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.
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 Spring '08
 Morrison
 Acceleration, Velocity, 8 m, 10 m, 6.8 kg

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