Advanced Physics, Unit 4 Vectors and Trig Review
Worksheet Packet (See last page for answers)
Worksheet 1.
Adding Colinear Vectors
Colinear vectors
Problem 1
C
A
D
B
A box rests on the floor. Four forces (pushes) act on the package. Force A is 50 Newtons (N); Force
B is 75 N; Force C is 75 N and Force D is also 75 N, respectively. Find the net (overall) force on the
box.
Answer = _____________ N
Which direction? __________________
Problem 2. There are two velocities acting on the plane. The engine of the plane gives it a velocity of
150 mph north and a head wind acts against it with a velocity of 50 mph south. Draw the velocities
vectors on the plane and find the net (overall) velocity vector.
Answer = _____________________ mph (remember to give direction)
Problem 3.
The engine accelerates the helicopter upward with an acceleration of 20 m/s/s. However, the earth
pulls down on the helicopter with the acceleration due to gravity (9.8 m/s/s). Find the net (overall)
acceleration vector. Draw the vectors on the helicopter, too.
Answer = ______________________ m/s/s (remember to give direction, too)
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Worksheet 2. Adding Perpendicular Vectors. Use the mathematical treatment on these
(Pythagorean theorem to find the magnitude of the resultant and arctan(θ) to find the
angle.)
Problem 1. The softball is given an acceleration to the right (along the x axis) of 15 m/s/s. However
the acceleration due to gravity is –9.8 m/s/s (along the –y axis). Draw the two vectors along the
coordinate grid and find the resultant acceleration vector (magnitude and direction). Find the direction
using quadrants and angles.
y (90 degrees)
x (0, 360 degrees)
Answer = ___________________________
Problem 2. There are two kids trying to push a box. One kid pushes it with a force of 80 N northward;
the other pushes it west with a force of 50 N. Draw the two vectors along the coordinate grid and find
the resultant force vector (magnitude and direction). Find the direction using quadrants and angles and
also using compass headings.
y (90 degrees)
x (0, 360 degrees)
Answer = ___________________________
Problem 3.
The helicopter flies at 180 degrees with a velocity of 100 mph along the -x axis. It encounters a
y (90 degrees)
downdraft pushing on it with a velocity of 25
mph. Draw the two vectors along the
coordinate grid and find the resultant velocity
vector (magnitude and direction). Find the
direction using quadrants and angles.
x (0, 360 degrees)
Answer = ___________________________
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Worksheet 3: Vector Resolution
Break the one vector up into its x and y components using
trigonometry.
Ax = Acosθ
Ay = Asinθ
#1
Y, North
The wind blows at 100 mph at an angle of 35
degrees North of East. Find Ax and Ay.
Ax = ___________________ mph, East
X, East
Ay = ___________________ mph, North
#2. You walk 40 km at an angle of 160 degrees (or 20 deg North of West) from your house.
Find Ax and Ay.
y, North
Ax = ___________________ km, West
x
--x, West
Ay = _____________________ km, North
#3. The box is pushed with a force of 125 N (Newtons) at an angle of 240 degrees
(30 deg West of South). Find Ax and Ay.
y
--x, West
x
Ax = _______________________ N, West
Ay = _______________________ N, South
--y, South
#4. The spaceship accelerates at 10m/s2 at an angle of 350 degrees (10 degrees South of
East). Find Ax and Ay.
Ax = ______________________ m/s/s, East
y
--x
X, East
Ay = ______________________ m/s/s, South
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Worksheet 4: Putting it all together. Add vectors that are not colinear and are
not perpendicular.
Problem 1. We have two force vectors. One (Vector A) is a 50 N force acting at 30 degrees and the
other (Vector B) is a 70 N force vector acting at 70 degrees. Find the one vector that these two
represent (magnitude and direction).
y
x
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Worksheet 4 (continued)
Problem 2. We have two velocity vectors. One (Vector A) is a 225 mph velocity acting at 20 degrees
and the other (Vector B) is a 100 mph velocity vector acting at 246 degrees (3rd quadrant). Find the
one vector that these two represent (magnitude and direction).
y
x
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Worksheet 4 (continued)
Problem 3. We have two force vectors. One (Vector A) is a 45 N force acting at 115 degrees and the
other (Vector B) is a 50 N force vector acting at -32 degrees. Find the one vector that these two
represent (magnitude and direction).
y
x
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Worksheet 4 (continued)
Problem 4. We can do the same thing with more than two vectors. In this example, let’s say you are
going camping and will take three hikes (three displacement vectors). Starting from your original
campsite, on day one, you hike 15 miles at 120 degrees (vector A). Then you turn directly north (90
degrees, along the y axis) and hike 20 additional miles on day two (Vector B); finally, on the third day,
you hike 25 miles along a line directed –22 degrees (4th quadrant). Find your resultant displacement
vector from the original campsite to your final position. [Resultant: the one vector in which these three
combine (magnitude and direction)].
Vector C:
25 miles at -22
degrees
Vector B:
20 miles at 90 degrees
(directly North)
Vector A:
15 miles at 120
degrees
y
Resultant
displacement
vector
x
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Worksheet 5: Practice with the Law of Sines and the Law of Cosines
Law of Sines:
1) When the angle of elevation of the sun is 62o, a telephone pole tilted at an
angle of 7o away from the sun casts a shadow 30 feet long on the ground. Find
the length of the phone pole.
2) To find the distance between two points A and B that are on opposite
sides of a river, a surveyor measures a distance on the same side of the river
as point A. The distance to this point is 240 feet and call it point C. He
then measures the angles from A to B as 62o and measures the angle from C
to B as 55o. Find the distance from A to B.
Law of Cosines:
3) In order for Mary to travel to work from home she has to
get on her bike and go 20 km directly east, and then she has to
go 10 km in a direction that is directly north-east (45º north of
east). The county that she lives in is trying to decide whether
they should build a road that goes straight from the area where
she lives to the area where she works at. How many kilometers
of travel will Mary save on each trip if this road is built?
4. Two jet aircraft leave an airport at the same time. The course of
the first is 160o while the course of the second is 290o. If the first
travels 500 mph and the second 600 mph what is the
distance between them at the end of 3 hours?
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Vector Worksheet Answers
Worksheet 1
1. Answer = 25 N left (or –25 N)
2. Answer = 100 mph North 3. 10.2 m/s/s up
Worksheet 2
1. 17.9 m/s/s at –33.1 degrees (or 33.1 degrees S of E or 326.9 degrees) [4th quadrant angle]
2. 94.3 N at 122 degrees [2nd Quadrant] (or 58 degrees N of W or 32 degrees W of N
3 103 mph at 194 degrees [3rd Quadrant] (14 degrees S of W)
Worksheet 3
1. Ax = 81.9 mph (East); Ay = 57.4 mph (North)
3. Ax = -62.5 N (West); Ay = -108.3 N (South)
2. Ax = -37.6 km (West); Ay = 13.7 km (North)
4. Ax = 9.8 m/s/s (East); Ay = -1.7 m/s/s (South)
Worksheet 4
1. Resultant = 113 N at 53.5 degrees (first quadrant or you can say 53.5 degrees North of East)
2. Resultant = 171.3 mph at –4.8 degrees (fourth quadrant) or 4.8 degrees South of East or 355.2
degrees
3. Resultant = 27.4 N at 31.4 degrees (first quadrant)
4. Resultant = 28.3 miles at 56.4 degrees (first quadrant)
Worksheet 5
1. The other base angle is (90 - 7) = 83o The vertex angle is 180 - ( 62 + 83) = 180 - 145 = 35o
Use the law of sines to calculate the height of the pole!!
(Sin 62)/x = (Sin 35)/30
x Sin 35 = 30 Sin 62;
x = (30 Sin 62)/ Sin 35;
x = 46.2
The pole is about 46 feet
2. First, we need / B: 180 - (62 + 55) = 180 - 117 = 63o; We can now use the law of sines:
The side we need is side c
(Sin 55)/c = (Sin 63)/240;
c Sin 63 = 240 Sin 55;
c = (240 Sin 55)/ Sin 63
c = 220.6
The distance from A to B is 220.6 feet.
3. c2 = a2 + b2 - 2a.b.Cos(q)
c2 = (20 km)2 + (10 km)2 - 2.(10km)(20km)Cos (135º)
c2 = 500 km2 + 282.84 km2
c2 = 400 km2 + 100 km2 - 400 x (-0.7071) km2
c2 = 782.84 km2
Taking the square root of both sides we get: c = 28 km. Thus, the person will save about 2 kilometers
in each direction since without the alternate path she will be traveling 30 km.
4.
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