really confusing derivative application question!!!!!!?

Question

The position function of a particle is given by
s= t^3 - 6t^2 +9t +1, t is greater than 0
What is the total distance traveled in the first 4 s?

How would you calculate this?


the answer should be 12 m.

Answer 1

If you've played with the position function (i.e. plotted the position as a function of time), then you will already realize that the particle changes directions as it moves. The trick here is to add up the absolute values of the position changes over time. The easiest way to do this is to find the times at which the sign of the velocity changes from positive to negative (i.e., when the velocity is zero), and then add up the absolute values of the particle displacements between those times.

Let s(t) = t^3 - 6t^2 + 9t + 1

the velocity is given by:

ds(t)/dt = v(t) = 3t^2 - 12t + 9

v(t) = (3t-9)*(t-1)

V(t) is zero at t = 1 and t = 3.

The total displacement is then:

D = |s(1) - s(0)| + |s(3) - s(1)| + |s(4) - s(3)|

D = |5 - 1| + |1 - 5| + |5 - 1|

D = |4| + |-4| + |4|

D = 12

Answer 2

The position is not a monotonic function of t, so the total distance traveled is ∫s(t)dt [0 to 4].

(1/4)*t^4 - 3*t^3 + (9/2)*t^2 + t; evaluate at t=4

Answer 3

This is a calculus problem where you have to take the derivative of the equation right? If so then the second derivative gives you the answer 12m.

s(t)=t^3-6t^2+9t+1>0
s1(t)= 3t^2-12t+9>0
s2(t)= 6t-12>0
substitute in 4 for t .
6t-12
6(4)-12
24-12
12


I hope this helps. Good Luck!!!!!!!!!!!!!!!!!!!!!!!!