MA 3972-MA-Book April 11, 2018 15:14More Applications of Derivatives 2231- 1
- 2
1v(t)(^0) 23 45
2
v
t
Figure 10.7-2
- (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. 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/sec(c) Average velocity=
s(4)−s(0)
4 − 0
=384 − 640
4
=−64 ft/sec.(d) Sets(t)= 0 ⇒− 16 t^2 + 640 = 0 ⇒
16 t^2 =640 ort=± 2√
10.
Sincet≥0,t=+ 2√
10 or
t≈ 6 .32 sec.
(e)|v(2√
10)|=|−32(2√
10)|=
|− 64√
10 |ft/s or≈ 202 .39 ft/sec- (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/sec, 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 10.7-3.)
[
02
LeftStopped Stoppedof motionDirection Righttv(t) 00 –––––––– + ++++++Figure 10.7-3Step 4: Determine acceleration. (See
Figure 10.7-4.)[[ttt––––––––– 0 +++++++++ +v(t)+ 0 ––– 0 ++++++00[
0211 2Speeding
upSpeeding
upSlowing
downStopped StoppedMotion of
particlea(t)Figure 10.7-4Step 5: Draw the motion of the particle.
(See Figure 10.7-5.)
s(0)=1,s(1)=−1, ands(2)=−3.