Math Help - [SOLVED] Find the velocity of the particle at time t

1. [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)?

2. 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)

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

4. 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

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

$f^{\prime}(1) = e^{1}(4 - 1) = 3e \approx 8.1548$

5. 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.

6. 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?

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

8. 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
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So at t=0

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

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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