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.)