# Thread: [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:

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

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

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

and then

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

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

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