Finding x(t) from a time dependant force

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In summary, we are given a particle of mass m subject to a force F(t) = mae-bt, with initial position and speed both equal to zero. The task is to find x(t) using the equations of motion. Using integration, we can obtain the formula x(t)=(a0/b2)e-bt, where a0 is the acceleration given by the equation a(t) = a0e-bt.
  • #1
Keshroom
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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??
 
Last edited:
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  • #2
  • #3
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
Keshroom said:
A particle of mass m is subject to a force
F(t) = mae-bt
Keshroom said:
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
tiny-tim said:
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
Mathoholic! said:
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?
 
Last edited:
  • #8
Keshroom said:
wait..you can just integrate like that without the dt on the right hand side?

I simply skipped the integration steps :-p
 
  • #9
Mathoholic! said:
I simply skipped the integration steps :-p

alright ill take your word for it!
 
  • #10
Mathoholic! said:
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
berkeman said:
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
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
Hi Keshroom! :smile:

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

When you got down to here …
Keshroom said:
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:
 

Related to Finding x(t) from a time dependant force

What is x(t) and how is it related to time dependant force?

x(t) refers to the position of an object at a specific time t. It is related to time dependant force because the force acting on an object can cause its position to change over time.

What is the equation for finding x(t) from a time dependant force?

The equation for finding x(t) is x(t) = x0 + v0t + 1/2at2, where x0 is the initial position, v0 is the initial velocity, a is the acceleration, and t is the time.

Can x(t) be negative or zero?

Yes, x(t) can be negative or zero. This depends on the initial position and the direction of the time dependant force.

What are the units for x(t) and time dependant force?

The units for x(t) are meters (m) or any other unit of length, while the units for time dependant force are Newtons (N) or any other unit of force.

Can x(t) be determined for all types of time dependant forces?

Yes, x(t) can be determined for all types of time dependant forces as long as the acceleration is constant. If the acceleration is not constant, the equation for x(t) will be more complex and may require calculus to solve.

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