[SOLVED] Find the velocity of the particle at time t

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March 26th 2009, 07:53 PM
moonman

[SOLVED] Find the velocity of the particle at time t

For 4.95 seconds, a particle moves in a straight line according to the position function:
f(t) = (e^(t)) (5 - t) -5

Find the velocity of the particle at t = 0, t = 1.

Okay the book states "the velocity function is the derivative of the position function. So I'm assuming I have to take the derivative of the function f(t)?
March 26th 2009, 08:08 PM
mollymcf2009

Quote:

Originally Posted by moonman

For 4.95 seconds, a particle moves in a straight line according to the position function:
f(t) = (e^(t)) (5 - t) -5

Find the velocity of the particle at t = 0, t = 1.

Okay the book states "the velocity function is the derivative of the position function. So I'm assuming I have to take the derivative of the function f(t)?

You got it. Then just evaluate f'(0) & f'(1)
March 26th 2009, 08:15 PM
moonman

Thanks for help,
So for the derivative of f(t) I got -5 is that correct?
March 26th 2009, 08:16 PM
sinewave85

Quote:

Originally Posted by moonman

For 4.95 seconds, a particle moves in a straight line according to the position function:
f(t) = (e^(t)) (5 - t) -5

Find the velocity of the particle at t = 0, t = 1.

Okay the book states "the velocity function is the derivative of the position function. So I'm assuming I have to take the derivative of the function f(t)?

Right! So:

$f^{\prime}(t) = (e^{t})(5 - t)^{\prime} + (e^{t})^{\prime}(5 -t) + (-5)^{\prime}$

$f^{\prime}(t) = e^{t}(-1) + e^{t}(5 - t)$

$f^{\prime}(t) = e^{t}(4 - t)$

and then

$f^{\prime}(0) = e^{0}(4 - 0) = 4$

$f^{\prime}(1) = e^{1}(4 - 1) = 3e \approx 8.1548$
March 26th 2009, 08:31 PM
mollymcf2009

Quote:

Originally Posted by moonman

Thanks for help,
So for the derivative of f(t) I got -5 is that correct?

Nope that is not quite right. You need to use the product rule for your derivative. Look at sinewave's post.
March 26th 2009, 11:56 PM
moonman

Now find when the particle is at rest.

Now I need to find when the particle is at rest.
Okay, so the particle will be at rest when the velocity is zero right? v(t)=0

The velocity function I have is (e^(t)) * (4 - t)

So if I set this up like this:

(e^(t)) * (4 - t) = 0

and then solve for "t" I should get when the particle is resting?
March 27th 2009, 12:01 AM
moonman

So when t = 0 and when t = 4 ? Right?
March 27th 2009, 02:23 AM
ADARSH

Quote:

Originally Posted by moonman

So when t = 0 and when t = 4 ? Right?

Not exactly
at t= 0

e^{t} = e^{0} = 1

at t=0
4-t = 4
--------------------
So at t=0

(e^(t)) * (4 - t) = 1*4 = 4

_____________________________________________

But when t= 4 , 4-t =0
Hence

(e^(t)) * (4 - t) = 0 = (e^ 4 ) * 0 = 0

Hence t=4 is correct & not t= 0 (Happy)