5 Steps to a 5 AP Calculus BC 2019

More Applications of Derivatives 189

Step 2: Setv(t) anda(t)=0.

Setv(t)= 0 ⇒ 3 t^2 − 12 t+ 9 = 0 ⇒3(t^2 − 4 t+3)= 0

⇒3(t−1)(t−3)=0ort=1ort=3.

Seta(t)= 0 ⇒ 6 t− 12 = 0 ⇒6(t−2)=0ort=2.

Step 3: Determine the directions of motion. (See Figure 9.3-3.)

t

0 1

v(t)+–+++++++ 0 –––––––––––++++++0

3

Direction Right

of Motion

Stopped Stopped

Left Right

Figure 9.3-3

Step 4: Determine acceleration. (See Figure 9.3-4.)

t

v(t)

a(t)

+++++++

++ + + + + + + +++

00–––– –––––––– +++++

–––– ––––––– 0

1

1

03

t

0 2

t

032

Particle Slowingdown Speedingup

Stopped Stopped

Slowingdown Speedingup

Figure 9.3-4

Step 5: Draw the motion of the particle. (See Figure 9.3-5.)
s(0)=−1,s(1)=3,s(2)=1, ands(3)=− 1

t = 3

t = 0

t = 2

t = 1

–1 0 1 3

Position s(t)

Figure 9.3-5

Att=0, the particle is at−1 and moving to the right. It slows down and stops att=1 and
att=3. It reverses direction (moving to the left) and speeds up until it reaches 1 att=2.
It continues moving left but slows down and stops at−1att=3. Then it reverses direction
(moving to the right) again and speeds up indefinitely. (Note: “Speeding up” is defined as
when|v(t)|increases and “slowing down” is defined as when|v(t)|decreases.)