This preview shows pages 1–2. Sign up to view the full content.
SOLUTIONS FOR HW 1, 3, AND 4 GRADED PROBLEMS
HW 1:
graded problems were Ch. 2, #20 and #23
20.
With an average speed equal to the highway speed limit,
!
t
=
!
r
/
v
=
(5000 mi)/(65 mi/h)
=
76.9 h
after 1995, and
!
t
=
(5000 mi)/(55 mi/h)
=
90.9 h
before, a difference of 14.0 h.
23.
Interpret
This is a onedimensional kinematics problem involving finding the
velocity as a function of time, given position as a function of time. The object of interest
is the model rocket.
Develop
The instantaneous velocity
v
(
t
)
can be obtained by taking the derivative of
y
(
t
).
The derivative of a function of the form
bt
n
can be obtained by using Equation
2.3.
Evaluate
(a)
The instantaneous velocity is
v
(
t
)
=
dy
dt
=
b
!
2
ct
.
(b)
The velocity is zero when
b
=
2
ct
,
or
t
=
b
2
c
=
82 m/s
2(4.9 m/s
2
)
=
8.37 s.
Assess
As the model rocket is launched upward with an initial velocity
b
=
82 m/s,
its
altitude increases and then reaches a maximum, where the instantaneous velocity is zero.
The rocket then falls back to the Earth.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '08
 Morrison

Click to edit the document details