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Homework Problems Chapter 2 - My Solutions HEJ
Q21
A car moves along a straight section of road so that its velocity is varies with time as shown
in the top graph.
a.
B
Does the car ever go backwards? Explain.
For the car to go backwards the velocity would have to
become negative as at point D in the lower graph.
b.
Q32
2
At what labeled points is the acceleration the greatest?
Explain.
The definition of acceleration, a, is a = ∆v/∆t , or , the
slope of the v vs t curve. Both A and C have the same
Dt = 2 s. But ∆vA is larger (>) ∆vC . So aA is the
greatest.
C
A
6
4
B
A
∆t=2s
∆vA
2
C
∆vC
∆t=2s
4
6
D
A physics instructor walks with increasing speed across the front of the room then suddenly
reverses direction and walks backwards with constant speed. Sketch the graphs of the
velocity and acceleration consistent with this
description.
“Increasing speed across the front of the room” means
an increasing speed in one direction, hence increasing
velocity. An increasing velocity means that v vs t
curve must be rising, thus, have a positive slope. By
definition , acceleration, a, is a = ∆v/∆t , or , the slope
of the v vs t curve. So, the a vs t for this part of the
instructor’s trip must be positive. Here we only know
how to deal with constant acceleration. So, the a vs t
curve here is straight line above the t axis. And, the v
vs t must be a straight line with constant positive
slope.
a
1
trevers
trevers
“suddenly reverses direction and walks backwards with constant speed” means the
following:
(1) If the first direction was positive, then this new velocity is negative. So v vs t is
now a line below the t axis.
(2) Constant speed in one direction means constant velocity. So, here, the velocity is
a straight line with zero slope lying below the t axis. Is the v vs t curve has zero
slope, then the acceleration is zero .
E9
A car travels an average speed of 58 mph. What is this speed in km/h?
a.
If you use 1km = 0.62 mi
then you want do the following pseudo-code mi -> km.
The program , the numbers and checking are
Homework Problems Chapter 2 - My Solutions HEJ
58 mph = 58
km
1 km
1 0.62 = km
mi (
)
mi (
)
mi =
0.62 mi the check
0.62
mi
hr
hr
hr
mi ( )
The value is
b.
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58/0.62 km/hr = 93.55 km/hr.
If you use
1 mi = 5280 f, 1 f = 12 in, 1in = 2.54 cm, 1m = 100 cm, 1m = 1000 m
then the pseudo-code is
mi -> f -> in -> cm -> m -> km
Using the pseudo-code as a guide gives the program as
mi (
)
58
f
(
mi
in
m
( ) (
f
in
hr
)
)
km
m
.
Putting in the conversion numbers yields
mi (5280)
58
f
in
cm 1 m
1 km
(12) ( 2.54)
(
)
(
)
mi
f
in 100 cm 1000 m
hr
Checking the positions of the conversion numbers gives
mi (5280)
58
f
in
cm
1 = 100m 1 = 1000km
(12)
( 2.54)
(
)
(
)
= 5280mi
12 = f
2.54 = in 100
cm 1000
m
hr
Every ratio cancels as it must since every ratio equals 1.
The results is 58*5280*12*2.54/(100*1000) = 58*16093/100000 = 58*1.6093 =93.34 km/hr .
c.
If you use 1f = 12 in, 1m = 39.37in and 1km = 1000 m.
then the pseudo-code is
mi -> f -> in -> m -> km
Using the pseudo-code as a guide gives the program as
mi (
58
)
f
(
mi
in
m
( ) (
f
in
hr
)
Entering the conversion numbers gives
mi (5280)
58
f
in
1
m 1 km
(12) (
) (
)
mi
f 39.37 in 1000 m
hr
)
km
m
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Homework Problems Chapter 2 - My Solutions HEJ
You do the checking of the positions of the conversion numbers.
The moral of the story here is that you can use any set of information you
have available so long as you know the method.
E11
Starting from rest, a car accelerates at a rate of 4.2
m/s2 for a time of 5 seconds. What is its velocity at
the end of this time?
a ∆v=a∆t
By definition, the acceleration a = ∆v/∆t.
Hence by cross multiplication of ∆t
a ∆t = (∆v/∆t)∆t
Cancelling ∆t on the right side gives
∆v = a ∆t = 4.2m/s2 * 5 s = 21.0 m/s
1
∆t
t=5s
But
∆v = vfinal - vinitial ,
so
vfinal = ∆v + vinitial .
Starting from rest means that vinitial = 0.
Hence
vfinal = 21.0 m/s
.
t=5s
E14
A runner traveling with an initial velocity of 5 m/s accelerates at a constant rate of 1.2 m/s2
for a time of 2 seconds.
a.
What is the velocity at the end of this time?
b.
What distance does the car cover in this time?
v
a.
By definition, the acceleration a = ∆v/∆t.
Hence by cross multiplication of ∆t
a ∆t = (∆v/∆t)∆t
Cancelling ∆t on the right side gives
∆v = a ∆t = 1.2m/s2 * 2 s = 2.4 m/s
But
∆v = vfinal - v0 ,
so
vfinal = ∆v + v0 .
Starting from rest means that v0 = 5m/s.
Hence
vfinal = 2.4m/s + 5 m/s = 7.4 m/s.
.
b.
(a)
∆t
2
a =1.2m/s
(b)
v0 =
5m/s
1
∆v=a∆t
By definition, the velocity v = ∆x/∆t.
∆t
t=2s
So,
∆x = v∆t
But from Fig (a) this ∆x is an area element under the v
vs t curve. Hence the total ∆x total area under the v vs
t curve. In Fig. (b) we see that this area has two parts; (c)
one a rectangle due to v0 and a triangle due to the
acceleration a.
= v0t (rectangle) + ½ base * height (triangle)
t=2s
= v0t + ½ t*at = v0t + ½ at2
= 5m/s*2s = ½ *1.2m/s2 *(2s)2 = 10m + ½ *1.2m/s2*4s2 = 10m + 2.4m = 12.4 m .