- #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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- Thread starter Keshroom
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- #1

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A particle of mass m is subject to a force

F(t) = mae

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

how would i begin to start this??

Last edited:

- #2

tiny-tim

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- #3

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this is what i tried

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

F(t)e

∫F(t)e

∫F(t)e

∫F(t)e

[∫F(t)e

∫ [∫F(t)e

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

No, that makes no sense.

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

so F(t)

What *exactly* was the question?

- #5

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F(t)=ma(t) in which a(t)=a

If you integrate you get:

v(t)=v

And again:

x(t)=x

Since x

x(t)=(a

- #6

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You mean "a" is theacceleration?

No, that makes no sense.

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

so F(t)cannotbe mae^{-bt}.

Whatexactlywas the question?

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

Question was F(t) = ma

now i know why it is a

so in this case acceleration a = a

but i'm still stuck! How do i get to x(t)?

- #7

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F(t)=ma(t) in which a(t)=a_{0}e^{-bt}

If you integrate you get:

v(t)=v_{0}-(a_{0}/b)e^{-bt}

And again:

x(t)=x_{0}+v_{0}t+(a_{0}/b^{2})e^{-bt}

Since x_{0}and v_{0}are both zero.

x(t)=(a_{0}/b^{2})e^{-bt}

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

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- #8

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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:

- #9

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I simply skipped the integration steps :tongue:

alright ill take your word for it!

- #10

berkeman

Mentor

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F(t)=ma(t) in which a(t)=a_{0}e^{-bt}

If you integrate you get:

v(t)=v_{0}-(a_{0}/b)e^{-bt}

And again:

x(t)=x_{0}+v_{0}t+(a_{0}/b^{2})e^{-bt}

Since x_{0}and v_{0}are both zero.

x(t)=(a_{0}/b^{2})e^{-bt}

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

- #11

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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

- #12

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Ok i figured it out properly

F(t) = ma

m(dv/dt) = ma

∫mdv = ∫ma

mv = m(a

(dx/dt) = (a

∫dx = ∫(a

x(t) = a

- #13

tiny-tim

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Yes, that's fine

When you got down to here …

F(t) = ma_{0}e^{-bt}

m(dv/dt) = ma_{0}e^{-bt}

∫mdv = ∫ma_{0}e^{-bt}dt (from v=0 to v=v and t=0 to t=t)

mv = m(a_{0}e^{-bt})/-b

… although you

(to put it another way: always check your solutions just by

Try again.

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