A particle moves along the X-axis so that at time t its position is given by x(t)=t^3-6t^2+9t+11?

Question

During what time intervals is the particle moving to the left?

What is the total distance traveled by the particle from t=0 to t=2?

Answer 1

Since we are measuring total distance we should check to see where the particle changes directions. For that we have to take the derivative (to find velocity):

x'(t) = 3t^2-12t+9

and set it equal to 0 (velocity has to be 0 in order to turn around)

3t^2-12t+9 = 0

divide by 3 to make it easier to factor

t^2-4t+3 = 0
(t-1)(t-3) = 0
t=1,3

Since 3 is not in our interval we'll disregard it.

So we'll find out how far it went from 0 to 1
x(0)=11
x(1)=15
Distance traveled from 0-->1 = 4 units

Now distance from 1-->2
x(1)=15
x(2)=13
Distance traveled from 1-->2 = 2 units.
The fact that it went in the negative direction is irrelevant since we are looking at distance not displacement!

So the total distance from 0-->2 = 6 units

Answer 2

for t=0
x=11

for t=2
x=8-24+18+11
=13

total distance=13-11=2