d) When a particle is moving in the positive direction then the derivative of f (t) is positive.
(You might think that it is when f(t) is positive but that is only where the point is ahead of the starting point)
f '(t) = 3t2 - 24t + 36 = 0 Simplify
f '(t) = t2 - 8t + 12 = 0
Using the quadratic formula t = 2, 6
So either the particle is moving in the positive direction in 2 - 6 or everywhere except 2 - 6
To find this take a point in between and see if it is positive
f '(3) = 3 * 9 - 24 * 3 + 36 = -9
Since it is negative the particle is moving in the negative direction in this interval.
The answer is [0, 2) ; (6, infinity) the particle moves in the positive direction
(Remember that at time 2 and 6 the particle is stationary or not moving in the positive direction hence the parenthesis)
e)
To find the total distance traveled not the distance from the origin then both positive and negative movement needs to be included.
So the particle is moving in the positive direction from 0 to 2 and negative from 2 to 6 then positive from 6 to 8
We need to add these distances to find the total distance.
So the distance traveled from 0 to 2 is the position at 2 minus the distance at 0
f(2) - f(0)
= 23 − 12*22 + 36*2 -( 03 − 12*02 + 36*0 )
= 8 - 48 + 72 - 0 = 32 ft
For the negative direction
|f(3) - f(2)| absolute value because the distance will be negative
63 − 12*62 + 36*6 - ( 23 − 12*22 + 36*2)
216 - 432 + 216 - (32) = -32 ft or 32 ft traveled
32 + 32 = 64 ft traveled over the first 6 seconds
Again for positive
f(8) - f(6)
83 − 12*82 + 36*8 - (63 − 12*62 + 36*6)
512 -768 + 288 - (0)
32 - (0) = 32 ft traveled from 6 to 8 seconds
64 + 32 = 96 ft traveled over 8 seconds
f)
Acceleration is the second derivative of distance
f(t) = t3 − 12t2 + 36t
f '(t) = 3t2 - 24t + 36
f ''(t) = 6t -24
a(t) = 6t -24
Acceleration at 4 seconds
a(4) = 6 * 4 - 24 = 0
There is no acceleration at this time.
