Finding x(t) from a time dependant force

[Keshroom]

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??

 

[tiny-tim]

Hi Keshroom!

Begin with good ol' Newton's second law …

F = m d²x/dt² ​

 

[Keshroom]

yeah...

this is what i tried

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

F(t)e^btdt = mdv

∫F(t)e^btdt = ∫mdv (with bounds t=0 to t=t, and v=0 to v=v)

∫F(t)e^btdt = mv (only integrating right side yields)

∫F(t)e^btdt = m(dx/dt)

[∫F(t)e^btdt]dt = dx (integrating both sides again)

∫ [∫F(t)e^btdt]dt = x(t)

is this correct? can it be simplified further?

 

[tiny-tim]

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? ​

 

[Mathoholic!]

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

 

[Keshroom]

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)?

 

[Keshroom]

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?

 

[Mathoholic!]

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

I simply skipped the integration steps

 

[Keshroom]

I simply skipped the integration steps

alright ill take your word for it!

 

[berkeman]

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.

 

[Mathoholic!]

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

 

[Keshroom]

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]

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. ​