Honors 2201-4
Homework 9 Solutions
Fall 2003
Section 3.3
(2) We set the initial location s(0) = 0, the initial velocity v(0) = 0 and define the downward
direction to be positive. We know that the acceleration due to gravity is a(t) = g =
32.17f t/s2. Therefore,
dv
=g
v(0) = 0
dt
v = gt + c1 ⇒ c1 = 0
v(t) = gt
ds
= gt
s(0) = 0
dt
1
s = gt2 + c2 ⇒ c2 = 0
2
1
s(t) = gt2
2
We must find the time, tg , when the marble hits the ground, or s(tg ) = 4.
1
4 = gt2g
2
s
8
tg =
≈ 0.499
g
The marble takes about half a second to hit the ground. Its velocity when it hits is v(tg ),
v
s !
8
g
=g
s
8 q
= 8g ≈ 16
g
The marble is moving at 16f t/sec when it hits the ground.
(22)
dy
= 3t2
dt
y(2) = 5
y = t3 + c
5 = 23 + c ⇒ c = 5 − 8 = −3
y(t) = t3 − 3
1
(25)
dy
= 2t + 2e−2t
y(0) = 3
dt
y = t2 − e−2t + c
3 = 0 − e0 + c = c − 1 ⇒ c = 4
y(t) = t2 − e−2t + 4
(27)
dy
= 1 + 3t2 + 3e−t
y(0) = 1
dt
y = t + t3 − 3e−t + c
1= 0+0−3+c ⇒c =4
y(t) = t + t3 − 3e−t + 4
(35) We are given a(t) = 1.4f t/sec2 and v(0) = 0. We will set s(0) = 0 and assume the
direction of travel is positive. Therefore,
dv
= 1.4
v(0) = 0
dt
v = 1.4t + c1 ⇒ c1 = 0
v(t) = 1.4t
ds
= 1.4t
s(0) = 0
dt
1.4 2
t + c2 ⇒ c2 = 0
s=
2
s(t) = .7t2
where time, t, is measured in seconds. At the end of one minute the car is traveling at
v(60) = 1.4(60) = 84f t/sec ≈ 57mi/hr. During that time the car travels s(60) = .7(60)2 =
2520f t.
(39) We are told that the initial velocity is 285k/hr. Converting this to seconds we have
v(0) = (285/3600)k/sec and v(28) = 0, where time, t, is measured in seconds. Let s(0)=0
2
and assume that the direction of travel is the positive direction. We are also told that the
deceleration rate is proportional to time t. Therefore, the acceleration a(t) = −kt, where
k > 0 since the the plane is slowing down. Then,
dv
= −kt
dt
v(0) =
285
3600
k
v = − t2 + c1
2
k 2
285
285
= (0) + c1 ⇒ c1 =
3600
2
3600
k 2
285
v(t) = − t +
2
3600
We know that the plane stopped after 28 seconds, so v(28) = 0. Therefore,
285
k
0 = − (28)2 +
2
3600
285(2)
k=
≈ .0002
3600(28)2
Now, we must find the distance traveled in the first 28 seconds. First, we find the equation
for the location over time using the initial condition s(0) = 0.
s(28) = −
1
6
ds
k
285
= v(t) = − t2 +
dt
2
3600
285
k 3
t + c2 ⇒ c2 = 0
s=− t +
6
3600
285
k
t
s(t) = − t3 +
6
3600
!
285(2)
285
(28)3 +
(28) ≈ 1.48
2
3600(28)
3600
The plane travels approximately 1.5 kilometers before stopping.
3