Derivative (problem solving)

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  • #1
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Let s = s(t) be the position function of a particle moving in a straight line.
Suppose that the position of the particle is given by the formula
s (t) = t^2 e^-t ; t >= 0
where t is measured in seconds and s in meters.

(i) Find the velocity of the particle at time t.
* should i differentiate s(t) to get the velocity?

(ii) When is the particle at rest?
* when velocity = 0.. am i right?

(iii) Find the total distance traveled by the particle during the first two seconds.
*dont have any idea..
 

Answers and Replies

  • #2
Dick
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(i) yes. (ii) yes. (iii) you might have a more informed opinion about this after you do (i) and (ii). Get started.
 
  • #3
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(i) Find the velocity of the particle at time t.
* should i differentiate s(t) to get the velocity?

answer--> (2t e^-t) - (t^2 e^-t)

(ii) When is the particle at rest?
* when velocity = 0.. am i right?

answer-->
(2t e^-t) - (t^2 e^-t) = 0
(2t e^-t) = (t^2 e^-t)
(2e^-t) = (t e^-t)
*how to solve t?

(iii) Find the total distance traveled by the particle during the first two seconds.
*dont have any idea..

at t = 2
s (t) = t^2 e^-t
s (2) = 2^2 e^-2
.......= 0.54 meters

is my answer for (i) and (iii) is correct?
how am i going to solve (ii)?
 
  • #4
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Are you just trying to solve (ii) for t?
 
  • #5
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Are you just trying to solve (ii) for t?

yes. because i need the value of 't' when the particle at rest..
 
  • #6
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do you see anything you can factor out from both sides?
 
  • #7
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do you see anything you can factor out from both sides?

(2t e^-t) - (t^2 e^-t) = 0
(2t e^-t) = (t^2 e^-t)
(2e^-t) = (t e^-t)
e^-t(2) = e^-t (t)
(e^-t)/(e^-t) 2 = t

*(e^-t)/(e^-t) = 1?
 
  • #8
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(i) Find the velocity of the particle at time t.
* should i differentiate s(t) to get the velocity?

answer--> (2t e^-t) - (t^2 e^-t)
You should write an equation; namely v(t) = 2te^(-t) - t^2*e^(-t)
(ii) When is the particle at rest?
* when velocity = 0.. am i right?
Yes.
answer-->
(2t e^-t) - (t^2 e^-t) = 0
I'm going to cut in here because the rest of your work doesn't help you get where you need to go. It's not wrong, but it isn't helpful either.
s'(t) = v(t) = 2t e^(-t) - t^2 e^(-t) = 0
v(t) = 0 ==> 2t e^(-t) - t^2 e^(-t) = 0 = 0 ==> (2t - t^2)e^(-t) = 0
e^(-t) is never 0. When is 2t - t^2 = 0? Those are the times when v(t) = 0.

(2t e^-t) = (t^2 e^-t)
(2e^-t) = (t e^-t)
*how to solve t?

(iii) Find the total distance traveled by the particle during the first two seconds.
*dont have any idea..
Integrate the velocity between t = 0 and t = 2.
at t = 2
s (t) = t^2 e^-t
s (2) = 2^2 e^-2
.......= 0.54 meters

is my answer for (i) and (iii) is correct?
how am i going to solve (ii)?
 
  • #9
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for (iii) 4e^-2
am i got it right?
 
Last edited:
  • #10
HallsofIvy
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Yes, of course. You were told in the problem itself that s(t)= t2e-t. The answere to (iii) is just s(2).
 
  • #11
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Thank u for the confirmation of my answer.. =)
 

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