# Finding x(t) from a time dependant force

## 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)

## The Attempt at a Solution

how would i begin to start this??

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tiny-tim
Homework Helper
Hi Keshroom! Begin with good ol' Newton's second law

F = m d2x/dt2 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?

tiny-tim
Homework Helper
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. What exactly was the question? 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

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. What exactly was the question? 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)?

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

I simply skipped the integration steps :tongue:

I simply skipped the integration steps :tongue:

alright ill take your word for it!

berkeman
Mentor
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.

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 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

tiny-tim
Homework Helper
Hi Keshroom! Yes, that's fine except for the constants. 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! (to put it another way: always check your solutions just by looking at them … does this satisfy the initial condition of v = 0 ? )

Try again. 