HRW Hints for exercise. Chapter 02.pdf

(81 KB) Pobierz
C
HAPTER
2
H
INT FOR
P
ROBLEM
5
Let
d
be the total distance and
∆t
be the time for either of the one-way trips. The trip is
composed of two parts, with different speeds:
d
=
d
1
+
d
2
and
∆t
=
∆t
1
+
∆t
2
. The partial
displacements and times are related by
d
1
=
v
1
∆t
1
and
d
2
=
v
2
∆t
2
.
For (a),
∆t
1
=
∆t
2
so
d
= (v
1
+
v
2
)∆t
1
. Divide by
∆t
=
∆t
1
+
∆t
2
= 2∆t
1
.
For (b),
d
1
=
d
2
, so
d
= 2d
1
and
∆t
=
1
1
+
v
1
v
2
d
1
.
To compute the average speed, divide the expression for
d
by the expression for
∆t.
(c) To find the average speed over the round trip, divide the total distance 2d by the total
time. Let
s
avg
t
be the average speed to Houston and
s
avg
f
be the average speed from
Houston. Then, the time for the first one-way trip is
∆t
t
=
the time for the second one-way trip is
t
f
=
and the total time is
t
=
d
s
avg
f
,
d
s
avg
t
,
1
s
avg
t
+
1
s
avg
f
d.
(d) Now you must compute the average velocity, not the average speed. This means you
must divide the displacement for the round trip by the time for that trip. Note that you
begin and end the trip at the same place.
(e) Your graph will consist of two straight line segments with different slopes. When you
have drawn them, draw the line with a slope that represents the average velocity.
ans:
(a) 73 km/h; (b) 68 km/h; (c) 70 km/h; (d) 0
C
HAPTER
2
H
INT FOR
P
ROBLEM
8
First calculate the time until the trains meet. This is their original separation divided by the
sum of their speeds. Then calculate the distance the bird flys in this time. It is the product
of the bird’s speed and the time.
C
HAPTER
2
H
INT FOR
E
XERCISE
11
(a) Differentiate
x(t)
with respect to
t
to obtain an expression for the velocity at any time.
Substitute
t
= 1 s into this expression.
(b) If
v
is positive the particle is moving in the positive
x
direction, if
v
is negative it is
moving in the negative
x
direction.
(c) The speed is the magnitude of the velocity.
(d) Look at the expression for the velocity as a function of time and ask what happens as
t
increases from 1 s. You will find several answers: over the short term the speed is doing one
thing; at longer times it is doing another.
(e) Look at the expression for the velocity as a function of time and notice that the ini-
tial velocity is negative but that the velocity is increasing algebraically (the acceleration is
positive).
(f) Notice that the velocity is zero at
t
= 2 s and then continues to increase without change
in sign.
ans:
(a)
−6
m/s; (b) negative
x
direction; (c) 6 m/s; (d) speed decreases until
t
= 2 s,
then increases; (e) yes; (f) no
C
HAPTER
2
H
INT FOR
P
ROBLEM
12
(a) The average velocity is given by
v
avg
=
∆x
x
f
−
x
i
=
,
∆t
∆t
where
x
i
is the initial coordinate (x(t) evaluated for
t
= 2.00 s) and
x
f
is the final coordinate
(x(t) evaluated for
t
= 3.00 s).
(b), (c), and (d) Differentiate
x(t)
with respect to
t
to find an expression for the velocity as
a function of time. Evaluate the expression for the three given times.
(e) First you must find the coordinate of the midpoint. It is given by
x
m
=
x
f
+
x
i
.
2
Now, find the time the particle is at the midpoint: solve
x
m
= 9.75 + 1.50t
3
for
t.
This is
a cubic equation. Look up a technique for solving it or else use a trial and error method.
Systematically substitute various values of
t
into 9.75 + 1.50t
3
until you get an answer that
is close to
x
m
. Keep a record of the results to help you decide on your next trial value. Once
you have found a value for
t
substitute it into your expression for the velocity.
Notice that all these velocities are different. The average velocity is not the same as the
velocity at the midpoint in position nor is it the same as the velocity at the midpoint in
time. It also is not the average of the velocities at the end points. If the acceleration were
constant the average velocity over the interval, the average of the initial and final velocities,
and the instantaneous velocity at the midpoint in time, would all be the same and they
would be different from the instantaneous velocity at the midpoint in position.
(f) To construct the graph draw the appropriate lines, with slopes equal to the various
average and instantaneous velocities requested.
C
HAPTER
2
H
INT FOR
P
ROBLEM
18
(a) The average velocity over an interval of time is given by
v
avg
=
x
f
−
x
i
,
∆t
where
x
f
is the coordinate at the end of the interval (t = 8.00 min),
x
i
is the coordinate
at the beginning (t = 2.00 min), and
∆t
is the duration of the interval. You may place the
origin of the
x
axis at the initial position of the man. Then
x
i
= 0. At time
t
= 8.00 min the
man has walked for 3.00 min at 2.20 m/s. Use this information to calculate
x
f
. You must
convert 3.00 min to seconds.
(b) The average acceleration over an interval of time is given by
a
avg
=
v
f
−
v
i
,
∆t
where
v
f
is the velocity at the end of the interval and
v
i
is the velocity at the beginning. At
t
= 2.00 min the man is standing still and at
t
= 8.00 min he is moving at 2.20 m/s.
(c) and (d) Repeat the calculations for the new time interval.
(e) The graphs all consist of straight line segments. Remember that the average velocity
over an interval is the slope of a certain line on a graph of the coordinate as a function of the
time and the average acceleration over an interval is the slope of a certain line on a graph
of the velocity as a function of the time.

Plik z chomika:

Inne pliki z tego folderu:

Inne foldery tego chomika:

Zgłoś jeśli naruszono regulamin