Moving particle...?

Question

A particle moves along the x-axis in such a way that its position at time t greater than or equal to 0 is given by
x = (1/3) (t^3) - 3t^2 + 8t.

(a) Show that at time t = 0 the particle is moving to the right.

(b) Find all values of t for which the particle is moving to the left.

(c) What is the position of the particle at time t = 3?
ok is this part just asking for what (t,x) is when t=3?

(d) When t = 3, what is the total distance the particle has traveled?
ok so is this part asking for the distance between (0,0) and the answer from part (t,x) when t=3?

Answer 1

Hi,
a) The velocity is dx/dt.
So,
dx/dt = (1/3)3t² -6t +8
=t² -6t +8
So, at t = 0
dx/dt = 8. It's positive and positive is to the right.

b) We want:
t² -6t +8 < 0
Factor the equation:
(t-4)(t-2) <
So, we want:
(t-4)(t-2) <0
Let's just plot the signs of the factors and see when the product is negative.

t-4 ------------------0+++++++
t-2 --------0+++++++++++++
_______|_______|______
.............2.............4
(The dots on the bottom line mean nothing. I had to add them to get the 2 and 4 spaced properly on this word processor.

So, product is negative between the times 2 and 4.

c) Yes, that's correct. I'm sure you can handle that.

d) No, that's it's dispalcement, not how far it has travelled. According to our little graph, it travelled to the right from 0 to 2 and then moved back to the left from 2 to 4. So, you need to calculate the distance from 0 to 2 and then add the distance from 2 to 3.
Based on the insight you've shown into the problem, I'm sure you can handle that. Be careful with the signs. You want the total value, not the algebraic sum.

Hope this helps.
FE