Particle Motion in a Straight Line: Deriving Velocity and Total Distance

[naspek]

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..

 

[Dick]

(i) yes. (ii) yes. (iii) you might have a more informed opinion about this after you do (i) and (ii). Get started.

 

[naspek]

(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)?

 

[mg0stisha]

Are you just trying to solve (ii) for t?

 

[naspek]

Are you just trying to solve (ii) for t?

yes. because i need the value of 't' when the particle at rest..

 

[mg0stisha]

do you see anything you can factor out from both sides?

 

[naspek]

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?

 

[Mark44]

(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)?

 

[naspek]

for (iii) 4e^-2
am i got it right?

 

[HallsofIvy]

Yes, of course. You were told in the problem itself that s(t)= t²e^-t. The answere to (iii) is just s(2).

 

[naspek]

Thank u for the confirmation of my answer.. =)