More Applications of Derivatives 201
- (a) Att=t 2 , the slope of the tangent is
negative. Thus, the particle is moving
to the left.
(b)Att=t 1 , and att=t 2 , the curve is
concave downward⇒
d^2 s
dt^2
=
acceleration is negative.
(c)Att=t 1 , the slope>0 and thus the
particle is moving to the right. The
curve is concave downward⇒the
particle is slowing down.
- (a)Att=2,v(t) changes from positive to
negative, and thus the particle reverses
its direction.
(b)Att=1, and att=3, the slope of the
tangent to the curve is 0. Thus, the
acceleration is 0.
(c)Att=3, speed is equal to|− 5 |= 5
and 5 is the greatest speed.
- (a)s(4)=−16(4)^2 + 640 =384 ft
(b)v(t)=s′(t)=− 32 t
v(4)=−32(4) ft/s=−128 ft/s
(c)Average Velocity=
s(4)−s(0)
4 − 0
=
384 − 640
4
=−64 ft/s.
(d) Sets(t)= 0 ⇒− 16 t^2 + 640 = 0 ⇒
16 t^2 =640 ort=± 2
√
10.
Sincet≥0,t=+ 2
√
10 ort≈ 6 .32 s.
(e)|v(2
√
10)|=|−32(2
√
10)|=
|− 64
√
10 |ft/s or≈ 202 .39 ft/s
- (a)Att=5,s(t)= 1.
(b)For 3<t<4,s(t) decreases. Thus,
the particle moves to the left when
3 <t<4.
(c)When 4<t<6, the particle stays
at 1.
(d) When 6<t<7, speed=2 ft/s, the
greatest speed, which occurs wheres
has the greatest slope.
Part B Calculators are allowed.
- Step 1: v(t)= 3 t^2 − 6 t
a(t)= 6 t− 6
Step 2: Setv(t)= 0 ⇒ 3 t^2 − 6 t= 0 ⇒
3 t(t−2)=0, ort=0ort= 2
Seta(t)= 0 ⇒ 6 t− 6 =0ort=1.
Step 3: Determine the directions of
motion. (See Figure 9.8-3.)
[
02
Left
Stopped Stopped
Direction Right
of Motion
t
v(t) 00 –– –– –– –– + ++ ++++
Figure 9.8-3
Step 4: Determine acceleration. (See
Figure 9.8-4.)
[
[
t
t
t
––––
––––– 0 ++++++++++
v(t)+ 0 ––– 0 ++++++
0
0
[
0
2
1
1 2
Speeding
up
Speeding
up
Slowing
down
Stopped Stopped
Motion of
Particle
a(t)
Figure 9.8-4
Step 5: Draw the motion of the particle.
(See Figure 9.8-5.)s(0)=1,
s(1)=−1, ands(2)=−3.
t = 2 t = 1
t = 0
t > 2
s(t)
–3 –1 0 1
Figure 9.8-5
