# Derivative (problem solving)

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
Homework Helper

(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?

(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)?

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

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

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

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

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
Mentor
(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.
(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)?

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

Last edited:
HallsofIvy
Homework Helper
Yes, of course. You were told in the problem itself that s(t)= t2e-t. The answere to (iii) is just s(2).

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