Foundation Engineering Science – Dr. Daniel Nankoo
Tutorial Solutions
Solutions to Tutorial 1
1. Picture the Problem: You walk in both the
positive and negative directions along a straight
line.
Strategy: The distance is the total length of
travel, and the displacement is the net change in
position.
Solution: (a) Add the lengths:
( 0.75 + 0.60 mi ) + ( 0.60 mi ) = 1.95 mi
(b) Subtract xi from xf to find the
displacement.
∆ x = x f − xi = 0.75 − 0.00 mi = 0.75 mi
Insight: The distance travelled is always positive, but the displacement can be negative.
2. Picture the Problem: Player A walks in the positive direction
and player B walks in the negative direction.
Strategy: In each case the distance is the total length of travel,
and the displacement is the net change in position.
Solution: (a) Note the distance travelled
by player A:
5m
The displacement of player A is positive:
∆ x = x f − xi = 5 m − 0 m = 5 m
(b) Note the distance travelled by player B:
2m
The displacement of player B is negative.
Let the origin be at the initial position of
player A.
∆ x = x f − xi = 5 m − 7 m = −2 m
Insight: The distance travelled is always positive, but the displacement can be negative.
3. Picture the Problem: The ball is putted in the
positive direction and then the negative
direction.
Strategy: The distance is the total length of
travel, and the displacement is the net change
in position.
Solution: (a) Add the lengths:
(10 + 2.5 m ) + 2.5 m = 15 m
(b) Subtract xi from xf to find the
displacement.
∆ x = x f − xi = 10 − 0 m = 10 m
Insight: The distance travelled is always positive, but the displacement can be negative.
4. Picture the Problem: You walk in both the
positive and negative directions along a straight
line.
Strategy: The distance is the total length of
travel, and the displacement is the net change in
position.
Solution: (a) Add the lengths:
( 0.60 + 0.35 mi ) + ( 0.75 + 0.60 + 0.35 mi ) =
(b) Subtract xi from xf to find the
∆ x = x f − xi = 0.75 − 0.00 mi = 0.75 mi
2.65 mi
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Foundation Engineering Science – Dr. Daniel Nankoo
Tutorial Solutions
displacement.
Insight: The distance travelled is always positive, but the displacement can be negative.
5. Picture the Problem: The runner moves along
the oval track.
Strategy: The distance is the total length of
travel, and the displacement is the net change in
position.
Solution: 1. (a) Add the lengths:
(15 m ) + (100 m ) + (15 m ) = 130 m
2. Subtract xi from xf to find the
displacement.
∆ x = x f − xi = 100 − 0 m = 100 m
3. (b) Add the lengths
15 + 100 + 30 + 100 + 15 m = 260 m
4. Subtract xi from xf to find the
displacement.
∆ x = x f − xi = 0 − 0 m = 0 m
Insight: The distance travelled is always positive, but the displacement can be negative. The
displacement is always zero for a complete circuit, as in this case.
6. Picture the Problem: The pony walks around the
circular track.
4.5 m
A
Strategy: The distance is the total length of
travel, and the displacement is the net change in
position.
Solution: (a) 1. The distance travelled is half the d =
circumference:
2. The displacement is the distance from A to B:
1
2
B
( 2π r ) = π r = π ( 4.5 m ) = 14 m
∆x = x f − xi = 2r = 2 ( 4.5 m ) = 9.0 m
3. (b) The distance travelled will increase when the child completes one circuit, because the pony
will have taken more steps.
4. (c) The displacement will decrease when the child completes one circuit, because the
displacement is maximum when the child has gone halfway around, and is zero when the child
returns to the starting position.
5. (d) The distance travelled equals the
circumference:
d = 2π r = 2π ( 4.5 m ) = 28 m
6. The displacement is zero because the child has returned to her starting position.
Insight: The distance travelled is always positive, but the displacement can be negative. The
displacement is always zero for a complete circuit, as in this case.
7. Picture the Problem: The kangaroo hops in the forward direction.
Strategy: The distance is the average speed multiplied by the time elapsed. The time elapsed is the
distance divided by the average speed.
km 
1h 

Solution: 1. (a) Multiply the
d = s t =  65
 3.2 min ×
 = 3.5 km
h 
60 min 

average speed by the time elapsed:
2. (b) Divide the distance by the
average speed:
t=
d 0.25 km 60 min
=
×
= 14 s
s 65 km/h
1h
Insight: The instantaneous speed might vary from 65 km/h, but the time elapsed and the distance
travelled depend only upon the average speed during the interval in question.
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Tutorial Solutions
8. Picture the Problem: The radio waves propagate in a straight line.
Strategy: The time elapsed is the distance divided by the average speed. The distance to the Moon
is 2.39×105 mi. We must double this distance because the signal travels there and back again.
Solution: Divide the distance by the average
speed:
t=
5
2d 2 ( 2.39 × 10 mi )
=
= 2.57 s
s
1.86 × 105 mi/s
Insight: The time is slightly shorter than this because the given distance is from the centre of the
Earth to the centre of the Moon, but presumably any radio communications would occur between
the surfaces of the Earth and Moon. When the radii of the two spheres is taken into account, the
time decreases to 2.52 s.
9. Picture the Problem: The sound waves propagate in a straight line from the thunderbolt to your
ears.
Strategy: The distance is the average speed multiplied by the time elapsed. We will neglect the
time it takes for the light wave to arrive at your eyes because it is vastly smaller than the time it
takes the sound wave to travel.
Solution: Multiply the average speed by the
time elapsed:
d = s t = ( 340 m/s )( 3.5 s ) = 1200 m = 1.2 km
Insight: The speed of sound, 340 m/s, works out to approximately one mile every five seconds, a
useful rule of thumb for estimating the distance to an approaching thunderstorm!
10. Picture the Problem: The nerve impulses propagate at a fixed speed.
Strategy: The time elapsed is the distance divided by the average speed. The distance from your
finger to your brain is on the order of one meter.
Solution: Divide the distance by the average
speed:
t=
d
1m
=
= 0.010 s
s 1× 102 m/s
Insight: This nerve impulse travel time is not the limiting factor for human reaction time, which is
about 0.2 s.
11. Picture the Problem: The finch travels a short distance on the back of the tortoise and a longer
distance through the air, with both displacements along the same direction.
Strategy: First find the total distance travelled by the finch and then determine the average speed
by dividing by the total time elapsed.
Solution: 1. Determine the total
distance travelled:
d = s1 ∆t1 + s2 ∆t2
2. Divide the distance by the time
elapsed:
s=
d = ( 0.060 m/s )(1.2 min ) + (12 m/s )(1.2 min )  × 60 s/min
d = 870 m = 0.87 km
d
870 m
=
= 6.0 m/s
∆t 2.4 min × 60 s/min
Insight: Most of the distance travelled by the finch occurred by air. In fact, if we neglect the 4.3 m
the finch travelled while on the tortoise’s back, we still get an average speed of 6.0 m/s over the 2.4
min time interval! The bird might as well have been at rest.
12. Picture the Problem: You travel 8.0 km on foot and then an additional 16 km by car, with both
displacements along the same direction.
Strategy: First find the total time elapsed by dividing the distance travelled by the average and
divide by the total time elapsed to find the average speed. Set that average speed to the given value
and solve for the car’s speed.
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Tutorial Solutions
Solution: 1. Use the definition of average ∆t = d = 8.0 + 16 km = 1.1 h
sav
22 km/h
speed to determine the total time elapsed.
2. Find the time elapsed while in the car:
∆t2 = ∆t − ∆t1 = 1.1 h − 0.84 h = 0.3 h
3. Find the speed of the car:
s2 =
d 2 16 km
=
= 50 km/h
∆t 2
0.3 h
Insight: This problem illustrates the limitations that significant figures occasionally impose. If
you keep an extra figure in the total elapsed time (1.09 h) you’ll end up with the time elapsed for
the car trip as 0.25 h, not 0.3, and the speed of the car is 64 km/h. But the rules of subtraction
indicate we only know the total time to within a tenth of an hour, so we can only know the time
spent in the car to within a tenth of an hour, or to within one significant digit.
13. Picture the Problem: You travel in a straight line at two different speeds during the specified time
interval.
Strategy: Determine the average speed by first calculating the total distance travelled and then
dividing it by the total time elapsed.
Solution: 1. (a) Because the time intervals are the same, you spend equal times at 20 m/s and 30
m/s, and your average speed will be equal to 25.0 m/s.
2. (b) Divide the total
distance by the time
elapsed:
sav =
s1∆t1 + s2 ∆t2 ( 20.0 m/s )(10.0 min × 60 s ) + ( 30.0 m/s )( 600 s )
=
∆t1 + ∆t2
600 + 600 s
sav = 25.0 m/s
Insight: The average speed is a weighted average according to how much time you spend travelling
at each speed.
14. Picture the Problem: You travel in a straight line at two different speeds during the specified time
interval.
Strategy: Determine the average speed by first calculating the total distance travelled and then
dividing it by the total time elapsed.
Solution: 1. (a) The distance intervals are the same but the time intervals are different. You will
spend more time at the lower speed than at the higher speed. Since the average speed is a time
weighted average, it will be less than 25.0 m/s.
2. (b) Divide the total distance by the
time elapsed:
sav =
d1 + d 2
d + d2
20.0 mi
= 1
=
∆t1 + ∆t2 d1 d 2  10.0 mi 10.0 mi 
+
+
s1 s2  20.0 m/s 30.0 m/s 
sav = 24.0 m/s
Insight: Notice that in this case it is not necessary to convert miles to meters in both the numerator
and denominator because the units cancel out and leave m/s in the numerator.
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Tutorial Solutions
15. Picture the Problem: The given position function indicates the particle begins travelling in the
negative direction but is accelerating in the positive direction.
Strategy: Create the x-versus-t plot using a spreadsheet, or calculate individual values by hand and
sketch the curve using graph paper. Use the known x and t information to determine the average
speed and velocity.
Solution: 1. (a) Use a spreadsheet or
similar program
to create the plot:
2
2


∆x ( −5 m/s )(1.0 s ) + ( 3 m/s ) (1.0 s )  − [0.0 m ]
=
2. (b) Find the average velocity from t vav =
∆t
1.0 s
= 0 to t = 1.0 s:
= −2.0 m/s
3. (c) The average speed is the
magnitude of the
average velocity:
sav = vav = 2.0 m/s
Insight: Note that the average velocity over the first second of time is equal to the slope of a
straight line drawn from the origin to the curve at t = 1.0 s. At that time the position is −2.0 m.
16. Picture the Problem: The given position function indicates the particle begins travelling in the
positive direction but is accelerating in the negative direction.
Strategy: Create the x-versus-t plot using a spreadsheet, or calculate individual values by hand and
sketch the curve using graph paper. Use the known x and t information to determine the average
speed and velocity.
Solution: 1. (a) Use a spreadsheet to create the plot shown at right:
∆x
∆t
( 6 m/s )(1.0 s ) + ( −2 m/s 2 ) (1.0 s )2  − [ 0.0 m ]


1.0 s
vav = 4.0 m/s
2. (b) Find the
vav =
average velocity
from t = 0 to t = 1.0
s:
=
3. (c) The average
speed is the
magnitude of the
average velocity:
sav = vav = 4.0 m/s
Insight: Note that the average velocity over the first second of time is equal to the slope of a
straight line drawn from the origin to the curve at t = 1.0 s. At that time the position is 4.0 m.
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Tutorial Solutions
17. Picture the Problem: Following the motion specified
in the position-versus-time graph, the tennis player
moves left, then right, then left again, if we take left to
be in the negative direction.
Strategy: Determine the direction of the velocity from
the slope of the graph. The speed will be greatest for
the segment of the curve that has the largest slope
magnitude.
Solution: 1. (a) The magnitude of the slope at B is
larger than A or C so we conclude the speed is greatest
at B.
2. (b) Find the slope of the graph at A:
sav =
3. (c) Find the slope of the graph at B:
sav =
4. (d) Find the slope of the graph at C:
sav =
∆x
∆t
∆x
∆t
∆x
∆t
=
=
=
−2.0 m
2.0 s
2.0 m
1.0 s
= 2.0 m/s
−1.0 m
2.0 s
= 1.0 m/s
= 0.50 m/s
Insight: The speed during segment B is larger than the speed during segments A and C, as
predicted. Speeds are always positive because they do not involve direction, but velocities can be
negative to indicate their direction.
18. Picture the Problem: The given position function indicates the particle begins travelling in the
positive direction but is accelerating in the negative direction.
Strategy: Create the x-versus-t plot using a spreadsheet, or calculate individual values by hand and
sketch the curve using graph paper. Use the known x and t information to determine the average
speed and velocity.
Solution: 1. (a) Use a
spreadsheet to create the
plot:
3
3
3
3

 

∆x ( 2 m/s )( 0.45 s ) − ( 3 m/s ) ( 0.45 s )  − ( 2 m/s )( 0.35 s ) − ( 3 m/s ) ( 0.35 s ) 
2. (b)
=
vav =
Find the
∆t
0.10 s
average
= 0.55 m/s
velocity
from t =
0.35 to
t = 0.45 s:
3
3
3
3

 

∆x ( 2 m/s )( 0.41 s ) − ( 3 m/s ) ( 0.41 s )  − ( 2 m/s )( 0.39 s ) − ( 3 m/s ) ( 0.39 s ) 
3. (c)
vav =
=
Find the
∆t
0.41 − 0.39 s
average
= 0.56 m/s
velocity
from t =
0.39 to
t = 0.41 s:
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Tutorial Solutions
4. (d) The instantaneous speed at t = 0.40 s will be closer to 0.56 m/s. As the time interval becomes
smaller the average velocity is approaching 0.56 m/s, so we conclude the average speed over an
infinitesimally small time interval will be very close to that value.
Insight: Note that the instantaneous velocity at 0.40 s is equal to the slope of a straight line drawn
tangent to the curve at that point. Since it is difficult to accurately draw a tangent line, we usually
resort to mathematical methods like those illustrated above to determine the instantaneous velocity.
19. Picture the Problem: The given position function indicates the particle begins travelling in the
negative direction but is accelerating in the positive direction.
Strategy: Create the x-versus-t plot using a spreadsheet, or calculate individual values by hand and
sketch the curve using graph paper. Use the known x and t information to determine the average
speed and velocity.
Solution: 1. (a) Use a spreadsheet to
create the plot:
2. (b) Find the
average velocity
from t = 0.150 to
t = 0.250 s:
3. (c) Find the
average velocity
from t = 0.190 to
t = 0.210 s:
 ( −2 m/s )( 0.250 s ) + ( 3 m/s 3 ) ( 0.250 s )3  − 




3 
3
( −2 m/s )( 0.150 s ) + ( 3 m/s ) ( 0.150 s )  
∆x 


vav =
=
= −1.63 m/s
∆t
0.250 − 0.150 s
 ( −2 m/s )( 0.210 s ) + ( 3 m/s3 ) ( 0.210 s )3  − 




3 
3


−2 m/s )( 0.190 s ) + ( 3 m/s ) ( 0.190 s ) 
∆x 
(

vav =
=
= −1.64 m/s
∆t
0.210 − 0.190 s
4. (d) The instantaneous speed at t = 0.200 s will be closer to −1.64 m/s. As the time interval
becomes smaller the average velocity is approaching −1.64 m/s, so we conclude the average speed
over an infinitesimally small time interval will be very close to that value.
Insight: Note that the instantaneous velocity at 0.200 s is equal to the slope of a straight line drawn
tangent to the curve at that point. Since it is difficult to accurately draw a tangent line, we usually
resort to mathematical methods like those illustrated above to determine the instantaneous velocity.
20. Picture the Problem: The airliner accelerates uniformly along a straight runway.
Strategy: The average acceleration is the change of the velocity divided by the elapsed time.
Solution: Divide the change in velocity aav = ∆v = 173 − 0 mi/h × 0.447 m/s = 2.20 m/s 2
∆t
35.2 s
mi/h
by the time:
Insight: The instantaneous acceleration might vary from 2.20 m/s2, but we can calculate only
average acceleration from the net change in velocity and time elapsed.
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Tutorial Solutions
21. Picture the Problem: The airliner slows down uniformly along a straight runway as it travels
towards the east.
Strategy: The average acceleration is the change of the velocity divided by the elapsed time.
Assume that east is in the positive direction
Solution: 1. Divide the change in velocity a = ∆v = 0 − 115 m/s = 8.85 m/s 2
av
∆t
13.0 s
by the time:
2. We note from the previous step that the acceleration is negative. Since east is the positive
direction, negative acceleration must be towards the west.
Insight: In physics we almost never talk about deceleration. Instead, we call it negative
acceleration.
22. Picture the Problem: The car travels in a straight line due north, either speeding up or slowing
down, depending upon the direction of the acceleration.
Strategy: Use the definition of acceleration to determine the final velocity over the specified time
interval.
Solution: 1. (a) Evaluate v = v = v0 + at = 18.1 m/s + (1.30 m/s 2 ) ( 7.50 s ) = 27.9 m/s north
v0 + at directly:
2. (b) Evaluate v = v0 + at
directly:
v = v0 + at = 18.1 m/s + ( −1.15 m/s 2 ) ( 7.50 s ) = 9.48 m/s north
Insight: In physics we almost never talk about deceleration. Instead, we call it negative
acceleration. In this problem south is considered the negative direction, and in part (b) the car is
slowing down or undergoing negative acceleration.
23. Picture the Problem: Following the motion specified in
the velocity-versus-time graph, the motorcycle is
speeding up, then moving at constant speed, then
slowing down.
Strategy: Determine the acceleration from the slope of
the graph.
∆v 10 m/s
=
Solution: 1. (a) Find the slope at aav =
∆t
5.0 s
A:
= 2.0 m/s 2
∆v 0 m/s
=
= 0.0 m/s 2
2. (b) Find the slope of the graph aav =
∆
t
10.0
s
at B:
∆v −5.0 m/s
=
= −0.50 m/s 2
3. (c) Find the slope of the graph aav =
∆
t
10.0
s
at C:
Insight: The acceleration during segment A is larger than the acceleration during segments B and
C because the slope there has the greatest magnitude.
24. Picture the Problem: The car travels in a straight line in the positive direction while accelerating
in the negative direction (slowing down).
Strategy: Use the constant acceleration equation of motion to determine the time elapsed for the
specified change in velocity.
Solution: 1. (a) The time required to come to a stop is the change in velocity divided by the
acceleration. In both cases the final velocity is zero, so the change in velocity doubles when you
double the initial velocity. Therefore the stopping time will increase by a factor of two when you
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Tutorial Solutions
double your driving speed.
2. (b) Solve for time:
t=
v − v0 0 − 16 m/s
=
= 3.8 s
a
− 4.2 m/s 2
3. (c) Solve for time:
t=
v − v0 0 − 32 m/s
=
= 7.6 s
a
− 4.2 m/s 2
Insight: Note that the deceleration is treated as a negative acceleration in this problem and
elsewhere in the text.
25. Picture the Problem: The train travels in a straight line in the positive direction while accelerating
in the positive direction (speeding up).
Strategy: First find the acceleration and then determine the final velocity.
Solution: 1. Use the definition of acceleration: a = v − v0 = 4.7 − 0 m/s = 0.94 m/s 2
t
5.0 s
2. Evaluate directly, using the final speed from v = v0 + at = 4.7 m/s + ( 0.94 m/s 2 ) ( 6.0 s )
the first segment as the initial speed of the
v = 10.3 m/s
second segment:
Insight: Another way to tackle this problem is to set up similar triangles on a velocity-versus-time
graph. The answer would then be calculated as (4.7 m/s) × 11 s / 5 s = 10.3 m/s. Try it!
26. Picture the Problem: The particle travels in a straight line in the positive direction while
accelerating in the positive direction (speeding up).
Strategy: Use the constant acceleration equation of motion to find the initial velocity.
Solution: Solve for v0 :
v0 = v − at = 9.31 m/s − ( 6.24 m/s 2 ) ( 0.300 s ) = 7.44 m/s
Insight: As expected the initial velocity is less than the final velocity because the particle is
speeding up.
27. Picture the Problem: The car travels in a straight line toward the west while accelerating in the
easterly direction (slowing down).
Strategy: The average velocity is simply half the sum of the initial and final velocities because the
acceleration is uniform.
Solution: Calculate half the sum of the
velocities:
vav =
1
2
( v0 + v ) = 12 (12 + 0 m/s ) =
6.0 m/s
Insight: The average velocity of any object that slows down and comes to a stop is just half the
initial velocity.
28. Picture the Problem: The boat travels in a straight line with constant positive acceleration.
Strategy: The average velocity is simply half the sum of the initial and final velocities because the
acceleration is uniform.
Solution: 1. (a) Calculate half the sum
of the velocities:
vav =
2. (b) The distance travelled is the
average velocity multiplied by the time
elapsed:
d = vav t = ( 2.06 m/s )( 4.77 s ) = 9.83 m
1
2
( v0 + v ) = 12 ( 0 + 4.12 m/s ) =
2.06 m/s
Insight: The average velocity of any object that speeds up from rest is just half the final velocity.
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Tutorial Solutions
29. Picture the Problem: The cheetah runs in a straight line with constant positive acceleration.
Strategy: The average velocity is simply half the sum of the initial and final velocities because the
acceleration is uniform. The distance travelled is the average velocity multiplied by the time
elapsed.
Solution: 1. (a) Calculate half the sum of the vav =
velocities:
2. Use the average velocity to find the
distance:
1
2
( v0 + v ) = 12 ( 0 + 25.0 m/s ) = 12.5 m/s
d = vav t = (12.5 m/s )( 6.22 s ) = 77.8 m
3. (b) For a constant acceleration the velocity varies linearly with time. Therefore we expect the
velocity to be equal to 12.5 m/s after half the time (3.11 s) has elapsed.
4. (c) Calculate half the sum of the velocities: vav,1 =
1
2
( v0 + v ) = 12 ( 0 + 12.5 m/s ) =
5. Calculate half the sum of the velocities:
vav,2 =
1
2
( v0 + v ) = 12 (12.5 + 25.0 m/s ) = 18.8 m/s
6. (d) Use the average velocity to find the
distance:
d1 = vav,1 t = ( 6.25 m/s )( 3.11 s ) = 19.4 m
7. Use the average velocity to find the
distance:
d 2 = vav,2 t = (18.8 m/s )( 3.11 s ) = 58.5 m
6.25 m/s
Insight: The distance travelled is always the average velocity multiplied by the time. This stems
from the definition of average velocity.
30. Picture the Problem: The air bag expands outward with constant positive acceleration.
Strategy: Assume the air bag has a thickness of 1 ft or about 0.3 m. It must expand that distance
within the given time of 10 ms. Employ the relationship between acceleration, displacement, and
time to find the acceleration.
Solution: Solve for a:
a=
2 ( 0.3 m )
2 ∆x
1g
=
= 6000 m/s 2 ×
2
2
t
9.81 m/s 2
(10 ms × 0.001 s/ms )
600 g
Insight: The very large acceleration of an expanding airbag can cause severe injury to a small child
whose head is too close to the bag when it deploys. Children are safest in the back seat!
31. Picture the Problem: The bacterium accelerates from rest in the forward direction.
Strategy: Employ the definition of acceleration to find the time elapsed, and the relationship
between acceleration, displacement, and velocity to find the distance travelled.
v − v0 12 − 0 µ m/s
=
= 0.077 s
a
156 µ m/s 2
Solution: 1. (a) Solve for time:
t=
2. (b) Solve for displacement:
∆x =
2
2
v 2 − v02 (12 µ m/s ) − 0
=
= 0.46 µ m
2
2a
2 (156 µ m/s )
Insight: The accelerations are tiny but so are the bacteria! The average speed here is about 3 body
lengths per second if each bacterium were 2 µm long. If this were a human that would be 6 m/s or
13 mi/h, much faster than we can swim!
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Tutorial Solutions
32. Picture the Problem: The rocket accelerates straight upward.
Strategy: Employ the relationship between acceleration, displacement, and time to find the
acceleration. Because the rocket was at rest before blast off, the initial velocity v0 is zero, and so
is the initial position x0 . Once the acceleration is known, we can use the constant acceleration
equation of motion to find the speed.
Solution: 1. (a) Use the following:
x = x0 + v0 t + 12 at 2
2. Let x0 = v0 = 0 and solve for
acceleration:
a=
3. (b) Evaluate:
v = 0 + at = (17 m/s 2 ) ( 3.0 s ) = 51 m/s
2 x 2 ( 77 m )
=
= 17 m/s 2 upward
2
t2
3.0
s
(
)
Insight: Notice the simple relationship between distance, acceleration, and time if the initial
position and the initial velocity are zero.
33. Picture the Problem: You drive in a straight line and then slow down to a stop.
Strategy: Employ the relationship between acceleration, displacement, and velocity to find the
displacement. Note that no time information is given. In this case the acceleration is negative
because the car is slowing down.
2
2
2
2
2
2
12.0 m/s )
Solution: 1. (a) Solve for ∆x : ∆x = v − v0 = 0 − v0 = − v0 = − (
= 21 m
2a
2a
2a
2 ( −3.5 m/s 2 )
2. (b) Since velocity is proportional to the square root of displacement, cutting the distance in half
will reduce the velocity by 2 , not 2. Therefore the speed will be greater than 6.0 m/s after
travelling half the distance.
3. Solve for v:
v = v02 + 2a
 v2  v
∆x
12.0 m/s
= v02 + a  − 0  = 0 =
= 8.49 m
2
2
a
2
2


Insight: For constant acceleration, the velocity changes linearly with time but nonlinearly with
distance.
34. Picture the Problem: The bicycle travels in a straight line, slowing down at a uniform rate as it
crosses the sandy patch.
Strategy: Use the time-free relationship between displacement, velocity, and acceleration to find
the acceleration. The time can be determined from the average velocity and the distance across the
sandy patch.
2
Solution: 1. (a) Solve for acceleration:
a=
2
v 2 − v02 ( 6.4 m/s ) − ( 8.4 m/s )
=
= −2.1 m/s 2
2∆x
2 ( 7.2 m )
where the negative sign means 2.1 m/s2 due east.
2. (b) Solve for t:
t=
1
2
∆x
=
( v + v0 )
1
2
7.2 m
= 0.97 s
( 8.4 + 6.4 m/s )
3. (c) Examining v 2 = v02 + 2a∆x in detail, we note that the acceleration is negative, and that the
final velocity is the square root of the difference between v02 and 2a∆x . Since 2a∆x is constant
because the sandy patch doesn’t change, it now represents a larger fraction of the smaller v02 , and
the final velocity v will be more than 2.0 m/s different than v0 . We therefore expect a final speed
of less than 3.4 m/s.
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Tutorial Solutions
Insight: In fact, if you try to calculate v in part (c) you end up with the square root of a negative
2
number, because the bicycle will come to rest in a distance ∆x =
− ( 5.4 m/s )
02 − v02
=
= 6.9 m ,
2a
2 ( −2.1 m/s 2 )
less than the 7.2 m length of the sandy patch.
35. Picture the Problem: David Purley travels in a straight line, slowing down at a uniform rate until
coming to rest.
Strategy: Use the time-free relationship between displacement, velocity, and acceleration to find
the acceleration.
0.278 m/s 

02 − 173 km/h ×

v −v
1 km/h 

a=
=
2 ∆x
2 ( 0.66 m )
2
Solution: Solve for acceleration:
2
2
0
a = −1800 m/s 2 ×
1.00 g
= 180 g
9.81 m/s 2
Insight: Mr. Purley was lucky to escape death when experiencing an acceleration this large. A
large acceleration implies a large force, which in this case must have been applied to his body in
just the right way to produce a non-lethal injury.
36. Picture the Problem: The rocket accelerates straight upward at a constant rate.
Strategy: Since the initial and final velocities are known, the time can be determined from the
average velocity and the distance travelled. The constant acceleration equation of motion can then
be used to find the acceleration. Once that is known, the position of the rocket as a function of
time is given, and the velocity as a function of time is also given.
∆x
3.2 m
Solution: 1. (a) Solve for time:
t=
2. (b) Solve for acceleration:
a=
3. (c) Evaluate directly, with
x0 = v0 = 0 :
x = 12 at 2 =
4. Evaluate directly, with v0 = 0 :
v = 0 + at = (110 m/s 2 ) ( 0.10 s ) = 11 m/s
1
2
( v + v0 )
=
1
2
( 0 + 26.0 m/s )
= 0.25 s
v − v0 26.0 − 0 m/s
=
= 110 m/s 2 = 0.11 km/s 2
t
0.25 s
1
2
(110 m/s ) ( 0.10 s )
2
2
= 0.55 m
Insight: Model rockets accelerate at very large rates, but only for a very short time. Still, even
inexpensive starter rockets can reach 1500 ft in altitude and can be great fun to build and launch!
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Tutorial Solutions
37. Picture the Problem: The distance-versus-time plot
at right shows how the bicyclist can overtake his
friend by pedalling at constant acceleration.
Strategy: To find the time elapsed when the two
bicyclists meet, we must set the constant velocity
equation of motion of the friend equal to the constant
acceleration equation of motion of the cyclist. Once
the time is known, the displacement and velocity of
the bicyclist can be determined.
Solution: 1. (a) Set the two equations of
motion equal to each other. For the
friend, set x0 = 0 and for the cyclist, use
xfriend = xbicyclist
v f t = 0 + 0 + 12 ab ( t − 2 )
2
set x0 = 0 and v0 = 0 :
vfriend t = 12 ab ( t 2 − 4t + 4 )
2. Solve for t:

2 ( 3.5 m/s ) 


2v
0 = t 2 −  4 + friend  t + 4 = t 2 −  4 +
t + 4
a
2.4 m/s 2 

b


0 = t 2 − 6.92t + 4
3. Now use the quadratic formula:
t=
+6.92 ± 6.922 − 4 (1)( 4 )
2
= 6.3, 0.64 s
4. We choose the larger root because the time must be greater than 2.0 s, the time at which the
bicyclist began pursuing his friend. The bicyclist will overtake his friend 6.3 s after his friend
passes him.
5. (b) Find x:
x = v0 t = ( 3.5 m/s )( 6.3 s ) = 22 m
6. (c) Find v. Keep in mind that v0 = 0 v = 0 + a ( t − 2 ) = ( 2.4 m/s 2 ) ( 6.3 − 2.0 s ) = 10 m/s
and that the bicyclist doesn’t begin
accelerating until two seconds have
elapsed.:
Insight: Even a smaller acceleration would allow the bicyclist to catch up to the friend, because the
speed is always increasing for any nonzero acceleration, and so the bicyclist’s speed would
eventually exceed the friend’s speed and the two would meet.
38. Picture the Problem: The velocity-versus-time plots
of the car and the truck are shown at right. The car
begins with a positive position and a negative velocity,
so it must be represented by the lower line. The truck
begins with a negative position and a positive velocity,
so it is represented by the upper line.
Strategy: The distances travelled by the car and the
truck are equal to the areas under their velocity-versustime plots. We can determine the distances travelled
from the plots and use the known initial positions to
find the final positions and the final separation.
Solution: 1. Find the final
position of the truck:
xtruck = x0,truck + ∆xtruck = ( −35 m ) + 12 ( 2.5 − 0 s )(10 m/s ) = −22.5 m
2. Find the final position of xcar = x0,car + ∆xcar = (15 m ) + 12 ( 3.5 − 0 s )( −15 m/s ) = −11.25 m
the car:
3. Now find the separation: xcar − xtruck = ( −11.25 m ) − ( −22.5 m ) = 11.3 m
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Tutorial Solutions
Insight: The velocity-versus-time graph is a rich source of information. Besides velocity and time
information, you can determine acceleration from the slope of the graph and distance travelled from
the area under the graph. In this case, we can see the acceleration of the car (4.29 m/s2) has a
greater magnitude than the acceleration of the truck (−4.00 m/s2).
39. Picture the Problem: The cart slides
down the inclined track, each time
travelling a distance of 1.00 m along the
track.
1.00 m
θ
Strategy: The distance travelled by the cart is given by the constant-acceleration equation of
motion for position as a function of time, where x0 = v0 = 0 . The magnitude of the acceleration
can thus be determined from the given distance travelled and the time elapsed in each case. We can
then make the comparison with a = g sin θ .
Solution: 1. Find the
acceleration:
x = 0 + 0 + 12 at 2 ⇒ a =
2x
t2
a = g sin θ
2.00 m
= 1.71 m/s 2
2. Now find the values for θ a =
2
(1.08 s )
= 10.0°:
a = ( 9.81 m/s 2 ) sin10.0o = 1.70 m/s 2
2.00 m
= 3.37 m/s 2
3. Now find the values for θ a =
2
( 0.770 s )
= 20.0°:
a = ( 9.81 m/s 2 ) sin 20.0o = 3.35 m/s 2
2.00 m
= 4.88 m/s 2
4. Now find the values for θ a =
2
( 0.640 s )
= 30.0°:
a = ( 9.81 m/s 2 ) sin10.0o = 4.91 m/s 2
Insight: We see very good agreement between the formula a = g sin θ and the measured
acceleration. The experimental accuracy gets more and more difficult to control as the angle gets
bigger because the elapsed times become very small and more difficult to measure accurately. For
this reason Galileo’s experimental approach (rolling balls down an incline with a small angle) gave
him an opportunity to make accurate observations about free fall without fancy electronic
equipment.
40. Picture the Problem: The apple falls straight downward under the influence of gravity.
Strategy: The distance of the fall is estimated to be about 3.0 m (about 10 ft). Then use the timefree equation of motion to estimate the speed of the apple.
Solution: 1. Solve for v,
assuming the apple drops
from rest ( v0 = 0 ):
2. Let a = g and calculate v:
v = 0 + 2 a ∆x
v = 2 ( 9.81 m/s 2 ) ( 3.0 m ) = 7.7 m/s = 17 mi/h
Insight: Newton supposedly then reasoned that the same force that made the apple fall also keeps
the Moon in orbit around the Earth, leading to his universal law of gravity. One lesson we might
learn here is—wear a helmet when sitting under an apple tree!
41. Picture the Problem: The lava bomb travels upward, slowing down under the influence of gravity,
coming to rest momentarily before falling downward.
Strategy: Since the acceleration of gravity is known, the constant acceleration equation of motion
can be used to find the speed and velocity as a function of time. Let upward be the positive
direction.
Solution: 1. (a) Set a = −g:
v = v0 − gt = 28 m/s − ( 9.81 m/s 2 ) ( 2.0 s ) = 8.4 m/s
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Tutorial Solutions
v = v0 − gt = 28 m/s − ( 9.81 m/s 2 ) ( 3.0 s ) = −1.4 m/s
2. (b) Set a = −g:
3. We interpret the answer to (b) as a speed of 1.4 m/s but a velocity of −1.4 m/s, where the
negative sign means it is travelling downward.
Insight: We can see the lava bomb must have reached its peak between 2.0 and 3.0 seconds. In
fact, it reached it at t = ( 0 − 28 m/s ) ( −9.81 m/s 2 ) = 2.9 s .
42. Picture the Problem: The material travels straight upward, slowing down under the influence of
gravity until it momentarily comes to rest at its maximum altitude.
Strategy: Since the maximum altitude is known, use the time-free equation of motion to find the
initial velocity. Let upward be the positive direction, so that a = −1.80 m/s2.
Solution: Solve for v0 , setting
v=0:
v0 = v 2 − 2a∆x = 02 − 2 ( −1.80 m/s 2 )( 2.00 ×105 m ) = 849 m/s
Insight: On Earth that speed would only hurl the material to an altitude of 37 km, as opposed to
200 km on Io. Still, that’s a very impressive initial velocity! It is equivalent to the muzzle velocity
of a bullet, and is 2.5 times the speed of sound on Earth.
43. Picture the Problem: The balls fall straight down under the influence of gravity. The first ball
falls from rest but the second ball is given an initial downward velocity.
Strategy: Since the fall distance is known in each case, use the time-free equation of motion to
predict the final velocity. Let downward be the positive direction for simplicity.
Solution: 1. (a) The speed increases linearly with time but nonlinearly with distance. Since the
first ball has a lower initial velocity and hence a lower average velocity, it spends more time in the
air. The first (dropped) ball will therefore experience a larger increase in speed.
2. (b) First ball: Solve for
v, setting v0 = 0 :
v = 02 + 2 g ∆x = 2 ( 9.81 m/s 2 ) ( 32.5 m ) = 25.3 m/s
3. Second ball: Solve for v:
v = v02 + 2 g ∆x =
4. Compare the ∆v values:
∆v1 = 25.3 − 0 m/s = 25.3 m/s for the first ball and
(11.0 m/s )
2
+ 2 ( 9.81 m/s 2 ) ( 32.5 m ) = 27.5 m/s
∆v2 = 27.5 − 11.0 m/s = 16.5 m/s for the second ball.
Insight: The second ball is certainly going faster, but its change in speed is less than the first ball.
44. Picture the Problem: The arrow rises straight upward, slowing down due to the acceleration of
gravity.
Strategy: Since the position, time, and acceleration are all known, we can use the equation of
motion for position as a function of time and acceleration to find the initial velocity v0 . The same
equation could be used to find the time required to rise to a height of 15.0 m above its launch point.
Let the launch position x0 = 0 and let upward be the positive direction.
2
2
1
x − 12 at 2 30.0 m − 2 ( −9.81 m/s ) ( 2.00 s )
Solution: 1. (a) Solve for v0 : v0 =
=
= 24.8 m/s
t
2.00 s
2. (b) Solve with x = 15.0 m:
15.0 m = ( 24.8 m/s ) t − 12 ( 9.81 m/s 2 ) t 2
0 = ( − 4.905 m/s 2 ) t 2 + ( 24.8 m/s ) t − 15.0 m
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3. Now use the quadratic
formula:
t=
Tutorial Solutions
−b ± b 2 − 4ac −24.8 ±
=
2a
( 24.8)
2
− 4 ( −4.905 )( −15.0 )
−9.81
t = 0.702 s , 4.36 s
Insight: The second root of the solution to part (b) corresponds to the time when the arrow, after
rising to its maximum height, falls back to a position 15.0 m above the launch point.
45. Picture the Problem: The book accelerates straight downward and hits the floor of the elevator.
Strategy: The constant speed motion of the elevator does not affect the acceleration of the book.
From the perspective of an observer outside the elevator, both the book and the floor have an
initial downward velocity of 3.0 m/s. Therefore from your perspective the motion of the book is
no different than if the elevator were at rest. Use the position as a function of time and
acceleration equation to find the time, setting v0 = 0 and letting downward be the positive
direction. Then use velocity as a function of time to find the speed of the book when it lands.
2 (1.2 m )
2x
=
= 0.49 s
g
9.81 m/s 2
Solution: 1. (a) Solve for t:
t=
2. (b) Find v:
v = v0 + gt = 0 + ( 9.81 m/s 2 ) ( 0.49 s ) = 4.8 m/s
Insight: The speed in part (b) is relative to you. Relative to the ground the speed of the book is
4.8 + 3.0 = 7.8 m/s.
46. Picture the Problem: The rocket rises straight upward, accelerating over a distance of 26 m and
then slowing down and coming to rest at some altitude higher than 26 m.
Strategy: Use the given acceleration and distance and the time-free equation of motion to find the
velocity of the rocket at the end of its acceleration phase, when its altitude is 26 m. Use that as the
initial velocity of the free fall stage in order to find the maximum altitude. Then apply the same
equation once again to find the velocity of the rocket when it returns to the ground. The given and
calculated positions at various stages of the flight can then be used to find the elapsed time in each
stage and the total time of flight.
Solution: 1. (a) Find the
velocity at the end of the
boost phase
vboost = v02 + 2 g ∆x = 02 + 2 (12 m/s 2 ) ( 26 m ) = 25 m/s
v2
2
2. Find the height change
02 = vboost
− 2 g ∆xboost ⇒ ∆xboost = boost
2g
during the boost phase and a
final speed of zero:
2
3. Now find the overall
maximum height:
hmax = 26 m +
2
( 25 m/s )
vboost
= 26 m +
= 26 + 32 m = 58 m
2g
2 ( 9.81 m/s 2 )
2
2
4. (b) Solve again between v = vboost − 2 g ∆x
the end of the boost phase
2
and the point where it hits the v = vboost − 2 g ∆x =
ground:
5. (c) First find the duration
of the boost phase. Use the
known positions:
tboost =
6. Now find the time for the
t =
rocket to reach its maximum up
altitude from the end of the
boost phase:
1
2
∆xboost
=
( v0 + vboost )
∆xup
1
2
(v
boost
+ vtop )
=
( 25 m/s )
1
2
2
− 2 ( 9.81 m/s 2 ) ( −26 m ) = 34 m/s
26 m
= 2.1 s
( 0 + 25 m/s )
32 m
1
2
( 25 + 0 m/s )
= 2.6 s
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7. Now find the time for the
tdown =
rocket to fall back to the
ground:
8. Sum the times to find the
time of flight:
Tutorial Solutions
∆xdown
1
2
( vtop + vground )
=
58 m
1
2
( 0 + 34 m/s )
= 3.4 s
ttotal = tboost + tup + tdown = 2.1 + 2.6 + 3.4 s = 8.1 s
Insight: Notice how knowledge of the initial and final velocities in each stage, and the distance
travelled in each stage, allowed the calculation of the elapsed times using the relatively simple, as
opposed to the quadratic equation. Learning to recognize the easiest route to the answer is an
important skill to obtain.
47. Picture the Problem: Nut A is dropped
from rest. When it has fallen 2.5 m, nut
B is thrown downward with an initial
speed vB,0. Both nuts land at the same
time after falling 10.0 m.
Nut B
thrown
vB,0 = ?
branch
2.5 m
Nut A
Strategy: First find the time it takes for
nut A to fall 2.5 m using the equation of
motion for position as a function of time
and acceleration. Also find the time
required for nut A to fall the entire 10.0
m. Subtract the first time from the
second to find the time interval over
which nut B must reach the ground in
order to land at the same instant as nut
A. Then find the initial velocity vB,0
required in order for nut B to reach the
ground in that time.
Both land
ground
2 ( 2.5 m )
2 ∆x
=
= 0.714 s
g
9.81 m/s 2
Solution: 1. Find the time it takes
for nut A to fall 2.5 m by solving
for t and setting vA,0 = 0.
tA,1 =
2. Find the time it takes for nut A
to fall the entire
10.0 m:
tA,total =
3. Subtract the times to find the
time over which nut B must reach
the ground:
4. Solve for vB,0:
10.0 m
2∆x
=
g
2 (10.0 m )
9.81 m/s 2
= 1.428 s
tB,total = tA,total − tA,1 = 1.428 − 0.714 s = 0.714 s
vB,0 =
2
∆x − 12 gtB,total
tB,total
=
10.0 m − 12 ( 9.81 m/s 2 ) ( 0.714 s )
2
0.714 s
vB,0 = 10.5 m/s ⇒ 11 m/s
Insight: In this problem we kept an additional significant figure than is warranted in steps 1, 2, and
3 in an attempt to get a more accurate answer in step 4. However, if you choose not to do so,
differences in rounding will lead to an answer of 10 m/s. The specified 2.5 m drop distance for nut
A limits the answer to two significant digits, and since the answer is right between 10 and 11 m/s,
it could correctly go either way.
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Tutorial Solutions
Solutions to Tutorial 2
y
1. Picture the Problem: The vector
direction is measured anticlockwise
from the +x axis.
r
r
r
r
35.0°
Strategy: In each case find the
vector components.
Solution: 1. (a) Find the x and y
components:
y
65.0°
x
(a)
x
(b)
rx = r cos θ = ( 75 m ) cos 35.0° = 61 m
ry = r sin θ = ( 75 m ) sin 35.0° = 43 m
2. (b) Find the x and y components: rx = r cos θ = ( 75 m ) cos 65.0° = 32 m
ry = r sin θ = ( 75 m ) sin 65.0° = 68 m
Insight: Resolving vectors into their components is an important skill for solving physics
problems.
2.
Picture the Problem: The water molecule forms a triangle with the positions
of the oxygen and hydrogen nuclei as shown.
Strategy: Break the triangle up into two right triangles and use the sine
function to find the distance between the hydrogen nuclei. The angle θ is half
of the 104.5° bond angle, or θ = 52.25°.
d
Solution: 1. Use the sine function to find the
sin θ =
distance d:
0.96 Å
2. The distance between hydrogen nuclei is
2d:
2d = 2 ( 0.96 Å ) sin ( 52.25° ) = 1.5 Å
Insight: Identifying right triangles and manipulating the trigonometric functions are important
skills to learn when solving physics problems.
3.
Picture the Problem: The given vector components correspond to the vector
r
r as drawn at right.
y
Strategy: Use the inverse tangent function to determine the angle θ. Then
r
use the Pythagorean Theorem to determine the magnitude of r .
Solution: 1. (a) Use the inverse tangent
function to find the distance angle θ :
14 m
r
r
 −9.5 
 = −34° or
 14 
34° below the +x axis
θ = tan −1 
2. (b) Use the Pythagorean Theorem to
r
determine the magnitude of r :
r = rx2 + ry2 =
3. (c) If both rx and ry are doubled, the
direction will remain the same but the
magnitude will double:
θ = tan −1 
(14 m )
2
+ ( −9.5 m )
x
−9.5
m
2
r = 17 m
 −9.5 × 2 
 = −34°
 14 × 2 
r=
( 28 m )
2
2
+ ( −19 m ) = 34 m
Insight: Any vector can be resolved into two components. The ability to convert a vector to and
from its components is an essential skill for solving many physics problems.
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4.
Tutorial Solutions
Picture the Problem: The given vector components correspond to the
r
vector r as drawn at right.
XII
Strategy: Determine the angle θ from our knowledge of analogue clocks.
The given component rx together with the angle θ will allow us to calculate
the length of r and the component ry .
Solution: 1. (a) Find the angle θ :
1
θ = × 360° = 30°
12
2. Find the length of r:
rx = r cos θ ⇒ r =
I
y
II
r
r
θ
3.0 cm
x
III
rx
3.0 cm
=
= 3.5 cm
cos θ cos 30o
3. (b) The components rx and ry are only equal when θ =45°. Since in this case θ =30°, the
component ry will be
less than rx or 3.0 cm.
4. (c) Find ry :
ry = r sin θ = ( 3.5 cm ) sin 30o = 1.7 cm
Insight: Any vector can be resolved into two components. The ability to convert a vector to and
from its components is an essential skill for solving many physics problems.
5.
r
r
Picture the Problem: The two vectors A (length 50 units) and B (length 120 y
units) are drawn at right.
r
Strategy: Resolve B into its x and y components to answer the questions.
Bx = (120 units ) cos 70° = 41 units
Solution: 1. (a) Find Bx:
r
2. Since the vector A points entirely in the x direction, we can see that Ax = 50
r
units and that vector A has the greater x component.
3. (b) Find By:
r
A
x
70°
r
B
Bx = (120 units ) sin 70° = 113 units
r
r
4. The vector A has no y component, so it is clear that vector B has the greater y component.
Insight: Any vector can be resolved into two components. The ability to convert a vector to and
from its components is an essential skill for solving many physics problems.
6.
Picture the Problem: The four possible locations of
the treasure are labelled A, B, C, and D in the figure at
right. The position vector for location A is also drawn.
North is up and east is to the right.
A
C
25.0 m
r
A
B
Strategy: Use the vector components to find the
magnitude and direction of each vector.
25.0 m
θA 12.0 m
5.00 m
D
palm tree
r
Solution: 1. Find the magnitude of A :
A=
r
2. Find the direction from north of A :
θ A = tan −1 
r
3. Find the magnitude of B :
B=
r
4. Find the direction from north of B :
θ B = tan −1 
r
5. Find the magnitude of C :
C=
( 25.0 m )
2
2
+ (12.0 + 5.00 m ) = 30.2 m
 25.0 m 
 = 55.8° west of north
 12.0 + 5.00 m 
( 25.0 m )
2
2
+ (12.0 − 5.00 m ) = 26.0 m
 25.0 m 
 = 74.4° west of north
 12.0 − 5.00 m 
2
( 25.0 m ) + (12.0 + 5.00 m )
2
= 30.2 m
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Foundation Engineering Science – Dr. Daniel Nankoo
Tutorial Solutions
 25.0 m 
 = 55.8° east of north
 12.0 + 5.00 m 
r
6. Find the direction from north of C :
θC = tan −1 
r
7. Find the magnitude of D :
D=
( 25.0 m )
2
2
+ (12.0 − 5.00 m ) = 26.0 m
 25.0 m 
 = 74.4° east of north
 12.0 − 5.00 m 
Insight: If you ever find a treasure map like this one, you’ll be glad you mastered vectors in
physics!
r
8. Find the direction from north of D :
7.
θ D = tan −1 
Picture the Problem: The whale dives
along a straight line tilted 20.0° below
horizontal for 150 m as shown in the
figure.
Strategy: Resolve the whale’s
displacement vector into horizontal and
vertical components in order to find its
depth ry and its horizontal travel distance
rx.
Solution: 1. (a) The depth is given by ry: ry = r sin θ = (150 m ) sin ( 20.0° ) = 51 m
2. (b) The horizontal travel distance is
given by rx:
rx = r cos θ = (150 m ) cos ( 20.0° ) = 140 m = 0.14 km
Insight: Note that both answers are limited to two significant figures, because although “20.0°”
has three, “150 m” has only two significant figures.
8.
r
r
Picture the Problem: The two vectors A (length 50.0 m) and B
(length 70.0 m) are drawn at right.
y r
B
r
A
r
C
50.0°
20.0°
x
r
r
Strategy: Add vectors A and B using the vector component method.
Solution: 1. (a) A sketch of the vectors and their sum is shown at right.
2. (b) Add the x
components:
Cx = Ax + Bx = ( 50.0 m ) cos ( −20.0° ) + ( 70.0 m ) cos ( 50.0° ) = 92.0 m
3. Add the y components: C y = Ay + By = ( 50.0 m ) sin ( −20.0° ) + ( 70.0 m ) sin ( 50.0° ) = 36.5 m
4. Find the magnitude of C = C x2 + C y2 =
C:
( 92.0 m )
2
2
+ ( 36.5 m ) = 99.0 m
C 
36.5 m 
5. Find the direction of C: θC = tan −1  y  = tan −1 
 = 21.7°
C
 92.0 m 
 x
Insight: Resolving vectors into components takes a little bit of extra effort, but you can get much
more accurate answers using this approach than by using a ruler and protractor to add the vectors
graphically.
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Foundation Engineering Science – Dr. Daniel Nankoo
9.
Tutorial Solutions
Picture the Problem: The vectors involved in the
problem are depicted at right. The control tower (CT)
is at the origin and north is up in the diagram.
r
r
Strategy: Subtract vector B from A using the
vector component method.
Solution: 1. (a) A sketch of the vectors and their
difference is shown at right.
2. (b) Subtract
the x
components:
Dx = Ax − Bx = ( 220 km ) cos (180 − 32° ) − (140 km ) cos ( 90 − 65° ) = −310 km
3. Subtract the y Dy = Ay − By = ( 220 km ) sin (180 − 32° ) − (140 km ) sin ( 90 − 65° ) = 57 km
components:
4. Find the
D = Dx2 + Dy2 =
magnitude of D:
5. Find the
direction of D:
( 310 km )
2
2
+ ( 57 km ) = 320 km = 3.2 × 105 m
 Dy 
−1  57 km 
 = tan 
 = −10° + 180° = 170° or 10° north of west
 −310 km 
 Dx 
θ D = tan −1 
Insight: Resolving vectors into components takes a little bit of extra effort, but you can get much
more accurate answers using this approach than by adding the vectors graphically. Notice,
however, that when your calculator returns −10° as the angle in step 5, you must have a picture of
the vectors in your head (or on paper) to correctly determine the direction.
10. Picture the Problem: The vectors involved in the problem are
depicted at right.
r
Strategy: Deduce the x and y components of B from the
r
r
information given about A and C . Use the known components
r
to estimate the length and direction of B as well as calculate
them precisely.
Solution: 1. (a) A sketch of the vectors is shown at right.
r
2. (b) The vector B must have an x component of −75 m so that
r
when it is added to A the x components will cancel out. It must
r
also have a y component of 95 m because that is the length of C
r
r
and A has no y component to contribute. Therefore B must be
r
r
longer than either A or C and it must have an angle of greater
than 90°. I estimate that its length is about 120 m and that it
points at about 130°.
3. (c) Using the known components
r
B = (−75 m) 2 + (95 m)2 = 121 m
of B
we can find its magnitude:
r
 95 m 
θ B = tan −1 
4. Find the direction of B :
 = −52° + 180° = 128°
 –75 m 
r
Insight: Here the length and direction of B are determined by its x and y components, which are
r
r
determined from A and C . Learning to manipulate vector components will be a useful skill
when tackling many physics problems.
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Tutorial Solutions
11. Picture the Problem: The vectors involved in the
problem are depicted at right.
y
r r
B−A
Strategy: Use the vector component method of
addition and subtraction to determine the
r
r
components of each combination of A and B .
Once the components are known, the length and
direction of each combination can be determined
fairly easily.
Solution: 1. (a) Determine the
r r
components of A + B :
r r
2. Find the magnitude of A + B :
θ Ar − Br
10
r r
A+B
θ Ar + Br
r
A
r r
A−B
r
B
θ Br − Ar
O
5
x
r r
A + B = ( –5 ) yˆ + (10 ) xˆ = 10xˆ – 5yˆ
r r
A+B =
2
(10 ) + ( −5)
2
= 11 units
r r
 −5 
3. Determine the direction of A + B ,
θ r r = tan −1   = −27° or 333°
measured anticlockwise from the positive A + B
 10 
x axis.
r r
4. (b) Determine the components of
A − B = ( –5 ) yˆ − (10 ) xˆ = −10xˆ – 5yˆ
r r
A−B :
r r
r r
2
2
5. Find the magnitude of A − B :
A − B = ( −10 ) + ( −5 ) = 11 units
r r
 −5 
6. Determine the direction of A − B ,
θ r r = tan −1 
 = 27° + 180° = 207°
measured anticlockwise from the positive A − B
 −10 
x axis.
r r
7. (c) Determine the components of
B − A = (10 ) xˆ − ( –5 ) yˆ − = 10xˆ + 5yˆ
r r
B−A:
r r
r r
2
2
8. Find the magnitude of B − A :
B − A = (10 ) + ( 5 ) = 11 units
r r
 5
9. Determine the direction of B − A ,
θ r r = tan −1   = 27°
measured anticlockwise from the positive B − A
 10 
x axis.
r
r
Insight: This problem is simplified by the fact that A and B have only one component each, but
a similar approach will work even with more complicated vectors. Notice that you must have a
picture of the vectors in your head (or on paper) in order to correctly interpret the directions in
steps 3, 6, and 9.
12. Picture the Problem: The vectors involved in
the problem are depicted at right.
y
r r r
A+B+C
Strategy: Add the vectors using the component
method in order to find the components of the
vector sum. Use the components to find the
magnitude and the direction of the vector sum.
r
B
30°
θ
r
A
r r r
Solution: 1. (a) Make estimates A + B + C ≈ 20 m
from the drawing:
r
C
x
45°
θ ≈ 1.5°
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Foundation Engineering Science – Dr. Daniel Nankoo
Tutorial Solutions
r r r
A + B + C = 0 + ( 20.0 m ) cos 45° + ( 7.0 m ) cos ( −30° )  xˆ +
2. (b) Add the vector
components:
( −10.0 m ) + ( 20.0 m ) sin 45° + ( 7.0 m ) sin ( −30° )  yˆ
r r r 
A + B + C = ( 20.2 m ) xˆ + ( 0.64 m ) yˆ
3. Use the components to find
the magnitude:
4. Use the components to find
the angle:
r r r
A+B+C =
( 20.2 m )2 + ( 0.64 m )2
= 20.2 m
 0.64 m 
 = 1.8°
 20.2 m 
θ = tan −1 
Insight: Resolving vectors into components takes a little bit of extra effort, but you can get much
more accurate answers using this approach than by adding the vectors graphically. Notice,
however, that when your calculator returns the angle of 1.8° in step 4, you must have a picture of
the vectors in your head (or on paper) to correctly determine the direction.
13. Picture the Problem: The vector involved in the problem is depicted at
right.
y
Strategy: Determine the x and y components of and then express them in
terms of the unit vectors.
54 m
rx = ( 54 m ) cos ( − 42 ) = 40 m
Solution: 1. Find the x and y
r
components of ∆r :
x
42°
o
r
∆r
ry = ( 54 m ) sin ( − 42o ) = −36 m
r
2. Now express ∆r in terms of the unit ∆rr = ( 40 m ) xˆ + ( −36 m ) yˆ
vectors:
r
Insight: In general, an arbitrary two-dimensional vector A can always be written as the sum of a
vector component in the x direction and a vector component in the y direction.
r
14. Picture the Problem: The vector A has a length of 6.1 m and points in the negative x direction.
Strategy: In order to multiply a vector by a scalar, you need only multiply each component of the
vector by the same scalar.
r
Solution: 1. (a) Multiply each
A = ( − 6.1 m ) xˆ
r
r
component of A by −3.7:
−3.7 A = ( −3.7 )( − 6.1 m )  xˆ = ( 23 m ) xˆ so Ax = 23 m
r
2. (b) Since A has only one component, its magnitude is simply 23 m.
Insight: Multiplying both components of a vector by a scalar will change the length of the vector
but not its direction.
15. Picture the Problem: The vectors involved in the problem are
depicted at right.
y
Strategy: Determine the lengths and directions of the various
vectors by using their x and y components.
 −2.0 m 
 = –22°
 5.0 m 
Solution: 1. (a) Find
r
the direction of A
from its components:
θ Ar = tan −1 
2. Find the magnitude
r
of A :
A=
3. (b) Find the
r
direction of B from its
components:
θBr = tan −1 
r r
A+B
2.0 m
O
2.0 m
( 5.0 m )
2
θ Ar + Br
r
B
5.0 m
x
r
A
2
+ ( −2.0 m ) = 5.4 m
 5.0 m 
 = −68° + 180° = 110°
 –2.0 m 
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Foundation Engineering Science – Dr. Daniel Nankoo
Tutorial Solutions
4. Find the magnitude
r
of B :
B=
5. (c) Find the
r r
components of A + B :
r r
A + B = ( 5.0 − 2.0 m ) xˆ + ( −2.0 + 5.0 m ) yˆ = ( 3.0 m ) xˆ + ( 3.0 m ) yˆ
6. Find the direction of
r r
A + B from its
components:
θ Ar +Br = tan −1 
7. Find the magnitude
r r
of A + B :
r r
A+B =
( −2.0 m )
2
2
+ ( 5.0 m ) = 5.4 m
 3.0 m 
 = 45°
 3.0 m 
( 3.0 m )2 + ( 3.0 m )2
= 4.2 m
Insight: In the world of vectors 5.4+5.4 m can be anything between 0 m and 10.8 m, depending
upon the directions that the vectors point. In this case their sum is 4.2 m.
16. Picture the Problem: The vectors involved in the problem are
depicted at right.
y
Strategy: Determine the lengths and directions of the various
vectors by using their x and y components.
Solution: 1. (a)
Find the direction
r
of A from its
components:
θ Ar
= tan
θ Ar + Br
r r
A+B
−1  −12
m

 = –26°
 25 m 
O
12 m
2. Find the
r
magnitude of A :
A=
3. (b) Find the
r
direction of B from
its components:
θBr = tan −1 
4. Find the
r
magnitude of B :
B=
( 25 m )2 + ( –12 m )2
r
A
r
B
x
25 m
= 28 m
 15 m 
 = 82°
 2.0 m 
( 2.0 m )2 + (15 m )2
= 15 m
5. (c) Find the
components of
r r
A+B:
r r
A + B = ( 25 + 2.0 m ) xˆ + ( −12 + 15 m ) yˆ = ( 27 m ) xˆ + ( 3.0 m ) yˆ
6. Find the direction
r r
of A + B from its
components:
θ Ar +Br = tan −1 
7. Find the
magnitude of
r r
A+B:
r r
A+B =
 3.0 m 
 = 6.3°
 27 m 
( 27 m )2 + ( 3.0 m )2
= 27 m
Insight: In the world of vectors 28 + 15 m can be anything between 13 m and 43 m, depending
upon the directions that the vectors point. In this case their sum is 27 m.
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Foundation Engineering Science – Dr. Daniel Nankoo
Tutorial Solutions
17 Picture the Problem: The vectors involved in the problem
. are depicted at right.
Strategy:
Use the information given in the figure to
determine the components of each vector.
r
Solution A = (1.5 m ) cos ( 40° ) xˆ + (1.5 m ) sin ( 40° ) yˆ
r
: 1.
A = (1.1 m ) xˆ + ( 0.96 m ) yˆ
2.
3.
4.
r
B = ( 2.0 m ) cos ( −19° ) − ( 2.0 m ) sin ( −19° )
r
B = (1.9 m ) xˆ + ( − 0.65 m ) yˆ
r
C = (1.0 m ) cos (180 − 25° ) xˆ + (1.0 m ) sin (180 − 25° ) yˆ
r
C = ( − 0.91 m ) xˆ + ( 0.42 m ) yˆ
r
D = 0 xˆ + (1.5 m ) yˆ
Insight: Any vector can be resolved into two components. The ability to convert a vector to and
from its components is an essential skill for solving many physics problems.
18. Picture the Problem: The vectors involved in the
problem are depicted at right.
Strategy:
Use the information given in the figure
r r
r
to determine the components of vectors A, B, and C .
Then add the components.
Solution: 1.
Add the x
component of
each vector:
Ax = (1.5 m ) cos ( 40° ) = 1.1 m
Bx = ( 2.0 m ) cos ( −19° ) = 1.9 m
Cx = (1.0 m ) cos (180 − 25° ) = − 0.91 m
r r r
A + B + C = 2.1 m
(
2. Add the y
component of
each vector:
)
x
Ay = (1.5 m ) sin ( 40° ) = 0.96 m
By = ( 2.0 m ) sin ( −19° ) = − 0.65 m
C y = (1.0 m ) sin (180 − 25° ) = 0.42 m
r r r
A + B + C = 0.74 m
(
)
y
r r r
A + B + C = ( 2.1 m ) xˆ + ( 0.74 m ) yˆ
3. Express the sum in unit
vector notation:
Insight: In this problem the vector component method of addition is much quicker than the
graphical method.
19. Picture the Problem: The displacement vectors are depicted at
right. North is in the y direction and east is in the x direction.
Strategy: Sum the components of the vectors in order to
r r
determine A + B . Multiply that vector by −1 in order to reverse
its direction. Then find the magnitude and direction of the
reversed vector.
r r
Solution: 1. (a) Add the two A + B = ( −72 m ) xˆ + (120 m ) yˆ
displacement vectors:
r r
2. Multiply by −1 in order to − A
+ B = ( 72 m ) xˆ + ( −120 m ) yˆ
(
)
r
B
y
120 m
θ
r r
− A+B
(
−72 m
)
r
A
O
x
25 of 65
Foundation Engineering Science – Dr. Daniel Nankoo
Tutorial Solutions
reverse the
direction of the net
displacement and bring the
cat back home:
r r
3. Find the magnitude of the − A + B =
desired displacement:
(
4. Find the direction of the
desired displacement:
)
( 72 m )
2
2
+ (120 m ) = 140 m
 −120 m 
θ = tan −1 
 = −59° = 59° south of east
 72 m 
5. (b) Vector addition is independent of the order in which the addition is accomplished. The initial
displacement is the same, so there is no change in the displacement for the homeward part of the
trip.
Insight: In this problem we could claim the cat’s initial displacement is a single vector with the
given components. The answers wouldn’t change, but it would simplify the solution a little bit.
20. Picture the Problem: You travel due west for 130 s at 27 m/s then due south at 14 m/s for 62 s.
Strategy: Find the components of the displacement vector. Once the components are known the
magnitude and direction can be easily found. Let north be the positive y direction and east be the
positive x direction.
Solution: 1. Find the westward
displacement:
rx = vx t = ( −27 m/s )(130 s ) = −3500 m
2. Find the southward
displacement:
ry = v y t = ( −14 m/s )( 62 s ) = −870 m
3. Find the direction of the
displacement:
θ = tan −1 
 ry
 rx

−1  −870 m 
 = tan 
 = 14° + 180° = 194°
 −3500 m 

or 14° south of west
4. Find the magnitude of the
displacement:
r=
( −3500 m )
2
2
+ ( −870 m ) = 3600 m = 3.6 km
Insight: The 14° refers to the angle below the negative x axis (west) because the argument of the
inverse tangent function is ry rx , or south divided by west.
21. Picture the Problem: The jogger runs at 3.25 m/s in a direction 30.0° above the positive x axis.
Strategy: Find the components of the velocity vector.
Solution: 1. (a) Find the x component of
r
v:
r
2. Find the y component of v :
vx = ( 3.25 m/s ) cos ( 30.0° ) = 2.81 m/s
v y = ( 3.25 m/s ) sin ( 30.0° ) = 1.63 m/s
3. (b) If the jogger’s speed is halved, the direction will remain unchanged but the x and y
components will be halved.
Insight: In this case the angle of 30.0° corresponds to the standard angle θ.
22. Picture the Problem: The ball rises straight upward, momentarily comes to rest, and then falls
straight downward.
Strategy: After it leaves your hand the only acceleration of the ball is due to gravity, so we expect
the answer to be 9.81 m/s2. To calculate the acceleration we need only consider the initial and
final velocities and the time elapsed. Because of the symmetry of the situation, the final velocity
downward will have the same magnitude as the initial velocity upward. Apply the relevant
equation, taking upward to be the positive direction.
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Foundation Engineering Science – Dr. Daniel Nankoo
Tutorial Solutions
r
r
v f − v i ( −4.5 m/s ) yˆ − ( 4.5 m/s ) yˆ
r
Solution: Apply equation: a av =
=
= −9.8 m/s 2 yˆ
∆t
0.92 s
Insight: A uniform acceleration will produce a symmetric trajectory, with the time to rise to the
peak of flight equalling the time to fall back down, and with equal initial and final speeds.
23. Picture the Problem: The skateboarder rolls down the ramp that is inclined 20.0° above the
horizontal.
Strategy: To calculate the acceleration we need only consider the initial and final velocities and
the time elapsed. Apply the relevant equation, taking the direction down the ramp to be the
positive direction.
Solution: 1. Apply equation: aav =
v f − vi
∆t
=
(10.0 m/s ) − ( 0 m/s )
3.00 s
= 3.33 m/s 2
g sin θ = ( 9.81 m/s 2 ) sin ( 20.0° ) = 3.36 m/s 2
2. Compare with g sin θ :
Insight: The two are equal to within rounding errors. Or perhaps there was a small amount of
friction between the skateboard wheels and the ramp.
24. Picture the Problem: The initial and final displacement vectors
are depicted at right.
Strategy: Use the given formulae to determine the components of
the initial and final positions. Then use those components to find
the displacement vector. Divide the displacement vector by the
elapsed time to find the velocity vector, and then determine its
magnitude and direction.
r
8
Solution: 1. (a) Find the ri = ( 3.84 × 10 m ) {cos [ 0] xˆ + sin [ 0] yˆ }
initial position vector:
= ( 3.84 × 108 m ) xˆ
2. Find the arguments of the sine ωt = ( 2.46 × 10−6 radians/s ) ( 7.38 d × 86400 s/d )
and cosine functions for t = 7.38
= 1.57 radians
days. Let ω = 2.46 ×10−6 radians/s :
3. Find the final position vector:
r
rf = ( 3.84 × 108 m ) {cos [ω t ] xˆ + sin [ω t ] yˆ }
= ( 3.84 × 108 m ) {cos [1.57 radians ] xˆ + sin [1.57 radians] yˆ }
r
rf = ( 3.06 × 105 m ) xˆ + ( 3.84 ×108 m ) yˆ ≅ ( 3.84 ×108 m ) yˆ
4. Find the displacement vector:
r
5. Find the vector v av :
r r r
∆r = rf − ri = ( 3.84 × 108 m ) yˆ − ( 3.84 × 108 m ) xˆ
8
8
r
∆r ( −3.84 × 10 m ) xˆ + ( 3.84 × 10 m ) yˆ
r
v av =
=
∆t
7.38 d × 86400 s/d
r
v av = ( − 602 m/s ) xˆ + ( 602 m/s ) yˆ
r
6. Find the magnitude of v av :
vav =
r
7. Find the direction of v av :
θ = tan −1 
( − 602 m/s )
2
2
+ ( 602 m/s ) = 852 m/s
 vav, y 
−1  602 m/s 
 = tan 
 = − 45° + 180° = 135°
v
 − 602 m/s 
av,
x


8. (b) The instantaneous speed of the Moon is greater than the average velocity because the distance
travelled is greater than the displacement in this case.
Insight: If the Moon had completed an entire orbit, instead of just one-quarter of an orbit, its
displacement and its average velocity would have been zero. Its speed remains constant, however,
at about 947 m/s using the data given in this problem. The given data correspond to a coordinate
system where the x direction always points towards the centre of the Sun even as the Earth orbits the
Sun.
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25 Picture the Problem: The initial and final velocity vectors are depicted at
. right.
Strategy: Use the given formulae to determine the components of the
initial and final velocities. Then use those components to find the change
in velocity vector. Divide the change in velocity vector by the elapsed
time to find the acceleration vector, and then determine its magnitude and
direction.
Solution: 1. (a) Find
the initial velocity
vector:
Tutorial Solutions
y
r
∆v
r
vf
r
v i = ( 945 m/s ) {− sin [ 0] xˆ + cos [ 0] yˆ }
945 m/s
r
vi
x
O
= ( 945 m/s ) yˆ
2. Find the arguments ωt = ( 2.46 × 10−6 radians/s ) ( 0.100 d × 86400 s/d )
of the sine and cosine
= 0.0213 radians
functions for t = 0.100
days. Let
ω = 2.46 ×10−6 radians/s
:
r
v f = ( 945 m/s ) {− sin [ωt ] xˆ + cos [ωt ] yˆ }
3. Find the final
position vector:
= ( 945 m/s ) {− sin [ 0.0213 radians ] xˆ + cos [ 0.0213 radians ] yˆ }
r
v f = ( −20.1 m/s ) xˆ + ( 945 m/s ) yˆ
4. Find the
acceleration vector:
r r
r v − v i ( −20.1 m/s ) xˆ − ( 945 m/s ) yˆ  − ( 945 m/s ) yˆ
=
= ( − 0.00233 m/s 2 ) x
a= f
∆t
0.100 d × 86400 s/d
5. (b) Find the
ωt = ( 2.46 ×10−6 radians/s ) ( 0.0100 d × 86400 s/d )
arguments of the sine
= 0.00213 radians
and cosine functions
for t = 0.0100 days:
r
v f = ( 945 m/s ) {− sin [ωt ] xˆ + cos [ωt ] yˆ }
6. Find the final
position vector:
= ( 945 m/s ) {− sin [ 0.00213 radians ] xˆ + cos [ 0.00213 radians ] yˆ }
r
v f = ( −2.01 m/s ) xˆ + ( 945 m/s ) yˆ
7. Find the
acceleration vector:
r r
r v − v i ( −2.01 m/s ) xˆ − ( 945 m/s ) yˆ  − ( 945 m/s ) yˆ
a= f
=
= ( − 0.00233 m/s 2 ) x
∆t
0.0100 d × 86400 s/d
Insight: The two answers ended up being the same because both time intervals are fairly small. If
r
instead we had examined an interval of 1.00 days there would have been a −yˆ component of ∆v and
a slightly different acceleration. The given data in this problem correspond to a coordinate system
where the x direction always points towards the centre of the Sun even as the Earth orbits the Sun.
26. Picture the Problem: The vectors involved in this problem are
depicted at right.
r
Strategy: Let v pg = plane’s velocity with respect to the
r
ground, v ap = attendant’s velocity with respect to the plane,
r
and add the vectors to find v ag = attendant’s velocity with
respect to the ground.
r
v ag
r
v ap
r
v pg
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Solution: 1. Apply
equation:
Tutorial Solutions
r
r
r
v ag = v ap + v pg = ( −1.22 m/s ) xˆ + (16.5 m/s ) xˆ = (15.3 m/s ) xˆ
vag = 15.3 m/s
Insight: If the attendant were walking towards the front of the plane, her speed relative to the
ground would be 17.7 m/s, slightly faster than the airplane’s speed.
27. Picture the Problem: The vectors involved in this problem are
depicted at right.
r
Strategy: Let v yw = your velocity with respect to the walkway,
r
v wg = walkway’s velocity with respect to the ground, and add the
r
vectors to find v yg = your velocity with respect to the ground.
Then find the time it takes you to travel the 85-m distance.
r
v yw
r
v wg
r
v yg
Solution: 1.
Find your
velocity with
respect to the
walkway:
r
 ∆x 
 85 m 
v yw =   xˆ = 
 xˆ = (1.25 m/s ) xˆ
 ∆t 
 68 s 
2. Apply equation
to find your
velocity with
respect to the
ground:
r
r
r
v yg = v yw + v wg = (1.25 m/s ) xˆ + ( 2.2 m/s ) xˆ = ( 3.45 m/s ) xˆ
3. Now find the time of
travel:
t=
∆x
85 m
=
= 25 s
vyg 3.45 m/s
Insight: The moving walkway slashed your time of travel from 68 s to 25 s, a factor of 2.7! Note
that we bent the significant figures rules a little bit by not rounding vr yw to 1.3 m/s. This helped us
avoid rounding error.
28. Picture the Problem: The vectors involved in this
problem are depicted at right.
r
Strategy: Let v pg = velocity of the plane relative to
r
the ground, v pa = velocity of the plane relative to the
r
air, and v ag = velocity of the air relative to the
ground. The drawing at right depicts the vectors
r
r
r
added, v pg = v pa + v ag . Determine the angle of the
triangle from the inverse sine function.
Solution: 1. (a) Use the inverse sine
function to find θ:
 vag
v
 pa
θ = sin −1 

−1  65 km/h 
 = sin 
 = 11° west of north
 340 km/h 

2. (b) The drawing above depicts the vectors.
3. (c) If the plane reduces its speed but the wind velocity remains the same, the angle found in part
(a) should be increased in order for the plane to continue flying due north.
Insight: If the plane’s speed were to be reduced to 240 km/h, the required angle would become
16°.
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Tutorial Solutions
29. Picture the Problem: The vectors involved in this problem are
depicted at right.
r
Strategy: Let v pf = the passenger’s velocity relative to the ferry,
r
r
v pw = the passenger’s velocity relative to the water, and v fw = the
ferry’s velocity relative to the water. Apply the relevant equation
r
r
and solve for v fw . Once the components of v fw are known, its
magnitude and direction θ can be determined.
Solution: 1. Solve equation for vr pw = vr pf + vr fw
r
r
r
r
v fw :
v fw = v pw − v pf
N
r
v fw
E
θ
r
v pw
30°
r
v pf
2. Determine the components
r
of v fw :
r
v fw =  − ( 4.50 m/s ) sin 30° xˆ + ( 4.50 m/s ) cos 30° yˆ  − (1.50 m/s ) yˆ
r
v fw = ( −2.25 m/s ) xˆ + ( 2.40 m/s ) yˆ
r
3. Find the direction of v fw :
θ = tan −1 
 vfw, x 
2.25 m/s 
 = tan −1 
 = 43° west of north
 vfw, y 
2.40 m/s 



r
2
2
4. Find the magnitude of v fw : vfw = vfw, x + vfw, y =
2
( −2.25 m/s ) + ( 2.40 m/s )
2
= 3.29 m/s
r
Insight: If the person were to walk even faster with respect to the ferry, then v fw would have to be
shorter and point more in the westerly direction.
30. Picture the Problem: The situation is similar to that depicted
in the figure at right, except the boat is supposed to be a jet ski.
Strategy: Place the x-axis perpendicular to the flow of the
river, such that the river is flowing in the negative y-direction.
r
r
Let v bw = jet ski’s velocity relative to the water, v bg = jet
r
ski’s velocity relative to the ground, and v wg = water’s
r
velocity relative to the ground. Find the vector v bw , and then
determine its magnitude.
r
r
r
r
r
r
Solution: 1. Solve for v bg = v bw + v wg ⇒ v bw = v bg − v wg
r
v bw :
2. Find the
r
components of v bg :
r
v bg = ( 9.5 m/s ) cos 20.0° xˆ + ( 9.5 m/s ) sin 20.0° yˆ
3. Subtract to find
r
v bw :
r
r
r
v bw = v bg − v wg = (8.9 m/s ) xˆ + ( 3.2 m/s ) yˆ  − ( −2.8 m/s ) yˆ
= ( 8.9 m/s ) xˆ + ( 3.2 m/s ) yˆ
= ( 8.9 m/s ) xˆ + ( 6.0 m/s ) yˆ
4. Find the magnitude vbw =
r
of v bw :
( 8.9 m/s )
2
2
+ ( 6.0 m/s ) = 11 m/s
Insight: Note that the 35° angle is extraneous information for this problem. If we work backwards
r
to find the angle from the components of v bw we get θ = tan −1 ( 6.0 8.9 ) = 34° , not exactly 35° due
to rounding errors.
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Tutorial Solutions
31. Picture the Problem: The situation is depicted in the figure at
right, except the boat is supposed to be a jet ski.
Strategy: Place the x-axis perpendicular to the flow of
the river, such that the river is flowing in the negative yr
direction. Let v bw = jet ski’s velocity relative to the
r
water, v bg = jet ski’s velocity relative to the ground, and
r
v wg = water’s velocity relative to the ground. Set the y
r
r
component of v bw equal to the magnitude of v wg so that
r
they cancel, leaving only an x component of v bg . Then
determine the angle θ.
Solution: 1. (a) Set vbw, y + vwg, y = 0 and vbw, y = vbw sin θ = −vwg, y
solve for θ:
 − ( −2.8 m/s ) 
 −v

θ = sin −1  wg, y  = sin −1 
 = 13°
 vbw 
 12 m/s 
2. (b) Increasing the jet ski’s speed relative to the water will increase vbw and therefore decrease
the angle θ.
Insight: Aeroplanes must also make heading adjustments like the jet ski’s in order to fly in a
certain direction when there is a steady wind present.
32. Picture the Problem: The vectors for Jet Ski A
r
and B are depicted at right. Note that v bw has the
same magnitude for each jet ski but if you inspect
the diagram for Jet Ski B you can see that
vbw > vbg,B .
Strategy: Use the x components of the velocities
of each jet ski relative to the ground to determine
the time required for each jet ski to cross the
river.
r
v bw
Jet Ski A
Jet Ski B
r
v bg,A
r
v bw
r
v wg
r
v wg
35°
r
v bg,B
Solution: 1. (a) The time required for each jet ski
to cross the river equals the width of the river
divided by the x component of the jet ski’s
velocity relative to the ground. From the
diagrams you can see that the x component of
r
r
v bg,A equals v bw , but that the x component
r
r
of v bg,B is shorter than v bw . Therefore Jet Ski A has a higher velocity in the x direction relative
to the ground, and will cross the river first.
2. (b) Find the ratio of the times:
∆xriver
∆t A vbg, A, x vbg, B, x vbw cos 35°
=
=
=
= 0.82
∆t B ∆xriver vbg, A, x
vbw
vbg, B, x
Insight: The ratio is less than one, so ∆t A < ∆t B and Jet Ski A reaches the opposite shore first.
Remember this the next time you race jet skis across a flowing river!
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Tutorial Solutions
Solutions to Tutorial 3
1. Picture the Problem: The vector involved in this problem is
depicted at right.
r
Strategy: Separate v into x- and y-components. Let north be along
the x-axis, west along the y axis. Find the components of the
velocity in each direction and use them to find the distances
travelled.
W
r
v
42°
Solution: 1. (a) Find vx vx = ( 3.2 m/s ) sin 42° = 2.1 m/s
and v y :
v y = ( 3.2 m/s ) cos 42° = 2.4 m/s
2. Find the westward
distance travelled:
y = v y t = ( 2.4 m/s )( 25 min × 60 s/min )
3. (b) Find the
northward distance
travelled:
x = vx t = ( 2.1 m/s )( 25 min × 60 s/min )
vx
vy
N
= 3600 m = 3.6 km
= 3200 m = 3.2 km
Insight: The northward and westward motions can be considered separately. In this case they are
each described by constant velocity motion.
2. Picture the Problem: The vectors involved in this problem are
depicted at right.
N
Strategy: Let north be along the y-axis and east along the x axis.
Find the components of the velocity in each direction, and use
them to find the times elapsed.
Solution: 1. (a) Find the x
r
component of v :
vx = (1.60 m/s ) cos15.0°
= 1.55 m/s
r
v
15°
vx
vy
E
20.0 m
x
= 12.9 s
2. Find the time elapsed to travel t = =
v
1.55
m/s
x
east 20.0 m:
3. (b) Find the y component of
r
v:
v y = (1.60 m/s ) sin15.0° = 0.414 m/s
y
30.0 m
=
= 72.4 s
4. Find the time elapsed to travel t =
v
0.414
m/s
y
north 30.0 m:
Insight: The northward and eastward motions can be considered separately. In both cases the
actual distance travelled is greater than 20.0 or 30.0 m, respectively. For instance, in the second
case you must actually travel 30.0 m sin15.0° = 116 m to change your displacement by 30.0 m
north.
3. Picture the Problem: The car moves up the 5.5° incline with constant acceleration, changing both
its horizontal and vertical displacement simultaneously.
Strategy: Find the magnitude of the displacement along the incline, and then independently find
the horizontal and vertical components of the displacement.
Solution: 1. (a) Find the magnitude of ∆r = 0 + v t + 1 at 2 = 0 + 0 + 1 2.0 m/s 2 (12 s )2 = 140 m
)
0
2
2(
the displacement along the incline:
2. Find the horizontal component of
∆r :
x = d cos θ = (140 m ) cos 5.5° = 140 m
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Foundation Engineering Science – Dr. Daniel Nankoo
Tutorial Solutions
3. (b) Find the vertical component ∆r : y = d sin θ = (140 m ) sin 5.5° = 14 m
Insight: The horizontal and vertical motions can be considered separately. In this case they are
each described by constant acceleration motion, but the vertical acceleration is less than the
horizontal. The two accelerations would be equal if the angle of the incline were 45°.
4.
Picture the Problem: The motion of the electron is depicted at right.
Strategy: Use the given information to independently write the equations
of motion in the x and y directions. There will be a pair of equations for
r
the position of the particle only the acceleration will not be the same as g .
Use the equations of motion to find the requested time and position
information. Let the x direction correspond to horizontal, and the y
direction to vertical.
Solution: 1. (a) The horizontal motion is
characterized by constant velocity.
2. (b) Use the time to find the vertical
deflection , again using except
r
r
substituting a for g.
t=
x − x0
6.20 − 0 cm
=
= 2.95 × 10 –9 s
vx
2.10 ×109 cm/s
y = y0 + v0 y t + 12 a y t 2
= 0 cm + 0 cm + 12 ( 5.30 × 1017 cm/s 2 ) (2.95 × 10−9 s)2
= 2.31 cm
Insight: This problem is very much akin to projectile motion, with uniform acceleration in one
direction and constant velocity the perpendicular direction. The only difference is the acceleration
is 5.30×1017 cm/s2 upward instead of 9.81 m/s2 downward.
5. Picture the Problem: The paths of the two canoeists are
shown at right.
Strategy: Canoeist 1’s 45° path determines an isosceles
right triangle whose legs measure 1.0 km. So canoeist
2’s path determines a right triangle whose legs measure
1.0 km north and 0.5 km west. Use the right triangle to
find the angle θ. Find the distance travelled by each
canoeist and set the times of travel equal to each other to
determine the appropriate speed of canoeist 2:
Solution: 1. (a) Find
the angle θ from the
right triangle of
canoeist 2:
2. (b) Set the travel
times equal to each
other:
3. Use the resulting
ratio to find the
appropriate speed of
canoeist 2:
 0.5 km 
 = 27°
 1.0 km 
θ = tan −1 
t1 = t2 ⇒
d
v2 = 2 v1 =
d1
∆r1 ∆r2
=
v1
v2
( 0.5 km )
2
2
+ (1.0 km ) 
m
1.35  = 1.1 m/s
2
2 
s 
(1.0 km ) + (1.0 km ) 
Insight: There are other ways to solve this problem. For instance, because the motions are
independent, we could set the time it takes canoeist 1 to travel 1.0 km horizontally equal to the time
it takes canoeist 2 to travel 0.5 km horizontally.
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Tutorial Solutions
6. Picture the Problem: The arrow falls below the
target centre as it flies from the bow to the target.
r
v0
0.52 m
Strategy:
Treat the vertical and horizontal
motions separately. First find the time required for
the arrow to drop straight down 52 cm from rest.
Then use that time together with the horizontal
target distance to find the horizontal speed of the
arrow. That must also equal the initial speed
because the arrow was launched horizontally.
2 ( 0.52 m )
2 ∆y
=
= 0.326 s
g
9.81 m/s 2
Solution: 1. Find the time to drop 52 cm:
∆t =
2. Find the speed of the arrow
from the horizontal distance and
time elapsed:
v0 x =
∆x
15 m
=
= 46 m/s
∆t 0.326 s
Insight: We had to bend the significant figures rules a bit to obtain an accurate answer. Another
way to solve this problem and avoid the rounding error is to solve for velocity:
v0 = g x 2 2 ( h − y ) . In this case we would set h = 0.52 m , y = 0.0 m, and x = 15 m.
7. Picture the Problem: The diver falls down along a parabolic arc, maintaining her horizontal
velocity but gaining vertical speed as she falls.
Strategy: Find the vertical speed of the diver after falling 3.00 m. The horizontal velocity remains
constant throughout the dive. Then find the magnitude of the velocity from the horizontal and
vertical components.
Solution: 1. Use to find v y :
v y2 = v02y − 2 g ∆y = 0 − 2 ( 9.81 m/s 2 ) ( 0 m – 3.00 m ) = 58.9 m 2 /s 2
2
2
2. Use the components vx and v y to v = vx + v y =
find the speed:
(1.75 m/s )
2
+ 58.9 m 2 /s 2 = 7.87 m s
Insight: Projectile problems are often solved by first considering the vertical motion, which
determines the time of flight and the vertical speed, and then considering the horizontal motion.
8. Picture the Problem: The rock falls down along a parabolic arc, maintaining its horizontal
velocity but gaining vertical speed as it falls.
Strategy: There is no initial component of velocity in the y-direction. Therefore we can solve for
the acceleration of gravity on the planet Zircon.
Solution: 1. Solve for g:
g=
2. Find the acceleration of gravity
on Zircon:
g=
2v02 ( h − y )
x2
2
2 ( 6.95 m/s ) (1.40 − 0 m )
(8.75 m )
2
= 1.77 m/s 2
Insight: Another way to solve this problem is to first find the time required to travel the horizontal
distance of 8.75 m, knowing the horizontal speed is 6.95 m/s. Use that time (1.26 s) together with
the vertical fall distance of 1.40 m to find the acceleration of gravity.
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Tutorial Solutions
9. Picture the Problem: The trajectory of the clam is indicated in
the figure at right.
Strategy: Analyze the motion of the clam as it is launched
horizontally and falls to the rocks below. Let upward be the
positive y direction.
Solution: 1. (a) The horizontal velocity remains constant at 2.70
m/s.
2. (b) The vertical
speed increases due
to the acceleration of
gravity:
v y = − gt = − ( 9.81 m/s 2 ) ( 2.10 s )
= –20.6 m s
3. (c) If the speed of the crow were to increase, the speed of the
clam in the x-direction would increase, but the speed in the ydirection would stay the same.
Insight: The speed of the crow determines vx and gravity determines v y . As in most of the other
problems in this section, the answers are only true as long as we neglect the effects of air friction.
10. Picture the Problem: The trajectory of the climber
is indicated in the figure at right.
Strategy: The − 45° direction of motion indicates
that, just prior to landing, the climber is falling
with a speed equal to his horizontal speed. Use this
fact (since the initial velocity is horizontal) to find
the height difference of the crevasse and the
landing point of the climber.
Solution: 1. (a) Use
the fact that v y = v0
to find the time of
flight:
2. Find the vertical drop
during the flight, which
is also the height
difference between the
two sides of the
crevasse:
3. (b) Find the
horizontal distance
travelled:
7.8 m/s
=−
g
9.81 m/s 2
= 0.80 s
t=−
vy
h − y = 12 gt 2 =
1
2
( 9.81 m/s ) ( 0.80 s )
2
2
= 3.1 m
x = vx t = ( 7.8 m/s )( 0.80 s ) = 6.2 m
4. The climber lands 6.2 – 2.8 m = 3.4 m beyond the far edge of the w = 2.8 m wide crevasse.
Insight: The climber impacts the other side of the crevasse at about 25 mi/h (verify this for
yourself!). It would be much safer to cross the crevasse with a ladder!
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Tutorial Solutions
11. Picture the Problem: The sparrow falls down along a parabolic arc, maintaining its horizontal
velocity but gaining vertical speed as it falls.
Strategy:
Find the time it takes the sparrow to travel a horizontal distance of 0.500 m given
that its horizontal velocity remains unchanged at 1.80 m/s. Then find the distance the sparrow falls
during that time interval.
Solution: 1. (a) Find the time to travel
x 0.500 m
t= =
= 0.2778 s
0.500 m horizontally.
1.80 ms
vx
2. Find the vertical drop distance:
h − y = 12 gt 2 =
1
2
(9.81 m/s ) ( 0.2778 s )
2
2
= 0.378 m
3. (b) If the sparrow’s initial speed increases, the time interval required for it to travel 0.500 m
horizontally decreases. The distance of fall decreases for a shorter time interval.
Insight: The speed of the sparrow determines vx and gravity determines v y . Flying faster will
increase vx but not v y .
12. Picture the Problem: The pumpkin’s trajectory is depicted in
the figure at right.
Strategy: Because the pumpkin’s initial velocity is horizontal,
we can find the required initial speed of the pumpkin.
Solution: Solve for
v0 :
v0 =
g x2
=
2(h − y )
( 9.81 m/s ) ( 3.5 m )
2
2
2 ( 9.0 − 0 m )
= 2.6 m/s
Insight: Another way to solve this problem is to first find the
time of fall using the height of the tower (1.36 s). Then
calculate the horizontal speed required to hit the target during
the time of fall.
13 Picture the Problem: The stuffed animal’s trajectory is depicted in
. the figure at right.
Strategy: Determine the average speed of the riders on the Ferris
wheel by dividing the circumference of the wheel by the time to
complete a revolution. This becomes the initial speed of the stuffed
animal that is launched horizontally. The initial speed and height of
the stuffed animal determines the location it lands.
Solution: 1. (a) Find the
circumference of the
Ferris wheel:
C = 2π r = 2π ( 5.00 m ) = 31.4 m
2. Find the average speed of a
rider:
v=
3. (c) Solve equation 4-8 for x
x=
given that
h = 2 ( 5.00 m ) + 1.75 m = 11.75 m :
C 31.416 m
=
= 0.982 m/s
∆t
32.0 s
2v02 ( h − y )
g
2
=
2 ( 0.982 m/s ) (11.75 − 0 m )
9.81 m/s 2
= 1.52 m
Insight: Another way to solve this problem is to first find the time of fall (1.55 s) using the h =
11.75 m. Then calculate the horizontal distance travelled given a horizontal velocity of 0.982 m/s.
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Tutorial Solutions
14. Picture the Problem: The swimmer’s trajectory is depicted
in the figure at right.
Strategy: Because the swimmer’s initial velocity was
horizontal we can find the initial height h.
Solution: 1. (a)
Solve for h:
h= y+
( 9.81 m/s2 ) (1.68 m )
gx 2
=
0
+
2
2v02
2 ( 3.62 m/s )
2
= 1.06 m
2. (b) It takes the same time to reach the water. Gravity and
the vertical distance, not the horizontal speed, determine the
time of flight.
Insight: In order to increase the time of flight the swimmer
should launch herself at least partly in the upward direction.
15. Picture the Problem: The basketball’s trajectory is
depicted at right.
Strategy: Find the x and y positions of the basketball
as a function of time. Use the right triangle formed by
the floor and the basketball’s release and landing
points to write a ratio that allows us to calculate the
time of flight and therefore the initial height.
Solution: 1. Find the y position as a
function of time:
y = h − 12 gt 2 = 0
h = 12 gt 2
2. Find the x position as a function of x = v0 t
time:
h 12 gt 2
gt
=
=
x
v0 t
2v0
3. Use the tangent function for the
right triangle:
tan θ =
4. Now solve for the flight time t:
t=
5. Find the initial height:
h = 12 gt 2 =
2v0 tan θ 2 ( 4.20 m/s ) tan 30.0°
=
= 0.495 s
g
9.81 m/s 2
1
2
( 9.81 m/s ) ( 0.495 s )
2
2
= 1.20 m
Insight: If the basketball player throws the ball from the same height but with a higher initial
speed, the 30.0° angle will decrease. For instance, v0 = 8.40 m/s produces an angle of 16.1°.
Dropping the ball from rest makes the angle 90.0°.
16. Picture the Problem: The ball falls down along a parabolic arc, maintaining its horizontal velocity
but gaining vertical speed as it falls.
Strategy: The ball is accelerated only by gravity. The initial height of the ball determines the
vertical component of its final velocity, which together with the final speed can be used to find the
horizontal component of the velocity. The initial speed equals the horizontal component because it
remains the same throughout the flight.
Solution: 1. (a) The acceleration is due
only to gravity:
r
a = 9.81 m/s 2 downward
2
2. (b) Find the vertical component of the v y = −2 g ∆y
final velocity:
v y = −2 ( 9.81 m/s 2 ) ( – 0.75 m ) = 3.84 m/s
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3. Find the horizontal component of the
ball’s velocity, which is also its initial
speed:
Tutorial Solutions
2
2
= 1.1 m/s
2
2
= 3.2 m/s
vx = v 2 − v y2 =
( 4.0 m/s ) − ( 3.84 m/s )
4. (c) Repeat for a final speed of 5.0 m/s: vx = v 2 − v y2 =
( 5.0 m/s ) − ( 3.84 m/s )
Insight: The vertical motion of an object dropped from rest or launched horizontally is determined
solely by the initial height and the acceleration of gravity. That is why the vertical component of
the final velocity is the same in part (c) as it is in part (b).
17. Picture the Problem: The cork travels along a parabolic arc, maintaining its horizontal velocity
but changing its vertical speed due to the constant downward acceleration of gravity.
Strategy:
Find the horizontal component of the initial velocity by dividing the horizontal
distance travelled by the time of flight. Then use the cosine function to find the initial speed of the
cork.
Solution: 1. Find the horizontal speed of
the cork:
vx =
x 1.30 m
=
= 1.04 m/s = v0 x
t 1.25 s
v
1.04 m/s
= 1.27 m/s
2. Use the cosine function to find the initial v0 = 0 x =
cos
θ
cos 35.0°
speed:
Insight: Because gravity acts only in the vertical direction, the horizontal component of the cork’s
velocity remains unchanged throughout the flight.
18. Picture the Problem: The ball travels along a parabolic arc, maintaining its horizontal velocity but
changing its vertical speed due to the constant downward acceleration of gravity.
Strategy: The given angle of the kick allows us to calculate the vertical component of the initial
velocity by using the sine function. The time it takes the acceleration of gravity to slow down the
vertical speed, bring it to zero, and speed it up again to its initial value is the same as the time the
ball is in the air.
Solution: 1. Find the y component of the initial v0 y = v0 sin θ = ( 9.50 m/s ) sin 25.0° = 4.01 m/s
velocity:
2. Let v y = −v0 y and find the time of flight:
t=
v y − v0 y
−g
=
− 4.01 − ( 4.01 m/s )
−9.81 m/s 2
= 0.818 s
Insight: Another way to solve this problem is to find the time it takes gravity to bring v0 y to zero.
That time corresponds to the peak of flight, and since the flight is symmetric we need only double
that time to find the time of flight.
19. Picture the Problem: The trajectory of the ball is
shown at right.
Strategy: Find the time the ball is in the air. Then use
the known initial speed and angle to find the horizontal
speed of the ball, which together with the time of flight
can be used to find the horizontal distance travelled.
In this case y = 0 corresponds to the point where the
ball leaves the forward’s hands, and the floor
corresponds to y = − 0.80 m.
Solution: 1. Find the time:
y = ( v0 sin θ ) t − 12 gt 2
− 0.80 m = ( 4.3 m/s ) sin ( −15° ) t − 12 ( 9.81 m/s 2 ) t 2
0 = 0.80 m + ( −1.1 m/s ) t + ( − 4.905 m/s 2 ) t 2
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Tutorial Solutions
2
1.1 ±
2. Apply the quadratic formula: t = −b ± b − 4ac =
2a
( −1.1)
2
− 4 ( − 4.905 )( 0.80 )
2 ( − 4.905 )
t = − 0.53, 0.31 s
3. Find the horizontal distance
travelled:
x = v0 x t = ( v0 cos θ ) t = ( 4.3 m/s ) cos ( –15° )( 0.31 s ) = 1.3 m
Insight: Another way to solve this problem is to use v y2 = v02 y − 2 g ∆y to find v y , then use v y , v0 y
and the acceleration of gravity to find the time of flight and hence the horizontal distance travelled.
Such an approach requires an extra step but avoids the quadratic formula.
20. Picture the Problem: The trajectories of the snowballs are
depicted at right.
Strategy: Find the vertical component of the final velocity of
each snowball, as well as the horizontal component of each
velocity. Use the known components to determine the landing
speed.
Solution: 1. (a) The landing speed of snowball A is the same
as that of snowball B, because the landing speed is independent
of launch angle.
2. (b) Find the vertical
component of the final velocity
for snowball A:
v y = ± v02 sin 2 θ − 2 g ∆y
2
= ± (13 m/s ) sin 2 ( −90° ) − 2 ( 9.81 m/s 2 ) ( −7.0 m )
= −18 m/s ( The snowball is traveling downward )
3. Find the horizontal
component of the velocity for
snowball A:
vx = v0 x = v0 cos ( −90° ) = 0 m/s .
4. Find the landing speed of
snowball A:
v = v y2 + vx2 =
5. Find the vertical component
of the final velocity for snowball
B:
v y = ± v02 sin 2 θ − 2 g ∆y
( −18 m/s )
2
+ 02 = 18 m/s
2
= ± (13 m/s ) sin 2 ( 25° ) − 2 ( 9.81 m/s 2 ) ( −7.0 m )
= −13 m/s ( The snowball is traveling downward )
6. Find the horizontal
component of the velocity for
snowball B:
vx = v0 x = (13 m/s ) cos ( 25° ) = 12 m/s
7. Find the landing speed of snowball
B:
v = v y2 + vx2 =
2
( −13 m/s ) + (12 m/s )
2
= 18 m/s
Insight: Conservation of mechanical energy provides an even clearer explanation of why the
landing speeds of the snowballs are the same regardless of launch angle.
21. Picture the Problem: The golf ball travels along a parabolic arc, landing at the same level from
which it was launched.
Strategy: The maximum range of a projectile launched from level ground occurs when the launch
angle is 45°. Predict the range of the golf ball when launched at 45°. The minimum speed of the
ball will occur when the ball reaches the peak of its flight. At that point the vertical component of
the velocity is zero and the speed equals the horizontal component of the velocity, which remains
unchanged throughout the flight.
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Tutorial Solutions
 ( 30.0 m/s )2 

sin
2
θ
=

 sin 90° = 91.7 m

2
 9.81 m/s 

Solution: 1. (a) Find the range of the
ball when it is launched at 45°:
 v2
R= 0
g
2. (b) Find the x component of the
ball’s velocity, which corresponds to
the minimum speed during the flight:
v0 x = v0 cos θ = ( 30.0 m/s ) cos 45° = 21.2 m/s
Insight: The maximum range will occur at θ =45° only in the absence of air resistance. In the
presence of the atmosphere you must launch the ball at a lower angle than that in order to
maximize the range of the ball.
22. Picture the Problem: The golf ball travels along a parabolic arc, landing at the same level from
which it was launched.
Strategy: The highest tree the ball in the previous problem could clear would be a tree that is just
shorter than the maximum altitude achieved by the ball. Use the maximum height formula to find
the maximum height of a ball launched at 45° above the horizontal with a speed of 30.0 m/s.
Solution: 1. Find the maximum height ymax =
of the ball:
( v0 sin θ )
2g
2
2
( 30.0 m/s ) sin 45° 
=
= 22.9 m
2 ( 9.81 m/s 2 )
Insight: To hit a hole-in-one is amazing enough, but to clear a 23m tree along the way is almost
unbelievable!
23. Picture the Problem: The path of the ball is depicted at
right.
Strategy: Use the horizontal component of the ball’s velocity
together with the horizontal distance d to find the time
elapsed between the hit and its collision with the wall. Then
use the time to determine the vertical position h of the ball
when it collides with the wall.
Solution: 1. (a) Find
the time:
t=
3.8 m
d
=
v0 cos θ (14 m/s ) cos 34°
y
h
r
v0
θ
x
d
= 0.33 s
2. (b) Find h:
h = ( v0 sin θ ) t − 12 gt 2
2
= (14 m/s ) sin 34° ( 0.33 s ) − 12 ( 9.81 m/s 2 ) ( 0.33 s ) = 2.0 m
Insight: In many cases the vertical motion determines the time of flight, but in this case it is the
horizontal distance between the point where the ball is struck and the wall that limits the time of
flight.
24. Picture the Problem: The trajectory of the girl
is depicted at right.
Strategy: Use the given time of flight, initial
speed, and launch angle to determine the initial
height of the girl at the release point.
Solution: Find the initial
height of the girl at the
release point. If we let the
release height correspond to y
= 0, then the landing height
is:
y = ( v0 sin θ ) t − 12 gt 2
= ( 2.25 m/s )( sin 35.0° )( 0.616 s ) − 12 ( 9.81 m/s 2 ) ( 0.616 s )
2
y = −1.07 m
In other words, she was 1.07 m above the water when she let go
of the rope.
Insight: The girl’s speed upon impact with the water is 5.10 m/s.
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Tutorial Solutions
25. Picture the Problem: The projectile travels along a parabolic arc, maintaining its horizontal
velocity but changing its vertical speed due to the constant downward acceleration of gravity.
Strategy: At the projectile’s highest point, v y = 0. Therefore the v0 4 corresponds to the
horizontal component of the velocity v0x , which remains constant throughout the flight. Use the
cosine function together with v0 and v0x to determine the launch angle θ.
Solution: Use the known v0 and v0x to v0 x = v0 cos θ
v
find θ:
θ = cos −1  0 x
 v0

−1  v0 4 
−1  1 
 = cos 
 = cos   = 75.5°
v
4

 0 
Insight: If the launch angle were to increase, the minimum speed would decrease until it becomes
zero when the launch angle is 90° (straight upward). When the launch angle is 0° (horizontal) then
the minimum speed is v0 x = v0 .
26. Picture the Problem: The dolphin travels along a parabolic arc as it glides through the air and
lands in the water.
Strategy: Because the dolphin is travelling horizontally as it passes through the hoop, we conclude
that v y = 0 at that point and that the dolphin must be at the peak of its flight. Use the appropriate
formula to find how high the centre of the hoop is above the surface of the water.
2
2
Solution: Find the maximum height of
( v sin θ ) (12.0 m/s ) sin 40.0°
ymax = 0
=
= 3.03 m
the dolphin above the water:
2g
2 9.81 m/s 2
(
)
Insight: In order to pass through a higher hoop the dolphin must either increase the launch angle or
jump with a higher initial speed.
27. Picture the Problem: The ball travels along a parabolic arc, landing at the same level from which
it was launched.
Strategy: Because the ball lands at the same level from which it was launched, we can find the
launch angle θ. The time of flight can then be found from the horizontal component of the velocity
and the range.
1
2
2
 gR  1 −1  ( 9.81 m/s ) ( 4.6 m ) 

 = 32°
=
sin

2
2


( 7.1 m/s )
 v0  2
Solution: 1. (a) Solve for θ:
θ = sin −1 
2. (b) Find the time of flight:
t=
x
4.6 m
=
= 0.76 s
v0 cos θ ( 7.1 m/s ) cos 32°
Insight: Verify for yourself that the maximum height above the level of the throw and catch is 0.72
m, and that for a launch speed of 7.1 m/s the maximum range of the ball would be 5.1 m.
28. Picture the Problem: The ball travels along a parabolic arc, landing at the same level from which
it was struck.
Strategy: Because the ball lands at the same level from which it was struck, we can use the
relevant formulas to solve this problem. We can get a single formula for the range as a function of
time of flight and launch angle, and solve it for launch angle. Once the launch angle is known, the
magnitude of the initial velocity can be found.
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Solution: 1. (a) Combine:
Tutorial Solutions
 2v 2 
R =  0  sin θ cos θ
 g 
Eq. 4-12
and v02 =
g 2t 2
Eq. 4-11
4 sin 2 θ
 2   g 2t 2 
gt 2
R =  
 sin θ cos θ =
2
2 tan θ
 g  4sin θ 
2. Solve the equation for θ:
 ( 9.81 m/s 2 ) ( 4.30 s )2 
 g t2 
−1
 = 45.5°
θ = tan 
 = tan 
2 ( 92.2 m )


 2R 


3. (b) Solve for v0 :
v0 =
−1
( 9.81 m/s2 ) ( 4.30 s ) = 30.1 m/s
gt
=
2sin θ
2sin 44.5°
Insight: Another way to solve this problem is to figure out the x component of the velocity (21.4
m/s) from the range and the time of flight. The y component of the initial velocity (21.1 m/s) can
be determined from the fact that gravity reduces it to zero in half the time of flight (2.15 s). The
two components can be used to find the direction and speed.
29. Picture the Problem: The lava that reaches maximum altitude is hurled straight upward, comes to
rest momentarily, and falls straight downward again.
Strategy: Determine the launch speed required to achieve the maximum altitude.
Solution: 1. (a) Find v0 :
v y 2 = v0 y 2 − 2 g ∆y = 0 at the peak of flight
v0 y = 2 g ∆y = 2 (1.80 m/s 2 )( 2.00 × 105 m ) = 849 m/s
2. (b) If the launch speed remains the same, the maximum height of the ejected lava on Earth
would be less than it is on Io because the acceleration of gravity on Earth is much greater than
1.80 m/s2.
Insight: On Earth a launch speed of 849 m/s would be 2.5 times the speed of sound—roughly
equal to the muzzle velocity of a rifle bullet! Tidal forces from Jupiter produce these violent
volcanic eruptions on Io.
30. Picture the Problem: The football travels along a parabolic arc, maintaining its horizontal velocity
but changing its vertical speed due to the constant downward acceleration of gravity.
Strategy: Find the components of the football’s velocity as a function of time. The components can
then be used to find the magnitude and direction of the velocity at each instant. The horizontal
component of the velocity remains the same throughout the flight.
Solution: 1. (a) Find vx = v0 x = v0 cos θ = (10.2 m/s ) cos 25.0° = 9.24 m/s
the horizontal
component of the
velocity:
2
2. Find v y at 0.250 v y = v0 sin θ − gt = (10.2 m/s ) sin 25.0° − ( 9.81 m/s ) ( 0.250 s ) = 1.86 m/s
s:
3. Find the magnitude v = vx2 + v y2 =
of the velocity:
( 9.24 m/s )
2
2
+ (1.86 m/s ) = 9.43 m/s
−1  1.86 m/s 
4. Find the direction θ = tan 
 = 11.4° or 11.4° above horizontal
 9.24 m/s 
of the velocity:
5. (b) Find v y at
0.500 s:
v y = v0 sin θ − gt = (10.2 m/s ) sin 25.0° − ( 9.81 m/s 2 ) ( 0.500 s ) = − 0.589 m/s
6. Find the magnitude v = vx2 + v y2 =
of the velocity:
2
( 9.24 m/s ) + ( − 0.589 m/s )
2
= 9.26 m/s
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Foundation Engineering Science – Dr. Daniel Nankoo
Tutorial Solutions
−1  − 0.589 m/s 
7. Find the direction θ = tan 
 = −3.68° or 3.68° below horizontal
 9.24 m/s 
of the velocity:
8. (c) The ball is at its greatest height before 0.500 s because at 0.500 s the vertical component of
its velocity is already negative, or downward.
Insight: You can verify for yourself that the ball reaches its maximum altitude of 0.947 m at 0.440
s after the kick.
31. Picture the Problem: The soccer ball travels along a parabolic arc, maintaining its horizontal
velocity but changing its vertical speed due to the constant downward acceleration of gravity.
Strategy: Find the initial vertical velocity of the ball from the time interval between the kick and
the peak of the flight. Then use the known magnitude of the initial velocity together with its
vertical component in order to find the direction of the kick.
Solution: 1. When the
ball is at the peak of
flight its vertical speed is
zero:
2. Solve the equation for θ:
v y = 0 = v0 sin θ − gt
 ( 9.81 m/s 2 ) ( 0.750 s ) 
 gt 
−1
 = 46.2°
 = sin 
10.2 m/s


 v0 
θ = sin −1 
Insight: You can verify for yourself that with the new angle of the kick the maximum height is
2.76 m and the downfield range is 10.6 m.
32. Picture the Problem: The golf ball travels along a parabolic arc, landing at the same level from
which it was struck.
Strategy: Because the ball lands at the same level from which it was struck, we can use the range
formula to solve this problem. We can find the ball’s downfield range, and utilize the fact that the
range is proportional to sin ( 2θ ) in order to determine the second angle that will produce the same
range.
2
Solution: 1. (a) Find the rang:
R=
v0 2
( 42.0 m/s )
sin 2θ =
sin 70.0° = 169 m
g
( 9.81 m/s2 )
2. (b) Because the range depends
upon sin ( 2θ ) , there are two launch angles that θ 2 = 90° − θ1 = 90° − 35.0° = 55.0°
will produce the same range: θ and 90°-θ.
Insight: You can verify for yourself that sin ( 70° ) = sin (110° ) and that is why the two ranges are
the same for θ =35.0° and θ =55.0°.
33. Picture the Problem: The hay bale travels along a parabolic arc, maintaining its horizontal
velocity but changing its vertical speed due to the constant downward acceleration of gravity.
r
Strategy: The initial velocity of the bale is given as v 0 = (1.12 m/s ) xˆ + ( 8.85 m/s ) yˆ . Use the fact
that the horizontal component of the bale’s velocity never changes throughout the flight in order to
find the vertical component of the velocity when the total speed is 5.00 m/s. Then find the time
elapsed between the initial throw and the instant the bale has that vertical speed. For part (b) set
the vertical speed equal to the (constant) horizontal speed in magnitude but negative in direction
(pointing downward). Find the time elapsed between initial throw and the instant the bale has that
new vertical speed.
Solution: 1. (a) Determine the y
v 2 = v02x + v y2
component of the velocity when the
v y = ± v 2 − v02x = ±
total speed is 5.00 m/s:
( 5.00 m/s )
2
2
− (1.12 m/s ) = ±4.873 m/s
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Foundation Engineering Science – Dr. Daniel Nankoo
Tutorial Solutions
2. Use the positive value of v y since
v y − v0 y 4.87 m/s − 8.85 m/s
=
= 0.405 s
the bale is rising when the speed first t =
−g
− 9.81 m/s 2
equals 5.00 m/s. Find the time
elapsed to this point from equations
4-6:
3. (b) If the bale’s velocity points
45.0° below the horizontal then we
know the vertical velocity:
v y = −v0 x = −1.12 m/s
4. Find the time elapsed:
t=
v y − v0 y
−g
=
−1.12 m/s − 8.85 m/s
= 1.02 s
− 9.81 m/s 2
r
5. (c) If v 0 is pointed straight upward then the initial vertical velocity component will be larger, so
it will rise higher and its time in the air will increase.
Insight: The bale will have a speed of 5.00 m/s again after 1.40 s has elapsed. If it were thrown
straight upward with the same initial speed (8.92 m/s) it would rise to a height of 4.06 m, as
opposed to 3.99 m as in the original case.
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Tutorial Solutions
Solutions to Tutorial 4
1.
Picture the Problem: The image shows a wave with
the given wave dimensions.
Strategy: Set the wavelength equal to the horizontal
crest-to-crest distance, or double the horizontal crest-totrough distance. Set the amplitude equal to the vertical
crest-to-midline distance, or half the vertical crest-totrough distance.
Solution: 1. (a) Double the horizontal crest-totrough distance:
λ = 2 ( 26 cm ) = 52 cm
2. (b) Halve the vertical crest-to-trough
distance:
A=
1
2
(11 cm ) =
5.5 cm
Insight: Note the difference in wavelength and amplitude. The wavelength is the entire distance
from crest to crest, but amplitude is only from the equilibrium point to the crest.
2.
Picture the Problem: A surfer measures the frequency and length of the waves that pass her.
From this information we wish to calculate the wave speed.
Strategy: Write the wave speed as the product of the wavelength and frequency.
 1 min 
v = λ f = ( 34 m )(14 /min ) 
 = 7.9 m/s
 60 sec 
Solution: Multiply wavelength by
frequency:
Insight: The wave speed can increase by either an increase in wavelength or an increase in
frequency.
3.Picture the Problem: The image shows
water waves passing to a shallow region
where the speed decreases. We need to
calculate the wavelength in the shallow
area.
Strategy: Calculate the frequency in the deep water. Then use the constant frequency and the speed
in the shallow water to calculate the new wavelength.
Solution: 1. Calculate the
frequency:
f =
2. Calculate the new wavelength:
λ2 =
v1
λ1
=
2.0 m/s
= 1.333 Hz
1.5 m
v2
1.6 m/s
=
= 1.2 m
f 1.333 Hz
Insight: Note that decreasing the speed, with constant frequency, will decrease the wavelength.
4.
Picture the Problem: The speed and wavelength of a tsunami are given and we wish to calculate
the frequency.
Strategy: Solve for the frequency.
Solution: Calculate the frequency:
f =
v
λ
=
( 750 km/h ) 
1h 
−4

 = 6.7 × 10 Hz
310 km  3600 s 
Insight: Although the tsunami has a very high speed, the long wavelength gives the tsunami a low
frequency.
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Tutorial Solutions
5.Picture the Problem: A wave of known amplitude, frequency, and wavelength travels along a string.
We wish to calculate the distance travelled horizontally by the wave in 0.5 s and the distance
travelled by a point on the string in the same time period.
Strategy: Multiply the time by the wave speed, where the wave speed is given, to calculate the
horizontal distance traveled by the wave. A point on the string travels up and down a distance four
times the amplitude during each period. Calculate the fraction of a period by dividing the time by the
time of a full period. Set the period equal to the inverse of the period and multiply by four times the
amplitude to calculate the distance traveled by a point on the string.
Solution: 1. (a) Calculate the
horizontal distance:
d w = vt = ( λ f ) t = ( 27 ×10 −2 m ) ( 4.5 Hz )( 0.50 s ) = 0.61 m
2. (b) Calculate the vertical
distance:
t
dk = ( 4 A) 
T

−2
 = 4 Aft = 4 (12 × 10 m ) ( 4.5 Hz )( 0.50 s ) = 1.1 m

3. (c) The distance travelled by a wave peak is independent of the amplitude, so the answer in part (a)
is unchanged. The distance travelled by the knot varies directly with the amplitude, so the answer in
part (b) is halved.
Insight: A point on the string travels four times the wave amplitude in the same time that the crest
travels one wavelength.
6.Picture the Problem: Using the equation for the speed of deep water waves given in the problem we
want to calculate the speed and frequency of the waves.
Strategy: Insert the given data into the equation v = g λ 2π to solve for the speed of the waves.
Then calculate the wave frequency.
Solution: 1. (a) Insert the frequency
into the deep water velocity equation:
v = g λ / ( 2π ) =
(9.81 m/s ) ( 4.5 m ) / ( 2π ) =
2
2.65 m s
2.651 m/s
= 0.59 Hz
4.5 m
Insight: Since the velocity is proportional to the square-root of the wavelength, the frequency is
inversely proportional to the square-root of the wavelength. Increasing the wavelength by a factor of
four will double the wave speed and cut the frequency in half.
2. (b) Solve for the frequency:
7.
f =
v
λ
=
Picture the Problem: The speed of shallow water waves is proportional to the square-root of the
water depth. We wish to calculate the speed and frequency of some shallow water waves.
Strategy: Use the speed equation v = gd given in the problem, where d is the water depth, to
calculate the wave speed. Then calculate the wave frequency.
Solution: 1. (a) Calculate the wave
speed:
v = gd =
( 9.81 m/s ) ( 0.026 m ) =
2
0.51 m s
0.505 m/s
= 67 Hz
λ 0.0075 m
Insight: As the wave approaches shallower water, with constant frequency, its wavelength
decreases. In this problem, if the depth drops to 1.3 cm, the wavelength will decrease to 0.59 cm.
2. (b) Calculate the wave frequency:
8.
f =
v
=
Picture the Problem: The string tension is changed until the wave speed doubles.
Strategy: The speed of a wave on a string is given. Solve the equation for the tension in the
string. Then use a ratio to find the factor by which the tension increases.
Solution: 1. Solve for the tension:
v = F µ ⇒ F = v2 µ
2. Divide the tension at higher velocity by the
initial tension:
F2 v22 µ  v2   32 m/s 
=
=
=
 =4
F v 2 µ  v   16 m/s 
2
2
The tension increases by a factor of 4.
Insight: The tension increases by a factor equal to the square of the fractional increase in velocity.
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Tutorial Solutions
9. Picture the Problem: The image shows two
people talking on tin can telephone. The cans
are connected by a 9.5-meter-long string
weighing 32 grams. We wish to calculate the
time it takes for a message to travel across the
string.
Strategy: Set the time equal to the distance
divided by the velocity, where the velocity is
given. The linear mass density is the total
mass divided by the length.
Solution: 1. Set
the time equal to
the distance
divided by
velocity:
2. Substitute µ = m d and insert
numerical values:
t=
d
µ
=d
v
F
t=d
m/d
=
F
md
=
F
( 0.032 kg )( 9.5 m )
8.6 N
= 0.19 s
Insight: The message travels the same distance in the air in 0.028 seconds, about 7 times faster.
10. Picture the Problem: The image shows two
people talking on tin can telephone. The cans
are connected by a 9.5-meter-long string
weighing 32 grams. We wish to determine
how the tension in the string affects the time
for the message to travel across the string.
Strategy: In problem 9, we found that the
travel time across the string is given by
t = md / F . Use this equation to calculate
the time for the different tensions.
Solution: 1. (a) Since the time is inversely related to the tension, increasing the tension will result
in less time.
2. (b) Set the tension equal to 9.0 N:
t=
( 0.032 kg )( 9.5 m ) / ( 9.0 N ) =
3. (c) Set the tension equal to 10.0 N:
t=
( 0.032 kg )( 9.5 m ) / (10.0 N ) =
0.18 s
0.17 s
Insight: As predicted, increasing the tension decreases the time for the message to travel the
string.
11. Picture the Problem: Sound takes 0.94 seconds to travel across a wire of known length and
density. We want to calculate the tension in the wire.
Strategy: Solve for the tension in the wire. The velocity is given by the length of the wire
divided by the time for the sound to travel across it. The linear mass density is the mass divided by
the length.
Solution: 1. (a) Solve for the tension:
v=
F
µ
2
F = µ v2 =
m L
mL
  = 2
Lt 
t
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Tutorial Solutions
2. Insert the given mass, length and time:
F=
( 0.085 kg )( 7.3 m )
=
2
( 0.94 s )
0.70 N
3. (b) The mass is proportional to the tension (if L and t remain constant). So increased mass
means increased tension.
4. (c) Solve with a mass of 0.095 kg.
F=
( 0.095 kg )( 7.3 m )
=
2
( 0.94 s )
0.78 N
Insight: A heavier string requires greater tension for a wave to travel across it in the same time.
12. Picture the Problem: Waves travel down two strings, made of the same material and having the
same length, but having different diameters and tensions. We wish to calculate the ratio of the
wave speeds on these two strings.
Strategy: Calculate the ratio of the velocities. Set the linear mass densities equal to the density
of steel times the cross-sectional area of the wires.
Solution: 1. (a) Write the ratio of the velocities:
vA
=
vB
FA / µA
=
FB / µB
FA µ B
FB µ A
2. Write the linear mass density in terms of density
and area:
vA
=
vB
FA ρ AB
=
FB ρ AA
FA AB
FB AA
3. Write the area in terms of the diameter:
vA
=
vB
FA π ( d B 2 )
=
FB π ( d A 2 )2
4. Insert the given tensions and diameters:
vA
410 N  1.0 mm 
=

 = 2 = 1.4
vB
820 N  0.50 mm 
2
FA d B
FB d A
Insight: The ratio of the velocities is proportional to the square root of the tensions and inversely
proportional to the diameters.
13 Picture the Problem: The dolphin sends a signal to the ocean floor and hears its
. echo.
Strategy: We want to calculate the time the elapses before the dolphin hears the
echo and the wavelength of the sound in the ocean. The wave must travel to the
ocean floor and back before it is heard. So the distance travelled is twice the
distance to the floor. Divide this distance by the speed of sound in water to
calculate the time. Calculate the wavelength from equation 14-1.
Solution: 1. (a) Divide the distance
by the
speed of sound in water:
t=
2. (b) Solve for the wavelength:
λ=
2d 2 ( 75 m )
=
= 0.098 s
v 1530 m/s
v 1530 m/s
=
= 28 ×10 −3 m = 28 mm
f
55 kHz
Insight: In air the wavelength would be 6.2 mm. The wavelength is longer in the water because the
wave travels faster in water, while the frequency is the same.
14. Picture the Problem: We need to calculate the wavelength of sound in air from its frequency.
Strategy: Solve for the wavelength, using 343 m/s for the speed of sound in air.
Solution: 1. (a) Solve for the
wavelength:
λ=
v 343 m/s
=
= 0.807 m
f
425 Hz
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Tutorial Solutions
2. (b) Examine the relationship between
wavelength and frequency:
Wavelength is inversely related to frequency so,
if the frequency increases the wavelength
decreases.
3. (c) Calculate the wavelength at 450
Hz:
λ=
343 m/s
= 0.722 m
475 Hz
Insight: As predicted, an increase in frequency corresponds to a decrease in wavelength.
15. Picture the Problem: The figure represents you dropping a rock
down a well and listening for the splash. From the time lapse
between dropping the rock and hearing the splash we want to
calculate the depth of the well.
Strategy: The time to hear the splash, t = 1.2 s, is the sum of the
time for the rock to fall to the water, t1, and the time for the sound
of the splash to reach you, t2. Solve the free-fall equation (equation
2-13) for the time to fall and displacement at constant velocity to
calculate the time for the sound to return. Set the sum of these
times equal to the time to hear the splash and solve for the distance.
2d
g
Solution: 1. (a) Solve for the falling time:
t1 =
2. Solve for the time for the sound to travel up
the well:
t2 =
3. Sum the two times to equal the total time:
t = t1 + t2 =
4. Rewrite as a quadratic
equation in terms of the
variable d :
0=
d
vs
1
vs
( d)
2d d
+
g vs
2
1


0=

 343 m/s 
5. Solve for d using the
quadratic formula and square the
result:
2
d −t
g
+
( d)
2


2
+ 
 d − 1.2s
2 
 9.81 m/s 
d = 3.2537 m ⇒ d = 10.587 m = 11 m
6. (b) The time to hear the sound would be less then 3.0 seconds because, although the sound travel
time would double, the fall time would less than double.
Insight: The time to hear the sound for a 21-meter-deep well is 2.1 s, which is indeed less than 3.0
s.
16. Picture the Problem: The figure shows a person throwing a rock
down an 8.80-m deep well. The sound of the splash reaches the
person’s ear 1.20 seconds after the rock is thrown. We want to
calculate the speed of the rock.
Strategy: Solve for the initial velocity of the rock, where the fall
time is equal to the total time minus the sound travel time. The sound
travel time is the depth of the well divided by the speed of sound.
d 8.80 m
=
= 0.0257 s
v 343 m/s
Solution: 1. Calculate ts :
ts =
2. Subtract ts from the total
time:
tf = 1.2 s − 0.0257 s = 1.1743 s
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3. Solve for v0 :
Tutorial Solutions
1
y = y0 + v0 tf + atf 2
2
y − y0 1
v0 =
− atf
tf
2
−8.8 m 1
− ( −9.81 m/s 2 ) (1.1743 s ) = −1.7 m s
1.1743 s 2
The initial velocity of the rock is 1.7 m/s downward
Insight: Even though the speed of sound is much larger than the speed of the rock, the time for the
sound to travel up the well is significant. If the sound travel time was not included, the initial
velocity of the rock would incorrectly be calculated as 1.4 m/s, which is 15% off of the actual
velocity.
=
17. Picture the Problem: We are given the intensity level in a truck and need to calculate the sound
intensity.
Strategy: Convert the sound intensity level to sound intensity.
Solution: Solve for the sound intensity:
 I 

 I0 
β = 10 log 
I = I 0 10β /10
= (10−12 W/m 2 )(1092 /10 ) = 1.6 mW m 2
Insight: Note that doubling the sound intensity from 1.6 mW/m2 to 3.2 mW/m2 increases the
intensity level from 92 dB to 95 dB.
18. Picture the Problem: We want to determine how tripling the distance to the sound source affects
the sound intensity and the intensity level.
Strategy: Find the ratio of the final sound intensity to initial intensity. Insert the new intensity to
calculate the change in intensity level.
2
2
Solution: 1. (a) Calculate the ratio of
intensities:
I2 P 4π r22  r1   r1 
1
=
=  =  =
I1 P 4π r12  r2   3r1 
9
2. (b) Insert the new intensity
I 2 = 19 I1 :
β 2 = 10 log 
3. (c) Solve for the change in intensity
levels:
β 2 = 10 log 
 I2
 I0

 19 I1 
 = 10 log 


 I0 
 I1 
 + 10 log
 I0 
1
  = β1 + 10 log
9
1
 
9
1
β 2 − β1 = 10 log   = −9.54
9
Insight: If the initial intensity was 1.0 µW/m2, the initial intensity level would be 60.0 dB. The
intensity three times the distance away from the source would be 0.11 µW/m2 and the intensity
level would be 50.5 dB.
19. Picture the Problem: We are given a sound intensity and asked to calculate the sound intensity
which would create a intensity level 2.5 dB higher than that produced by the first sound intensity.
Strategy: Calculate the intensity level for Sound 1. Add 2.5 dB to the intensity level and then
solve for the intensity of Sound 2.
 I1 
 38.0 W/m 2 
= 135.80 dB
 = 10 log  −12
2 
 10 W/m 
 I0 
Solution: 1. Calculate the intensity level
for sound 1:
β1 = 10log 
2. Add 2.5 dB to the intensity level:
β 2 = β1 + 2.5 dB = 135.80 dB + 2.5 dB = 138.30 dB
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Tutorial Solutions
 I2 

 I0 
3. Invert to calculate the intensity:
β 2 = 10 log 
I 2 = I 0 10 β2 /10 = (10−12 W/m 2 )10138.30 /10 = 67.6 W/m 2
Insight: An increase of 2.5 dB increases the intensity by a factor of 102.5 /10 = 1.78.
20.Picture the Problem: The intensity level at a distance of 2.0 meters is given. We want to find the
intensity levels at 12 m and 21 meters. We also want to find the distance for which the intensity
level is 0, the furthest point at which the siren can be heard.
Strategy: Create an equation for intensity level in relation to distance. Use this relationship to
calculate the intensity levels at 12 m and 21 m. To calculate the furthest distance at which the siren
can be heard, set the intensity level to zero, and solve the relation for distance.
Solution: 1. Create equation:
 I2
 I0
β = 10 log 
 r  2  I  

 1   1  
=
10
log

 r2   I 0  

I 
r 
= 10 log  1  + 10 log  1 
 r2 
 I0 
2
2
 2.0 m 
 = 104 dB
 12 m 
2. (a) Insert β at r1 = 2.0 m and set
r2=12 m:
β = 120 + 10 log 
3. (b) Repeat for r2=21 m:
β = 120 + 10 log 
4. (c) Set the intensity level equal to
zero:
β = 10 log 
2
 2.0 m 
 = 99.6 dB
 21 m 
 I1
 I0

 r1 
 + 10 log  

 r2 
 2.0 m 
0 = 120 + 10 log 

 r 
5. Solve for r:
2
2
2
 2.0 m 
−10 log 
 = 120
 r 
2
2.0 m
 2.0 m 
−12
⇒ r=
= 2.0 × 106 m

 = 10
10− 6
 r 
Insight: This is a theoretical limit that could be realized in an ideal case. In a more realistic
scenario, ambient noise, as well as energy losses when the sound waves are reflected or absorbed by
surfaces, would prevent us from hearing the sound 2000 km away. Sometimes the real-world factors
we ignore make a huge difference!
21. Picture the Problem: We are given the sound intensity of one hog caller and are asked to
calculate how many hog callers are needed to increase the intensity level by 10 dB.
Strategy: Multiply the intensity by N callers, setting the intensity level to 120 dB and solve for N.
 NI 
 I 
 = 10 log ( N ) + 10 log  
I
 0 
 I0 
Solution: 1. Write the intensity level for N
callers:
β = 10 log 
2. Insert the intensity levels and solve for
N:
120 = 10 log ( N ) + 110
10 = 10 log ( N )
N = 1010 /10 = 10 callers
Insight: Increasing intensity level by 10 dB increases the intensity by a factor of 10. Therefore 10
callers, each with intensity level 110 dB, would produce a net intensity level of 120 db. 100 callers
(10 × 10 callers) would be needed to produce an intensity level of 130 dB (120 dB + 10 dB).
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Tutorial Solutions
22. Picture the Problem: Twenty identical violins play simultaneously.
Strategy: Write equation for the intensity level of 20 violins and, using the rules of logarithms
solve for the intensity level of one violin.
 20 I 
 I
 = 10 log 
I
 0 
 I0
= β1 + 10 log ( 20 )

 + 10 log ( 20 )

Solution: 1. (a) Write equation for 20
violins:
β 20 = 10 log 
2. Solve for the intensity level of one
violin:
β1 = β 20 − 10 log ( 20 ) = 82.5 − 13 = 69.5 dB
3. (b) Doubling the number of violins will increase the intensity level by 10 log ( 2 ) = 3.0 dB ,
therefore the intensity of 40 violins will be 82.5 dB + 3.0 dB = 85.5 dB, which is less than 165 dB.
Insight: The intensity is proportional to the number of instruments. However, the intensity level is
related to the logarithm of the intensity, so it is not proportional to the number of instruments.
23. Picture the Problem: We are given the size of the eardrum and the intensity of sound hitting the
eardrum. From this information we want to calculate the energy per second (or power) received by
the eardrum.
Strategy: Solve for the power in terms of the intensity and area. The area of the eardrum is the
area of a circle with radius given in the problem. The intensities for the threshold of hearing (10-12
W/m2) and pain (1 W/m2) are given.
2
Solution: 1. Calculate the area of the
eardrum:
A = π r 2 = π ( 4.0 × 10−3 m ) = 5.0 × 10−5 m 2 .
2. (a) Solve for the power at
the threshold of hearing:
W

P = 10−12 2  ( 5.0 × 10−5 m 2 ) = 5.0 × 10−17 W
m 

3. (b) Solve for the power at
the threshold of pain:
 W
P = 1 2  ( 5.0 × 10−5 m 2 ) = 5.0 ×10 –5 W
 m 
Insight: The ear can receive acoustical power over a range of 12 orders of magnitude!
24 Picture the Problem: The image shows Brittany 12.5
. m north of a sound source and Phillip a distance RP
east of the sound source. The sound intensity at
Brittany is twice that at Phillip. From this information
we wish to calculate the distance RBP from Phillip to
Brittany.
Strategy: Calculate the distance from Phillip to the
sound source. Then use the Pythagorean Theorem to
calculate the distance between Brittany and Phillip.
2
Solution: 1. Solve for the distance to Phillip:
R 
IP =  B  IB
 RP 
RP = RB
2. Find the hypotenuse of the right triangle:
RBP =
IB
= (12.5 m ) 2 = 17.68 m
IP
(12.5 m )
2
2
+ (17.68 m ) =
21.7 m
Insight: Since the intensity at Brittany is greater than the intensity at Phillip, Brittany must be
closer to the source than Phillip.
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Tutorial Solutions
25 Picture the Problem: The train, a moving
. source, sounds its horn. We wish to calculate
the frequency heard by a person standing near
the tracks.
Strategy: Solve for the observed frequency,
using the negative sign since the train is moving
toward the observer.


 1 
1
f′=
 (136 Hz )
f =
1 − ( 31.8 m/s ) ( 343 m/s ) 
1– u v 
Solution: Insert the given data into
equation:
= 1.50 × 102 Hz
Insight: If the train were moving away from the observer, he would hear a frequency of 124 Hz.
26. Picture the Problem: A stationary person
sounds a 136-Hz horn as a train approaches
him at 31.8 m/s. We want to know at what
frequency a passenger on the train hears the
horn.
Strategy: This problem has a stationary
source and an approaching observer, the
passenger. Use equation with a plus sign for
the observed frequency.
Solution: Insert the given data into
equation:
f ′ = (1 + u v ) f
= 1 + ( 31.8 m/s 343 m/s )  (136 Hz ) = 149 Hz
Insight: The frequency is slightly lower than the frequency found in problem 35, where the source
was moving.
27.Picture the Problem: Two jets, one moving close to the speed of sound and the other stationary,
emit the same frequency of sound. We wish to calculate the frequencies heard by the pilots.
Strategy: Assume that the moving jet is headed toward the stationary jet. Use the equation for the
frequency heard by the pilot in the moving jet, because he is a moving observer of a stationary
source. Use the equation for the frequency heard by the pilot in the stationary jet, because she is a
stationary observer of a moving source.
Solution: 1. (a) The pilot of the jet on the ground hears a greater Doppler shift, because Doppler
effects are greater with a moving source.
2. (b) Solve for the
f ′ = (1 + u v ) f = (1 + 0.825v v )( 495 Hz ) = (1.825 )( 495 Hz ) = 903 Hz
frequency heard by the
moving pilot:
3. (c) Solve for the
frequency heard by the
stationary pilot:
 1 


1
495 Hz
f′=
= 2.83 kHz
 f =
 ( 495 Hz ) =
1
−
u
v
1
−
0.825
v
v
0.175




Insight: Note that as the speed of the jet approaches the speed of sound, the frequency heard by the
moving pilot approaches 2f, while the frequency heard by the stationary pilot approaches ∞.
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Tutorial Solutions
28 Picture the Problem: The figure shows two cyclists
. approaching each other at the same speed.
Strategy: We want to calculate the frequency at which
Cyclist B hears Cyclist A’s horn. Both the source and
the observer are moving. Since they are approaching
each other, the plus sign is used in the numerator and
the minus sign in the denominator.
Solution: 1. (a) Solve for
the observed frequency:
1 + ( 8.50 m/s 343 m/s ) 
 1 + uo v 
f′=
 ( 315 Hz ) = 0.33 kHz
f =
1 − ( 8.50 m/s 343 m/s ) 
 1 − us v 
2. (b) Cyclist A speeding up will have the greatest effect on the Doppler shift. For equal changes
in speed, moving-source effects are greater than moving-observer effects.
Insight: Increasing the Cyclist B’s speed by 1.5 m/s results in an observed frequency of 332 Hz,
while increasing Cyclist A’s speed by 1.5 m/s results in an observed frequency of 333 Hz.
29. Picture the Problem: The image
shows two trains travelling in the same
direction at different speeds. The front
train sounds a horn. We want to
calculate the speed of the second train
from the frequency heard by a
passenger on the second train.
Strategy: Since both the source train and the observer train are moving, to solve for the speed of
the second train. The observer train is moving toward the source train, so use the plus sign in the
numerator. The source train is moving away from the observer train, so use the plus sign in the
denominator also.
Solution: 1. Write equation
with plus signs in numerator
and denominator:
 1 + uo v 
f′=
f
 1 + us v 
2. Solve for the observer
speed:
( f ′ f )(1 + us v ) = 1 + uo v
uo = v ( f ′ f )(1 + us v ) − 1
3. Insert the given data:
uo = ( 343 m/s ) (133 Hz 124 Hz )(1 + 35.8 m/s 343 m/s ) − 1
= 63.3 m/s
Insight: Since the observed frequency is greater than the source frequency, the two trains have to
be getting closer together. Therefore, the speed of the second train has to be greater than the speed
of the first train. After the second train passes the first, the observed frequency will drop to 91.5
Hz.
30. Picture the Problem: Two cars travel away from each
other at the same speed, as shown in the figure. We
want to calculate the speed of the cars based on the
difference in emitted and observed horn frequencies.
Strategy: Since both the observer and source are
moving, find the relationship between the source and
observed frequencies. Since both vehicles are moving
away from
each other, use the minus sign in the numerator and the plus sign in the dominator. Set the source
and observer speeds equal to each other and solve for the speed.
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Tutorial Solutions
Solution: 1. Write equation
14-11 with uo = us = u :
 1− u v 
f′=
f
 1+ u v 
f ′(v + u ) = f (v − u )
2. Solve for the speed u:
 f − f ′
 205 Hz − 192 Hz 
u=
v = 
 ( 343 m/s ) = 11.2 m s
′
f
+
f
 205 Hz + 192 Hz 


Insight: The relative speed of the two cars is 22.4 m/s. If the source car were stationary and the
observer car was moving away at 22.4 m/s, the observed frequency would be slightly less than 192
Hz. If the observer were stationary while the source moved away at 22.4 m/s, the observed
frequency would be slightly greater than 192 Hz.
31. Picture the Problem: As a train passes a crossing signal the observers on the train hear the pitch
drop to 2/3 of the pitch they heard as the train approached the signal.
Strategy: Assume the frequency emitted by the crossing signal remains constant. Use equation
when the source is stationary. Use the plus sign in the equation to calculate f, the observed
frequency as the train approaches the signal, and use the minus sign to calculate 2/3f, the frequency
as the train moves away from the signal. Combine the two equations to solve for the speed.
Solution: 1. Write the observed
frequency as the train approaches
the signal:
2. Write the observed frequency as the train leaves
the signal:
3. Divide the leaving equation by the approaching
equation:
4. Solve the expression from step 3 for u:
f = (1 + u v ) f 0
2
3
f = (1 − u v ) f 0
2
3
f
f
=
(1 − u v )
(1 + u v )
f0
2 1− u v
⇒ =
f0
3 1+ u v
2 (v + u ) = 3(v − u )
1
1
u = v = ( 343 m/s ) = 68.6 m s
5
5
Insight: The train is travelling about 150 mph.
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Tutorial Solutions
Solutions to Tutorial 5
1. Picture the Problem: This is a units conversion problem
Strategy: Use the appropriate unit conversions to measure the distance of one light year in
kilometres, the speed of light in light years per year, and the speed of light in feet per nanosecond.
Solution: 1. (a)
Express 1 ly in
km:
2. (b) Write the
speed of light in
lightyears per
year:
3. (c) Convert
the speed of
light to units of
feet per
nanosecond:
m

 86,400 s   1 km 
12
1.00 ly =  3.00 × 108  ( 365 d ) 

 = 9.46 ×10 km
s 
d


  1000 m 
c = 1.00 ly y
m   1 s   1 ft 

c =  3.00 × 108   9  
 = 0.984 ft/ns
s   10 ns   0.3048 m 

Insight: In solving problems involving the light year it is helpful to remember that the unit ly is a
unity of distance, where d = c t = c (1 y ) , the speed of light times one year.
2. Picture the Problem: The distance to Alpha Centauri is given in lightyears.
Strategy: Convert the distance from light years to meters.
Solution: Convert to meters:
 9.46 × 1015 m 
16
d = 4.3 ly 
 = 4.1× 10 m
1
ly


Insight: The light that we see from Alpha Centauri, left the star 4.3 years before we see it.
3. Picture the Problem: Radio waves travel from Mars to Earth at the speed of light.
Strategy: Multiply the speed of light by the time to calculate the distance to Mars.
60 s 
11
Solution: Calculate the distance: d = c t = ( 3.00 × 108 m/s ) (12 min ) 
 = 2.2 ×10 m
1
min


Insight: The distance between Earth and Mars varies between 7.8 ×1010 m and 3.8 × 1011 m , so
the time for a signal to reach Earth from Mars can vary between 4.3 minutes and 21 minutes.
4. Picture the Problem: The wavelength of light from a star that is travelling away from Earth is
Doppler-shifted to a longer wavelength.
Strategy: Replace the frequency with the wavelength given in c = fλ. Solve the resulting expression
for the ratio of the shifted wavelength to the unshifted wavelength.
Solution: 1. Write
the Doppler shift in
terms of
wavelength:
2. Solve for the ratio of
wavelength:
 u
f ′ = f 1 −  ⇒
 c
c
λ′
=
c
λ 
u
1− 
c
λ′
1
1
=
=
= 1.14
7
λ 1 − u c 1 − ( 3.65 ×10 m/s ) ( 3.00 × 108 m/s )
Insight: Note that the wavelength is increased when the star moves away from Earth and decreased
when the star moves toward Earth.
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Tutorial Solutions
5. Picture the Problem: The wavelength of light from a star that is moving toward Earth is Dopplershifted to a shorter wavelength.
Strategy: Replace the frequency with the wavelength given in c = fλ. Solve the resulting expression
for the shifted wavelength divided by the unshifted wavelength.
Solution: 1. (a) Because the star is moving toward Earth, the frequencies of the electromagnetic
waves are increased. However, the measured wavelengths are less than what they would be if the
star were at rest relative to us because the wavelengths are inversely proportional to the frequencies
2. (b) Write the Doppler shift in
terms of wavelength:
 u
f ′ = f 1 +  ⇒
 c
c c u
=
1+
λ ′ λ  c 
3. Solve for the ratio of wavelength:
λ′
1
1
=
=
= 0.892
λ 1 + u c 1 + ( 3.65 × 107 m/s ) ( 3.00 ×108 m/s )
Insight: The wavelength is shorter than it is when the star is at rest relative to Earth, as predicted.
6. Picture the Problem: The frequency of light from a distant galaxy is reduced by 15% because of
the Doppler effect.
Strategy: Solve for the speed of the galaxy when the frequency equals 85% of the emitted
frequency.
Solution: 1. (a) According to the Doppler effect, the frequency of electromagnetic radiation as
measured by an observer is less than what it was when emitted if the source is receding. Therefore,
the galaxy is moving away from Earth.
u
2. (b) Set the shift frequency to 85% initial f ′ = (1 − 0.15 ) f = 0.85 f = f 1 − 
c

frequency:
u
c
u = (1 − 0.85 ) c = 0.15c
3. Solve for the speed:
0.85 = 1 −
Insight: Note that the percent change in frequency is equal to fraction of the speed of light that the
galaxy is travelling.
7. Picture the Problem: Light travels
from one lantern a distance d and then
back to the first lantern. The distance
d is to be large enough that the
reaction time of 0.2 s will produce an
error less than 15% in the speed of
light.
Strategy: Set the speed of light equal to
twice the distance divided by the travel
time. Add the reaction time to the
total time and set the result equal to 85% of the speed of light. Solve the resulting equation for the
minimum distance.
Solution: 1.
Calculate the
travel time
without the
reaction time:
2d
= c ⇒ t = 2d c
t
2. Add the reaction
time and set the result
equal to 85% of the
speed of light:
2d
= (1 − 0.15 ) c = 0.85c
t + ∆t
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3. Eliminate the
variable t and solve for
the distance:
Tutorial Solutions
2d
= 0.85c
2 d c + ∆t
d=
(3.00 ×10
8
c ∆t
−1
2 ( 0.85 ) − 1


=
m/s ) ( 0.2 s )
−1
2 ( 0.85 ) − 1


= 2 × 108 m
Insight: This distance is about half way to the Moon!
8. Picture the Problem: The figure shows
Michelson’s eight-sided mirror for determining
the speed of light.
Strategy: Set the speed of light equal to twice
the distance to the mirror divided by the time to
rotate the mirror one eighth of a revolution, were
the time is the angular distance divided by the
angular speed.
Solution: 1. Write an
expression for c:
c=
2. Insert numerical
values:
c=
2d
2d
2d ω
=
=
∆t ∆θ / ω
∆θ
2 ( 35.5 × 103 m ) ( 528 rev/s )
1
8
rev
= 3.00 ×108 m/s
Insight: If the mirrors rotated at only 415 rev/s the distance to the fixed mirror would need to
increase to 45.2 km.
9. Picture the Problem: Radio signals travel from Earth to a distant spacecraft.
Strategy: Divide the distance by the speed of light to calculate the time for the signal to reach the
craft.
Solution: Calculate the time:
∆t =
d
4.5 × 1012 m
=
= 1.5 × 10 4 s
c 3.00 × 108 m/s
Insight: This time delay is 4 hours and 10 minutes. When NASA sends a signal to the craft it takes
8 hours and 20 minutes for NASA to receive a confirmation from the satellite.
10. Picture the Problem: Sound travels from the cricket bat to the daughter in the stands 115 m away
at the speed of sound. The sound also travels as a radio wave (at the speed of light) to the father a
distance 132 km away.
Strategy: Calculate the time at which each person hears the sound by dividing the distance by the
speed.
Solution: 1. Calculate the time for the
daughter:
∆tdaughter =
2. Calculate the time for the father:
∆tfather =
d
115 m
=
= 0.335 s
v 343 m/s
d
132 ×103 m
=
= 4.40 × 10−4 s
c 3.00 × 108 m/s
3. The father hears the six first.
Insight: In the time the sound took the reach the daughter in the stands, the radio signal could
have travelled around the world two and a half times!
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Tutorial Solutions
11. Picture the Problem: As a motorist approaches a yellow signal light, the motorist sees the light
Doppler shifted to green.
Strategy: Calculate the two frequencies from the wavelengths, and then insert the frequencies into
c = fλ to solve for the speed.
Solution: 1. (a) Calculate the f = c λ = ( 3.00 ×108 m/s ) ( 590 × 10 −9 m ) = 5.0847 × 1014 Hz
frequencies:
f ′ = c λ ′ = ( 3.00 × 108 m/s ) ( 550 ×10−9 m ) = 5.4545 ×1014 Hz
2. Solve for the speed:
f ′ = f (1 + u c )
 5.45 × 1014 Hz 
 f′ 
u = c  − 1 = 3.00 × 108 m/s 
− 1 = 2.2 × 107 m/s
14
 f

 5.08 × 10 Hz 
3. (b) The wavelength decreases if the motorist travels toward the traffic light.
Insight: If the motorist were travelling away from the light, it would have a wavelength of 640 nm
and appear red.
12. Picture the Problem: Light of frequency 5.000×1014 Hz is emitted from a galaxy that is receding
from Earth at a speed of 3325 km/s.
Strategy: Solve for the observed frequency, using the negative sign because the galaxy is receding.
Solution: Calculate the
observed frequency:
 3325 × 103 m/s 
 u
14
f ′ = f 1 −  = 5.000 × 1014 Hz  1 −
 = 4.945 ×10 Hz
8
3.00
×
10
m/s
 c


Insight: The observed frequency is lower than the emitted frequency because the galaxy is
receding.
13. Picture the Problem: The wavelength of light that is sent from the Enterprise to the Constitution
is shortened due to the Doppler effect as the two starships approach each other.
Strategy: Replace the frequency with the wavelength given. Solve the resulting equation for the
emitted wavelength:
Solution: 1. Write the Doppler shift f ′ = f (1 + u c ) ⇒
in terms of wavelengths:
c
λ′
=
c
λ
(1 + u c )
 722.5 × 103 m/s 
u
2. Solve for the emitted wavelength: λ = λ ′  1 +  = 670.3 nm  1 +
 = 671.9 nm
3.00 ×108 m/s 
 c

Insight: If the ships were receding at the same speed, the Constitution would observe a wavelength
of 673.5 nm.
14. Picture the Problem: As a ball travels toward a radar gun, it receives a Doppler-shifted frequency
and reflects that frequency back to the gun. The frequency is again shifted as the ball’s a moving
source.
Strategy: Solve for the change in frequency received by the ball. Double this change to calculate
the change in frequency received back by the radar gun.
Solution: 1.
Calculate the
change in
frequency
received by the
ball:
fu
 u
f ′ = f 1 +  = f +
c
 c
10.525
× 109 Hz ) ( 90.0 mph )  0.447 m/s 
(
fu
f ′− f =
=

 = 1410 Hz
c
3.00 × 108 m/s
 mph 
2. Double change in frequency: f ′′ − f = 2 ( f ′ − f ) = 2 (1410 Hz ) = 2.82 kHz
Insight: This represents only a 2.7×10−5 % change in frequency.
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Tutorial Solutions
15. Picture the Problem: As a car travels away from a radar gun, it receives a Doppler-shifted
frequency and reflects that frequency back toward the gun. The frequency is again shifted because
the car is a moving source.
Strategy: Solve for the change in frequency received by the car. Double this change to calculate
the change in frequency received back by the radar gun.
fu
 u
f ′ = f 1 −  = f −
c
 c
× 109 Hz ) ( 44.5 m/s )
8.00
(
fu
f ′− f = −
=−
= −1190 Hz
c
3.00 × 108 m/s
Solution: 1.
Calculate the change
in frequency
received by the car:
2. Double the change in frequency: f ′′ − f = 2 ( f ′ − f ) = 2 ( −1187 Hz ) = −2.37 kHz
Insight: This represents only a 3.0×10-5 % change in frequency.
16. Picture the Problem: The image
shows a spinning spiral galaxy that
is receding from Earth. The
spinning of the galaxy produces a
Doppler shift of the light emitted
by each arm.
Strategy: Solve for the observed frequency, where the speed is (a) u = V − v and (b) u = V + v . In
both cases, use the negative sign in because the galaxy is receding from Earth.
Solution: 1. (a)
Calculate the
Doppler shift for
the approaching
arm:
2. (b) Calculate
the Doppler shift
for the receding
arm:
 3.600 ×105 m/s − 6.400 ×105 m/s 
 u
f ′ = f 1 −  = 8.230 × 1014 Hz 1 −

3.00 × 108 m/s
 c


= 8.238 × 1014 Hz
 3.600 × 105 m/s + 6.400 × 105 m/s 
f ′ = 8.230 × 1014 Hz  1 −

3.00 ×108 m/s


14
= 8.203 × 10 Hz
Insight: The motion of stars in the galaxy will vary between the two limits V − v and V + v. This
results in the 8.230×1014 Hz frequency being observed as a range of frequencies between
8.203×1014 Hz and 8.238×1014 Hz.
17. Picture the Problem: As a car travels away from a radar gun, it receives a Doppler shifted
frequency and reflects that frequency back toward the gun. The frequency is again shifted because
the car is a moving source.
Strategy: Divide the total Doppler shift in half to calculate the change in frequency between the
radar gun and the frequency received by the speeding car. Insert this change in frequency and solve
for the speed of the car.
Solution: 1. (a) The car is moving away from the radar gun because observed frequencies
decrease for receding objects.
2. (b) Divide the change in
frequency in half:
3. Solve for the speed:
f ′− f =
1
2
( f ′′ − f ) = 12 ( −4.04 kHz ) = −2.02 kHz
fu
 u
f ′ = f 1 −  = f −
c
 c
 f ′− f 
 −2020 Hz 
8
u = −c 
 = −3.00 × 10 m/s 
 = 25.1 m/s
9
f
 24.150 × 10 Hz 


Insight: When the reflected wave interferes with the transmitted wave it creates a beat frequency
that determines the speed of the car. The speed of the car is equivalent to 56.2 mi/h.
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Tutorial Solutions
18. Picture the Problem: Dental X-rays have a wavelength of 0.30 nm.
Strategy: Solve for the frequency.
Solution: Calculate the frequency:
f =
c
λ
=
3.00 ×108 m/s
= 1.0 × 1018 Hz
0.30 × 10− 9 m
Insight: This frequency falls in the X-ray part of the spectrum.
19. Picture the Problem: Blue light has a wavelength of 460 nm.
Strategy: Solve for the frequency.
Solution: Calculate the frequency:
f =
c
λ
=
3.00 × 108 m/s
= 6.52 × 1014 Hz
460 ×10− 9 m
Insight: This frequency falls in the visible light region of the spectrum.
20. Picture the Problem: The frequency of a cell phone transmission is 1.25×108 Hz.
Strategy: Solve for the wavelength.
Solution: Calculate the wavelength:
λ=
c 3.00 ×108 m/s
=
= 2.40 m
f 1.25 × 108 Hz
Insight: We can see that this wavelength corresponds to a radio/TV wave.
21. Picture the Problem: The radiation emitted by humans has a wavelength of about 9.0 µm.
Strategy: Solve to calculate the frequency. Then compare the frequencies to the ranges given in
section 25-3 of the text.
Solution: 1. (a) Calculate the frequency:
f =
c
λ
=
3.00 ×108 m/s
= 3.3 × 1013 Hz
9.0 × 10−6 m
2. (b) This frequency falls in the infrared range (1012 Hz to 4.3 × 1014 Hz).
Insight: This frequency can be detected with the infrared goggles that are often used by hunters
and the military.
22. Picture the Problem: The ultraviolet range of the spectrum is subdivided into three regions.
Strategy: Calculate the frequencies of each range of ultraviolet radiation.
Solution: 1. (a) Calculate the minimum
frequency of UV-A:
f UVA-min =
3.00 × 108 m/s
= 7.50 ×1014 Hz
400 × 109 m
2. Calculate the frequency division between
UV-A and UV-B:
f UVA-UVB =
3.00 × 108 m/s
= 9.38 × 1014 Hz
320 × 109 m
3. Calculate the frequency division between
UV-B and UV-C:
f UVB-UVC =
3.00 × 108 m/s
= 1.07 × 1015 Hz
280 ×109 m
4. Calculate the maximum frequency of UV-C:
f UVC-max =
3.00 × 108 m/s
= 3.00 × 1015 Hz
100 × 109 m
5. UV-B falls in the range: 9.38 × 1014 Hz < f UV-B < 1.07 ×1015 Hz
6. (b) 7.9×1014 Hz falls into the UV-A range.
Insight: Frequencies smaller than 7.50×1014 Hz fall in the visible spectrum and frequencies
greater than 3.00×1015 Hz are X-rays.
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Tutorial Solutions
23. Picture the Problem: The very low frequency radio waves have a frequency of 10.0 kHz.
Strategy: Solve for the wavelength.
Solution: Calculate the wavelength:
λ=
c 3.00 × 108 m/s
=
= 30.0 km
f 10.0 × 103 Hz
Insight: The intensity of the wave as it enters water decreases exponentially on a scale inversely
proportional to the wavelength. The longer the wavelength, the deeper the signal will penetrate the
water. However, longer waves cannot carry information as quickly as shorter waves, so that
communication with submarines tends to be very slow.
24. Picture the Problem: As a wave enters a medium in which the speed of light decreases, the
frequency remains constant, but the wavelength changes.
Strategy: Write the ratio of the new wavelength in terms of the initial wavelength.
Solution: 1. (a) The wavelength of an electromagnetic wave is directly proportional to its speed.
Therefore, if the wave speed decreases, the wavelength decreases.
λ ′ 34 c f
3
=
=
λ
4
c f
2. (b) Write the ratio:
Insight: The speed of light in water is
3
4
of the speed of light in a vacuum, and its index of
refraction is .
4
3
25. Picture the Problem: Blue light has a wavelength of 470 nm and red light a wavelength of 680
nm.
Strategy: Calculate the frequencies of both colours.
Solution: 1. (a) Frequency varies inversely with wavelength, so that a shorter wavelength
corresponds to a higher frequency. Since the wavelength of violet light is shorter than that for red
light, violet light has the higher frequency.
2. (b) Calculate the frequency of blue
light:
f blue =
3. Calculate the frequency of red light:
f red =
c
λblue
c
λred
=
=
3.00 ×108 m/s
= 6.4 ×1014 Hz
470 ×10− 9 m
3.00 ×108 m/s
= 4.4 × 1014 Hz
680 ×10− 9 m
Insight: Violet light (λ = 420 nm) has a frequency 7.1×1014 Hz, even greater than the frequency of
blue light.
26. Picture the Problem: Ultra low frequency electromagnetic waves have a wavelength of 29
million kilometres.
Strategy: Find the period by taking the inverse of the frequency.
Solution: Calculate the period:
T=
1 λ
29 × 109 m
= =
= 97 s
f c 3.00 × 108 m/s
Insight: A sound wave with this same period would have a wavelength of 33 km.
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Tutorial Solutions
27. Picture the Problem: To obtain optimal reception of a 66.0-MHz TV signal, the length of the
antenna should be equal to half a wavelength.
Strategy: Set the length of the antenna equal to half a wavelength, where the wavelength is given
in terms of the frequency.
Solution: Calculate the length of the
antenna:
L=
c
1
3.00 × 108 m/s
λ=
=
= 2.27 m
2
2 f 2 ( 6.60 × 107 Hz )
Insight: To listen to an FM radio station at 107 MHz the antenna should be 1.40 m long.
28. Picture the Problem: A quarter-wavelength antenna is designed to broadcast an 880-kHz AM
radio station.
Strategy: Set the length of the antenna equal to one quarter of a wavelength, where the
wavelength is given in terms of the frequency.
Solution: Calculate the length of the
antenna:
L=
c
1
3.00 ×108 m/s
λ=
=
= 85.2 m
4
4 f 4 ( 880 ×103 Hz )
Insight: This is almost the length of a football field and is equivalent to the height of a 28-story
building.
29. Picture the Problem: A quarter-wavelength radio antenna is 112 m high.
Strategy: Set the wavelength equal to four times the antenna length and calculate the frequency.
Solution: Calculate the broadcast
frequency:
f =
c
λ
=
c
3.00 × 108 m/s
=
= 670 kHz
4L
4 (112 m )
Insight: For a higher frequency AM station (such as 1400 kHz) a quarter-wavelength antenna
would not need to be as tall (53.6 m).
30. Picture the Problem: Compare the difference in wavelengths for radio stations that differ in
frequency by 2 kHz.
Strategy: Subtract the two wavelengths to find the difference in wavelength, and write the
wavelength in terms of the frequency.
Solution: 1. (a)
Calculate the
wavelength
difference for the
50 kHz and 52 kHz
frequencies:
2. (b) Repeat for
the 500 kHz and
502 kHz
frequencies:
λ1 − λ2 =


c c
1
1
−
= ( 3.00 ×108 m/s ) 
−
 = 0.2 km
3
3
f1 f 2
50
×
10
Hz
52
×
10
Hz




1
1
−
= 2m
3
3
500
×
10
Hz
502
×
10
Hz


λ1 − λ2 = ( 3.00 × 108 m/s ) 
Insight: Another approach to solving this problem is to use the relation, ∆λ λ = ∆f f . For
instance, you could find that ∆λ = λ
8
3
∆f
c ∆f ( 3.00 × 10 m/s )( 2 × 10 Hz )
=
=
= 0.2 km as we
2
f
f f
( 51×103 Hz )
calculated in part (a) above.
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Tutorial Solutions
31. Picture the Problem: Compare the difference in frequencies for radio stations that differ in
wavelengths by 0.5 m.
Strategy: Subtract the two frequencies to find the difference in frequency, and write the frequency
in terms of the wavelength.
Solution: 1. (a) Calculate
c c
1
1


the frequency difference for f1 − f 2 = −
= ( 3.00 ×108 m/s ) 
−
 = 2 kHz
λ1 λ2
 300.0 m 300.5 m 
the 300.0-m and 300.5-m
wavelengths:
1 
2. (b) Repeat for the 30.0-m
 1
f1 − f 2 = ( 3.00 × 108 m/s ) 
−
 = 0.2 MHz
and 30.5-m wavelengths:
 30.0 m 30.5 m 
Insight: Another approach to solving this problem is to use the relation, ∆λ λ = ∆f f . For
instance, you could find that ∆ f = f
∆λ
λ
=
c ∆λ
λ λ
(3.00 ×10
8
=
m/s ) ( 0.5 m )
( 30.25 m )
2
= 0.2 MHz as we
calculated in part (b) above.
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Tutorial Solutions
Solutions to Tutorial 6
The solutions to Tutorial 6 can be found in Lecture 6: Introduction to Semiconductors.
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