MA 3972-MA-Book April 11, 2018 15:14
More Applications of Derivatives 217
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 10.3-3.)
t
0
[
1
v(t)++++++++ 0 –––––––––––– ++++++0
3
Direction Right
of motion
Stopped Stopped
Left Right
Figure 10.3-3
Step 4: Determine acceleration. (See Figure 10.3-4.)
t
v(t)
a(t)
+++++++
++ + + + + + + +++
00 –––––––––––– +++++
––––––––––– 0
1
1
03
t
0 2
t
032
Particle Slowingdown Speedingup
Stopped
[
[
[
Stopped
Slowingdown Speedingup
Figure 10.3-4
Step 5: Draw the motion of the particle. (See Figure 10.3-5.)
s(0)=−1,s(1)=3,s(2)=1 ands(3)=− 1
t = 3
t = 0
t = 2
t = 1
- 10 1 3
Position s(t)
Figure 10.3-5
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 that “speeding up” is defined
as when|v(t)|increases and “slowing down” is defined as when|v(t)|decreases.)