Making Connections
18. In the relay, the second, third, and fourth runners have already accelerated to vf before they handoff to the next runner.
Thus, the time is shorter than if you include acceleration time for all runners.
1.3 ACCELERATION DUE TO GRAVITY
PRACTICE
(Pages 32–33)
Understanding Concepts
1.
2.
The skydiver’s velocity is much greater than the diver’s velocity. Air resistance increases with velocity and cannot be
neglected for the skydiver.
The disadvantage is that what might seem logical or reasonable does not agree with what actually happens. One reason for
this is some events occur too rapidly for our senses to be able to observe slight differences. Aristotle’s reasoning that
heavy objects fell faster than lighter ones provides an example of the disadvantage of not using experimentation to
determine the dependency of one variable on another, in this case the dependency of the acceleration of a falling body on
the mass of the body.
Applying Inquiry Skills
3.
The experimental setup would require a vacuum chamber in which the coin and the feather are released simultaneously.
(This device is available commercially from scientific supply companies.)
Making Connections
4.
Since there is no atmosphere on the Moon, falling objects do not experience air resistance.
Case Study: Predicting Earthquake Accelerations
(Page 34)
(a) The map stretches from Northern California (40° north latitude) to a few kilometres north of Vancouver (50° north
latitude), and from the middle of Vancouver Island (125° west longitude) to just east of Trail, B.C. (about 117° west
longitude). The regions affected most severely, as depicted by deep reds, lie near the west coast of the continent,
especially in Northern California and Southern Oregon, as well as areas fairly close to Vancouver and Victoria. Areas
affected moderately, as depicted by yellows, stretch inland somewhat in British Columbia, Washington, and Oregon, and
a long way in California. Areas affected slightly, as depicted by blue-greens, are in the northern and eastern parts of
British Columbia, the eastern parts of Washington and Oregon, and the state of Idaho. There are no areas unaffected, as
depicted by white.
(b) Students have to combine the colours on the map with the colours in the legend, and they have to compare the contours
and locations on the map to a conventional atlas of the same region.
PRACTICE
(Page 35)
Understanding Concepts
5.
32
G
vi = 0
G
a = 9.80 m/s2 [down]
G
(a) ∆d = 5.00 m
G
vf = ?
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
G
G
G G
vf 2 = vi 2 + 2a ∆d
vf 2 = 0 + 2(9.80 m/s 2 [down])(5.00 m[down])
vf = 2(9.80 m/s 2 )(5.00 m)
G
vf = 9.90 m/s[down]
= 9.90 m/s ×
1km
× 3600 s/h
1000 m
G
vf = 35.6 km/h [down]
The diver’s velocity is 9.90 m/s [down] or 35.6 km/h [down].
G
(b) ∆d = 10.00 m
G
vf = ?
G
G
G G
vf 2 = vi 2 + 2a ∆d
vf 2 = 0 + 2(9.80 m/s 2 [down])(10.0 m[down])
vf = 2(9.80 m/s 2 )(10.0 m)
G
vf = 14.0 m/s [down]
= 14.0 m/s ×
1km
× 3600 s/h
1000 m
G
vf = 50.4 km/h [down]
The diver’s velocity is 14.0 m/s [down] or 50.4 km/h [down].
Applying Inquiry Skills
6. (a) A good guess would be between 150 ms and 250 ms. The actual answer depends on the student’s hand span.
(b) v yi = 0
a y = 9.8 m/s
2
∆y = 0.20 m (a typical hand span)
∆t = ?
1
∆y = v yi ∆t +
∆y =
∆t =
=
1
2
a y ∆t
2
a y ∆t
2
2
2 ∆y
ay
2(0.20 m)
9.8 m/s
2
∆t = 0.20 s
For a hand span of 0.20 m, the time interval is 0.20 s, or 200 ms.
(c) Comparisons will vary. Students can improve their skills by trying to estimate answers to problems before performing
calculations, and by practising estimating quantities in everyday transactions and activities.
G
7. (a) vi = 0
G
∆d = 1.55 m [down]
∆t = 0.600 s
G
a =?
G G
1G
2
Rearranging the equation ∆d = vi ∆t + a ( ∆t ) :
2
Copyright © 2003 Nelson
Chapter 1 Kinematics
33
G
G 2∆d
a=
2
( ∆t )
=
(
)
2 1.55 m [down ]
(0.600 s )
G
a = 8.61m/s 2 [down ]
Thus, the acceleration of the ball is 8.61 m/s2 [down].
(b) Plot the position-time graph, find the slopes of the tangents at three or more times, plot the corresponding velocity-time
graph, and calculate the slope of the line on that graph to find the acceleration.
G
(c) a = 9.81 m/s2 [down]
measured value − accepted value
% error =
× 100%
accepted value
2
=
8.61 m/s 2 − 9.81 m/s 2
9.81 m/s 2
×100%
% error = −12.2%
The percent error is 12.2 %.
(d) The high percent error results from a variety of possible errors, with air resistance likely the most crucial influence.
(Since the ball is very light, air resistance has a greater effect than if the ball were more massive but the same size.)
Another important source of error is measuring the time interval of the fall.
Making Connections
8.
Answers may vary. The change in the gravitational field strength at various altitudes is quite small, and deciding how
much adjustment would be required might be controversial. Furthermore, if records were adjusted downward for events
that are easier at higher altitudes, then calls would be made for records to be adjusted upward for events that are more
difficult in the rarified atmosphere at higher altitudes. Adjustments are not made for the effects of low winds, so they
would not likely be made for changes in altitude.
PRACTICE
(Pages 37–38)
Understanding Concepts
9. (a)
(b)
(c)
(d)
The time the ball takes to rise equals the time the ball takes to fall since air resistance is negligible.
The initial and final velocities would be equal in magnitude but opposite in direction.
The ball’s velocity at the top of the flight is zero.
The ball’s acceleration during the entire motion is the acceleration due to gravity (9.8 m/s2 [down]), even when it is at
the top of the flight.
(e)
34
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
10. We could solve Sample Problem 1(b) using the following equations:
(viy + vfy )
1
1
∆y = viy ∆t + a y ( ∆t )2
∆y = vfy ∆t − a y ( ∆t )2
∆y =
∆t
2
2
2
11. (a) viy = 0
a y = 9.80 m/s2
∆y = 12.5 m
vfy = ?
vfy 2 = viy 2 + 2 a y ∆y
2
vfy = 0 + 2(9.80 m/s 2 )(12.5 m)
vfy = 245 m 2 /s2
vfy = 15.6 m/s
The shellfish has a speed of 15.6 m/s at impact.
(b) viy = 0
a y = 9.80 m/s2
∆t = 3.37 s
vfy = ?
ay =
vfy − viy
∆t
vfy = viy + a y ∆t
= 0 + (9.80 m/s 2 )(3.37 s)
vfy = 33.0 m/s
The steel ball has a speed of 33.0 m/s at impact.
12. viy = −15.0 m/s
a y = 9.80 m/s2
∆y = 15.0 m
Using the equation ∆y = viy ∆t +
a y ( ∆t ) 2
1 
, rearrange as  a y  ( ∆t )2 + viy ∆t − ∆y = 0 , which is a form of the quadratic
2 
2
equation. Thus,
∆t =
−b ± b 2 − 4ac
2a
 ay 
−viy ± viy 2 − 4   ( −∆y )
 2 
=
a
 y
2 
 2 
∆t =
−viy ± viy 2 + 2a y ∆y
ay
(a) The initial velocity is up. Define +y as down.
∆t = ?
vfy = ?
∆t =
15.0 m/s ±
( −15.0 m/s )2 + 2(9.80 m/s2 )(15.0 m)
= 1.53s ± 2.33s
9.80 m/s 2
(only the positive root is applicable)
∆t = 3.86 s
Copyright © 2003 Nelson
Chapter 1 Kinematics
35
vfy = viy + a y ∆t
(
)
= −15.0 m/s + 9.80 m/s2 (3.86s )
vfy = 22.8 m/s
Thus, the total flight time is 3.86 s and the speed of impact is 22.8 m/s.
(b) The initial velocity is down. Define +y as down.
∆t = ?
vfy = ?
∆t =
−15.0 m/s ±
(15.0 m/s )
= −1.53s ± 2.33s
2
+ 2(9.80 m/s 2 )(15.0 m)
9.80 m/s 2
(only the positive root is applicable)
∆t = 0.794s
vfy = viy + a y ∆t
(
= 15.0 m/s + 9.80 m/s2
) (0.794 s )
vfy = 22.8 m/s
Thus, the total flight time is 0.794 s and the speed of impact is 22.8 m/s if the initial velocity is down. (Notice that the
answers are written to three significant digits in order to compare them to the answers in (a). Rules for rounding off
have not been followed exactly here.)
(c) The final velocity is independent of whether the ball is thrown up or down (at the same speed).
13. Let ∆y1 represent the distance the ball travels from t = 0.0 s to t = 1.0 s and let ∆y2 represent the distance the ball travels
from t = 1.0 s to t = 2.0 s. For ∆y1, the initial velocity viy,1 = 0
1
Using the equation ∆y = viy ∆t + a y ( ∆t )2 :
2
1
∆y1 = viy ,1∆t1 + a y ( ∆t1 )2
2
1
= 0 m/s (1.0 s − 0.0 s) + 9.8 m/s 2 (1.0 s − 0.0 s) 2
2
∆y1 = 4.9 m
For ∆y2, the initial velocity viy,2 is the velocity at t = 1.0 s. Thus,
viy ,2 = viy ,1 + a y ( ∆t1 )
(
)
= 0 m/s + 9.8 m/s2 (1.0 s − 0.0 s)
viy ,2 = 9.8 m/s
1
∆y2 = viy ,2 ∆t2 + a y ( ∆t2 )2
2
1
= 9.8 m/s (2.0s − 1.0s) + 9.8 m/s 2 (2.0s − 1.0s) 2
2
∆y2 = 14.7 m
∆y2 (14.7 m )
=
= 3.0
∆y1 (4.9 m )
Therefore, a ball travels three times as far from 1.0 s to 2.0 s as from 0.0 s to 1.0 s.
14. Define +y as up.
vfy = 0 m/s
4.2s
∆t =
= 2.1 s
2
ay = –9.80 m/s2
36
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
(a) viy = ?
ay =
vfy − viy
∆t
viy = vfy − a y ∆t
= 0 m/s − ( −9.80 m/s2 )(2.1s)
viy = 21m/s
Thus, the pitcher threw the ball with a velocity of 21 m/s [up].
(b) ∆y = ?
1
∆y = viy ∆t + a y (∆t ) 2
2
1
= 21 m/s (2.1 s) + (−9.80 m/s 2 )(2.1 s) 2
2
∆y = 22 m
The ball rises 22 m.
15. Define +y as down.
viy = −2.1 m/s
∆t = 3.8 s
ay = 9.80 m/s2
(a) ∆y = ?
a y (∆t ) 2
∆y = viy ∆t +
2
(9.80 m/s 2 )(3.8s)2
= (−2.1m/s)(3.8s) +
2
∆y = 63m
The balloon was 63 m when the ballast was released.
(b) vfy = ?
vfy − viy
ay =
∆t
vfy = viy + a y ∆t
= −2.1 m/s + (9.80 m/s 2 )(3.8 s)
vfy = 35 m/s
The velocity of the ballast at impact was 35 m/s [down].
16. Define +y as down.
∆d = 2.3 m
∆t = 1.7 s
viy = 0
G
(a) g = ?
∆y = viy ∆t +
∆y =
G
g=
=
1
g ( ∆t ) 2
2
1
g ( ∆t ) 2
2
2∆y
( ∆t ) 2
2(2.3 m)
(1.7 s)2
G
g = 1.6 m/s 2 [down]
The acceleration of gravity on the Moon is 1.6 m/s2 [down].
Copyright © 2003 Nelson
Chapter 1 Kinematics
37
(b)
G
g Earth
9.8 m/s 2
=
= 6.1
G
g Moon
1.6 m/s 2
G
G
Thus, the ratio of g Earth to g Moon is 6.1:1.
Applying Inquiry Skills
17. (a) Students can use graphing techniques or uniform acceleration equations to determine the acceleration. Sample
calculations using the final data points are shown below.
Let +y be down.
For mass 1:
viy = 0
∆y = 0.736 m
∆t = 0.40 s
ay1 = ?
∆y = viy ∆t +
∆y =
a y1 =
=
1
2
a y 1 ∆t
1
2
a y 1 ∆t
2
2
2 ∆y
∆t
2
2(0.736 m)
( 0.40 s )2
a y1 = 9.2 m/s
2
For mass 2:
viy = 0
∆y = 0.776 m
∆t = 0.40 s
ay2 = ?
1
∆y = viy ∆t +
∆y =
ay 2 =
=
1
2
2
a y 2 ∆t
a y 2 ∆t
2
2
2 ∆y
∆t
2
2(0.776 m)
( 0.40 s )2
a y 2 = 9.7 m/s
2
The accelerations of the two masses are 9.2 m/s2 and 9.7 m/s2, respectively.
(b) To find the percent difference:
difference in values
× 100%
% difference =
average of values
2
=
9.2 m/s − 9.7 m/s
1
2
(9.2 m/s
2
2
+ 9.7 m/s
2
)
× 100%
% difference = 5.3%
38
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
(c) Likely, the lower experimental acceleration, 9.2 m/s2, is attributable to the ticker-tape timer because friction between
the paper strip and the timer would reduce the acceleration.
Making Connections
18. Answers will vary. Walking, running, and going up stairways are just a few examples of the activities that would be more
difficult with a higher acceleration due to gravity. One advantage is that friction would be increased when it might be
helpful, for example for a construction worker on a slanted roof. Another advantage is that objects subject to toppling over
might be more stable.
PRACTICE
(Page 39)
Understanding Concepts
19. The two main factors that affect terminal speed are mass and surface area. (For solid spherical objects, this is equivalent
to saying that the terminal speed depends on the object’s density.) For objects of the same mass, an increased surface area
contacting the air causes the terminal speed to become lower. For objects with the same surface area, the greater the mass
the greater is the terminal speed.
20. There is no “terminal speed” related to air resistance on the Moon because the Moon does not have an atmosphere.
21. There are two terminal speeds in this case, a high
speed followed by a much lower terminal speed when
the parachute is opened. The graph is shown below.
The slope of the line near the beginning of the motion
is equal to the magnitude of the acceleration due to
gravity at that location. The terminal speed values are
found on page 39 of the text (Table 5).
Applying Inquiry Skills
22. (a) Examples of factors to consider in designing the package are the mass, dimensions, and contents of the package, the
materials of the packets within the package, the material of the package itself, the method of securing the outer layer of
the package, the height from which the package will be dropped, the forward speed of the aircraft that drops the
package, the type of terrain on which the package will land, and the ease with which the package can be opened when it
is found.
(b) Since there are so many factors that could be tested, it would be wise to predict what choices would be best for the
design, create a prototype, and then perform a controlled experiment by dropping the sample of the design from various
heights onto various surfaces to determine how well the package can survive the fall. Modifications and further testing
are part of the process.
Making Connections
23. There are two main factors to consider, terminal speed and stopping distance. Once the terminal speed of a body is
reached, the speed remains constant, or it may even drop as the air becomes more dense nearer the ground. Furthermore, a
person falling from a greater height can reduce the terminal speed by using the “spread-eagle” orientation. Increasing the
stopping or deceleration distance reduces the chances of death or serious injury. The distance can be increased if the
landing occurs on the down slope of a hill or if the fall is broken by trees, bushes, a layer of snow, or water.
Section 1.3 Questions
(Page 40)
Understanding Concepts
1.
2.
Air resistance increases dramatically with increased speed, so air resistance is negligible for most common objects that
fall for a short distance. For example, air resistance is negligible for all but the lowest density objects that can fall in a
classroom setting.
Aristotle believed that the central region of Earth was made up of four elements: earth, air, fire, and water. Each element
had its proper place, which was determined by its relative heaviness. Each element moved in a straight line to its proper
place where it could be at rest. For example, similar objects were attracted to each other. Thus, objects made from the
Copyright © 2003 Nelson
Chapter 1 Kinematics
39
earth were attracted down to Earth. Fire tended to rise from Earth. Aristotle also believed that heavier objects fell faster
than less massive objects of the same shape.
Galileo believed that all objects fall toward Earth at the same acceleration, regardless of their mass, size, or shape,
when gravity is the only force acting on them. Evidently, he proved his theory by dropping two objects of different mass
from the top floor of the Leaning Tower of Pisa.
3. In each case, we define +y as down.
(a) viy = 0
ay = 9.8 m/s2
∆y = 36 m
vfy = ?
vfy 2 = viy 2 + 2a y ∆y
vfy 2 = 0 + 2(9.8 m/s 2 )(36 m)
 27 m   1 km   3600 s 
vfy = 



 s   1000 m   1 h 
vfy = 97 km/h
Thus, the landing speed of the diver is 27 m/s or 97 km/h.
(b) viy = 0
ay = 9.8 m/s2
∆t = 3.2 s
vfy = ?
vfy − viy
ay =
∆t
vfy = viy + a y ∆t
= 0 + (9.8 m/s 2 )(3.2 s)
 1 km   3600 s 
= (31m/s) 


 1000 m   1 h 
vfy = 1.1× 102 km/h
4.
Thus, the landing speed of the stone is 31 m/s or 1.1 × 102 km/h.
Define +y as up.
viy = 5.112 m/s
vfy = 0
∆y = ?
In Java, g = −9.782 m/s2:
vfy 2 = viy 2 + 2a y ∆y
2
2a y ∆y = vfy − viy
∆y =
=
−viy
2
2
2a y
−(5.112 m/s)2
2( −9.782 m/s 2 )
∆y = 1.336 m
In London, g = −9.823 m/s2:
∆y =
=
−viy 2
2a y
−(5.112 m/s)2
2( −9.823 m/s 2 )
∆y = 1.330 m
In Java, the jumper achieved a height of 1.336 m; in London, the jumper achieved a height of 1.330 m.
40
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
5.
Let +y be the forward direction of the shuttle, assumed to be a straight line.
viy = 0
ay = 5(9.80 m/s2)
∆t = 1.0 min = 60 s (two significant digits)
vfy = ?
vfy − viy
ay =
∆t
vfy = a y ∆t
= 5(9.80 m/s 2 )(60 s)
 1 km   3600 s 
= (2.9 ×103 m/s) 


 1000 m   1 h 
vfy = 1.1× 10 4 km/h
The shuttle’s speed is 2.9 m/s or 1.1 × 104 km/h.
6. Let +y be up.
∆t = 2.6 s
ay = g = −9.80 m/s2
(a) ∆t = ?
The time for the ball to rise will be half of the total time.
2.6 s
∆t =
2
∆t = 1.3 s
Therefore, the time for the ball to rise is 1.3 s.
(b) vfy = 0
∆t = 1.3 s
viy = ?
vfy − viy
ay =
∆t
viy = vfy − a y ∆t
= 0 − ( −9.8 m/s2 )(1.3s)
viy = 13m/s
The velocity is 13 m/s [up] when the golf ball leaves the person’s hand.
(c) Again, +y is up.
viy = 13 m/s
ay = g = −3.8 m/s2
∆t = ?
vfy − viy
ay =
∆t
vfy − viy
∆t =
ay
=
7.
0 − 13m/s
−3.8 m/s 2
∆t = 3.4 s
The total time in flight would be 2(3.4 s) = 6.8 s.
Let +y be down.
∆t = 0.087 s
∆y = 0.80 m
ay = 9.8 m/s2
Copyright © 2003 Nelson
Chapter 1 Kinematics
41
1
∆y = vfy ∆t − a y (∆t ) 2
2
∆y 1
vfy =
+ a y ( ∆t )
∆t 2
0.80 m 1
=
+ 9.8 m/s 2 (0.087 s)
0.087 s 2
(
)
vfy = 9.6 m/s
8.
The velocity of the ball as it hits the floor is 9.6 m/s [down].
Let +y be down.
viy = 14 m/s
ay = 9.8 m/s2
(a) ∆y = 21 m
∆t = ?
ay
1
Using the equation ∆y = viy ∆t + a y ( ∆t )2 , rearrange as
( ∆t )2 + viy ∆t − ∆y = 0 , which is a form of the quadratic
2
2
equation. Therefore,
∆t =
−b ± b 2 − 4ac
2a
 ay 
−viy ± viy 2 − 4   ( −∆y )
 2 
=
 ay 
2 
 2 
=
=
−viy ± viy 2 + 2a y ∆y
ay
−14 m/s ± (14 m/s) 2 + 2(9.8 m/s 2 )(21m)
9.8 m/s 2
∆t = 1.1s or − 3.9s
The stone will take 1.1 s to reach the water below.
(b) The positive root is the actual answer when the stone is thrown vertically downward. The negative root is the time that
the stone would have travelled had the initial velocity been upward rather than downward (i.e., if vi = −14 m/s in this
case).
9.
10. Note: To make the solution more compact, units are omitted until the final answer is written.
Let +y be down, let F represent the flowerpot, and let B represent the ball.
Since the ball is thrown 1.00 s after the flowerpot is released, after some time ∆tF = ∆tB − 1.00, let us assume the ball will
pass the flowerpot. At this instant, the flowerpot will have travelled 28.5 m − 26.0 m = 2.5 m farther than the ball. Thus,
∆yF = ∆yB + 2.5.
But with uniform acceleration, ∆y = viy ∆t +
1
2
ay∆t 2 .
Combining these relationships, we have:
42
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson
∆yF = ∆yB + 2.5
viyF ∆tF +
1
2
a y ∆tF 2 = viyB ∆tB +
1
2
a y ∆tB 2 + 2.5
0 + 4.9∆t F 2 = 12 ( ∆t F − 1) + 4.9 ( ∆t F − 1) + 2.5
2
(
)
4.9 ∆t F 2 = 12 ( ∆t F − 1) + 4.9 ∆t F 2 − 2 ∆t F + 1 + 2.5
4.9 ∆t F 2 = 12 ∆t F − 12 + 4.9 ∆t F 2 − 9.8∆t F + 4.9 + 2.5
0 = 12 ∆t F − 12 − 9.8∆t F + 4.9 + 2.5
0 = 2.2∆t F − 4.6
∆t F =
4.6
2.2
∆tF = 2.09 s
After this time interval, the flowerpot has fallen:
1
2
∆y F = a y ∆t F
2
1
2
2
= 9.80 m/s ( 2.09 s )
2
∆yF = 21.4 m
(
)
The flowerpot is 28.5 m − 21.4 m = 7.10 m above the ground when the ball passes it.
11. The ranking from highest terminal speed to lowest is: pollen; a ping-pong ball; a basketball; a skydiver diving headfirst;
and a skydiver in spread-eagle orientation.
Applying Inquiry Skills
12. (a) 9.809 060 m/s2 has 7 significant digits, a possible error of ±0.000 000 5, and percent possible error of
±0.000 000 5 m/s 2
× 100% = ±0.000 005%.
9.809 060 m/s 2
(b) 9.8 m/s2 has 2 significant digits, a possible error of ±0.05 m/s2, and percent possible error of
±0.05 m/s 2
× 100% = ±0.5%.
9.8 m/s 2
(c) 9.80 m/s2 has 3 significant digits, a possible error of ±0.005 m/s2, and percent possible error of
±0.005 m/s 2
× 100% = ±0.05%.
9.80 m/s 2
(d) 9.801 m/s2 has 4 significant digits, a possible error of ±0.0005 m/s2, and percent possible error of
±0.0005 m/s 2
× 100% = ±0.005%.
9.801 m/s 2
(e) 9.8 × 10–6 m/s2 has 2 significant digits, a possible error of ±0.05× 10–6 m/s2, and percent possible error of
±0.05 ×10 −6 m/s 2
×100% = ±0.5%.
9.8 ×10−6 m/s 2
13. (a) Students may recall performing this activity in a previous grade. Hold a metre stick vertically while your partner holds
his or her separated index finger and thumb ready to catch the stick at a specific mark, such as the 50-cm mark.
Without warning, drop the stick and determine how far the stick falls before the partner catches it. Repeat several times
and take an average of the distances, ∆y. Use this value in the appropriate equation, as shown below.
Let +y be down.
Assume ∆y = 18 cm = 0.18 m
ay = 9.8 m/s2
viy = 0
∆t = ?
Copyright © 2003 Nelson
Chapter 1 Kinematics
43
∆y = viy ∆t +
∆y =
∆t =
=
1
2
a y ∆t
1
2
a y ∆t
2
2
2 ∆y
ay
2(0.18 m)
9.8 m/s 2
∆t = 0.19 s
(b) Talking on a cell phone would likely increase reaction time. Students can simulate this situation by engaging in
distracting conversation with the lab partner whose reaction time is being tested.
Making Connections
14. Aristotle and Galileo influenced the philosophy and scientific thought of their respective eras, and in both cases their
influence lasted long after they died. Students can find information about these science “giants” in books and
encyclopedias, or on the Internet. For example, an advanced word search on the Internet, entering only the words Aristotle
and Galileo, found more than 20 thousand hits, many of which featured discussions of the same topics featured in the text.
15. Deductive reasoning involves using theories to account for specific experimental results. Thus, deductive reasoning uses
ideas to explain observed phenomena. Inductive reasoning involves making and collecting observations, and then
developing general theories or hypotheses to account for the observations.
(a) Aristotle and other ancient scientists used deductive reasoning.
(b) Galileo used the process of inductive reasoning. When he observed that heavy objects fall with increasing speed, he
formed the hypothesis that the speed of the object was directly proportional to the distance the object fell. When this
hypothesis proved false, he hypothesized that the speed of a falling object is directly proportional to the time, not the
distance. Through experiments, he was able to verify his hypothesis.
(c) Various ways are used to contrast these types of reasoning. For example, in deductive reasoning, particular results are
inferred from a general law, whereas in inductive reasoning, a general law is inferred from particular results. Stated
another way for deduction, conclusions follow from premises, that is the reasoning goes from the general to the
specific. In induction, premises lead to the conclusions, or the reasoning goes from the specific to the general. In
mathematics, induction involves proving a theorem using a process in which the theorem is verified for a small value of
an integer, and then extending the verification to greater values of the integer.
16. Many sites can be found by doing an advanced word search on the Internet. For example, in entering the words Luis
Alvarez, Yucatan, and dinosaurs, more than 400 sites were found, many of them highly credible. Two examples are:
www.ceemast.csupomona.edu/nova/alvarez2.html
www.space.com/scienceastronomy/planetearth/deep_impact_991228.html
1.4 PROJECTILE MOTION
PRACTICE
(Page 46)
Understanding Concepts
1.
A projectile is an object that moves through the air without a propulsion system and follows a curved path. An airplane
has a propulsion system and does not follow a trajectory. Thus, an airplane is not a projectile.
2. The projectile experiences constant downward acceleration due to gravity (vertical acceleration) and the horizontal
component of acceleration is zero.
3. Let +x be to the right and +y be downward. The initial position is the position where the marble leaves the table.
(a) Horizontally (constant vix ):
vix = 1.93 m/s
∆x = ?
∆t = ?
44
Unit 1 Forces and Motion: Dynamics
Copyright © 2003 Nelson