = ma,

so F(t) cannot be mae-bt. :redface:

What exactly was the question? :confused:

touche sir. That does not make sense. ahh i see now

Question was F(t) = ma0e-bt
now i know why it is a0

so in this case acceleration a = a0e-bt

but i'm still stuck! How do i get to x(t)?
 
  • #7
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I think I get it:

F(t)=ma(t) in which a(t)=a0e-bt

If you integrate you get:

v(t)=v0-(a0/b)e-bt

And again:

x(t)=x0+v0t+(a0/b2)e-bt

Since x0 and v0 are both zero.

x(t)=(a0/b2)e-bt

wait..you can just integrate like that without the dt on the right hand side?
 
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  • #8
wait..you can just integrate like that without the dt on the right hand side?

I simply skipped the integration steps :tongue:
 
  • #9
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I simply skipped the integration steps :tongue:

alright ill take your word for it!
 
  • #10
berkeman
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I think I get it:

F(t)=ma(t) in which a(t)=a0e-bt

If you integrate you get:

v(t)=v0-(a0/b)e-bt

And again:

x(t)=x0+v0t+(a0/b2)e-bt

Since x0 and v0 are both zero.

x(t)=(a0/b2)e-bt

Please do not do the student's homework for them. That is against the PF rules.
 
  • #11
Please do not do the student's homework for them. That is against the PF rules.

I'm sorry, I let myself go :shy:
It won't happen again :approve:
 
  • #12
25
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haha yeah stop doing my homework for me!

Ok i figured it out properly

F(t) = ma0e-bt

m(dv/dt) = ma0e-bt

∫mdv = ∫ma0e-btdt (from v=0 to v=v and t=0 to t=t)

mv = m(a0e-bt)/-b

(dx/dt) = (a0e-bt)/-b

∫dx = ∫(a0e-bt)/-b dt (from x=0 to x=x and t=0 to t=t)

x(t) = a0e-bt/b2
 
  • #13
tiny-tim
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Hi Keshroom! :smile:

Yes, that's fine except for the constants. :wink:

When you got down to here …
F(t) = ma0e-bt

m(dv/dt) = ma0e-bt

∫mdv = ∫ma0e-btdt (from v=0 to v=v and t=0 to t=t)

mv = m(a0e-bt)/-b

… although you said "from … t=0 to t=t", you didn't do it! :rolleyes:

(to put it another way: always check your solutions just by looking at them … does this satisfy the initial condition of v = 0 ? :wink:)

Try again. :smile:
 

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