CHAPTER 3: KINEMATICS IN TWO
NAME _________________________ PER ___
DIMENSIONS; VECTORS
3-1 VECTORS AND SCALARS
1. VECTOR - a quantity which has direction as well as magnitude
EX: velocity, displacement, force, and momentum
2. SCALAR – a quantity with magnitude without direction.
EX: mass, time, and temperature
On a diagram, each vector is represented by an arrow .
The arrow is always drawn so that it: points in the direction of the vector it represents.
The length of the arrow is drawn: proportional to the magnitude of the vector.
3-2 ADDITION OF VECTORS  GRAPHICAL METHODS
Simple arithmetic can be used for adding vectors if: they are in the same direction.
Draw the resultant vectors for the following diagrams, determining their magnitude and direction:
14 km east
8
10
12
8 km east
14
km
8
10
12
14
km
3. The sum of the two vectors is called the:
net
or
resultant
vector.
But arithmetic can’t be used if: the two vectors are not along the same line.
Displacements can be represented on a graph in which the positive y axis points north and the positive
x axis points east .
Draw the resultant vector for the following diagram, determining the magnitude and direction
(km) 6
a2 + b2 = c2
4
(12.0 km)2 + (4.0 km)2 = c
c = 12.6 km = 13 km
2
tan−1 o/a = θ
tan−1 4/12 = 18.4° = 18°
(km)
N of E
0
2
4
6
8
10 12 14 16
You can only use the Pythagorean theorem when the vectors are: perpendicular to each other.
The resultant displacement vector (D) is the sum of the vectors: DR = D1 + D2 (Vector Equation)
GENERAL RULES FOR GRAPHICALLY ADDING VECTORS (AKA:
Tail-to-Tip Method )
1. On a diagram: draw one of the vectors, V1, to scale.
2. Draw the second vector, V2, to scale , placing: its tail at the tip of the first vector in the correct
direction.
3. Draw an arrow from the tail of the first vector to the tip of the second vector. This represents
the: resultant of the two vectors.
4. The tail-to-tip method can be extended to more than two vectors. The resultant is drawn from the: tail
of the first vector to the tip of the last one added.
A second way to add two vectors is the: parallelogram method.
The two vectors are drawn starting from a: common origin.
A parallelogram is constructed using these two vectors as: adjacent sides.
The resultant is the: diagonal drawn from the common origin.
3-3 SUBTRACTION OF VECTORS, AND MULTIPLICATION OF A VECTOR BY A SCALAR
5. The negative of a vector has the same magnitude, but: opposite direction.
The magnitude of every vector is positive .
A minus sign tells us about: its direction.
The difference between two vectors is equal to: the sum of the first plus the negative of the second.
V2 - V1 = V2 + (-V1)

=
+
=
A vector, V, can be multiplied by a scalar, c.
cV has the same direction as V , and the magnitude of cV .
If the scalar is negative, the magnitude of cV is the same, but the direction of cV is: the opposite direction.
3-4 ADDING VECTORS BY COMPONENTS
6. Any vector can be expressed as the sum of two other vectors, called: the components of the original vector.
These are chosen to be along: two perpendicular directions.
7. The process of finding the components is know as: resolving the vector into components.
The vector (V) is resolved into its components by:
dashed arrows along the x and y axes, from the
origin to the tip of the vector.
Vy
These component vectors are written as: Vx and Vy .
 = 30 north of east
Draw, and label, the component vectors on the diagram.
Vx
In order to add vectors using th method of components, we need to use: the trigonometric functions
sine (sin), cosine (cos), and tangent (tan).
The longest side of the triangle is called the: hypotenuse which we label h .
The side opposite the angle  is labeled o .
h=V
o = Vy
The side adjacent to the angle  is labeled a .
We let h, o, and a represent the: lengths of their respective sides.

a = Vx
sin  = o
h
cos  = a
h
tan  = o
a
Example: If V represents a displacement of 500.0 m at 30.0 north of east, what are the magnitudes
of the component vectors?
sin 30.0 = Vy / 500.0 m
Vy = (sin 30.0) (500.0 m) = 250. m north
cos 30.0 = Vx / 500.0 m
Vx = (cos 30.0) (500.0 m) = 433 m east
Adding Vectors Using Components
V1 + V2 = V
Vx = V1x + V2x
Vy = V1y + V2y
Where the magnitude of the components can be
found using trigonometric functions.
Vx
Vy
V
V2
V2y
V1
V1y
V2x
V1x
Example 3-1: A rural mail carrier leaves the post office and drives 22.0 km in a northerly direction to the
next town. She then drives in a direction 60.0 south of east for 47.0 km to another town. What is
her displacement?
(HINT: Draw a diagram using the y axis as north/south and the x axis as east/west.)
D2x
D1y = 22.0 km north
D1x = 0
60.0
D1
D2
D2y
Dx

D
Dy
sin 60.0 = D2y / - 47.0 km
D2y = (sin 60.0) (- 47.0 km)
cos 60.0 = D2x / 47.0 km
D2x = (cos 60.0) (47.0 km)
D2y = - 40.7 km
D2x = 23.5 km
Dy = D1y + D2y = 22.0 km - 40.7 km = -18.7 km
Dx = D1x + D2x = 0 km + 23.5 km = 23.5 km
D = Dx2 + Dy2
D = (23.5 km)2 + (-18.7 km)2 = 30.0 km
tan  = o/a = Dy / Dx = - 18.7 km / 23.5 km = - 0.796
 = tan-1 (- 0.796) = - 38.5 Negative indicates below the x axis.
= 38.5 south of east
D2 = Dx2 + Dy2
Example 3-2: An airplane trip involves three legs, with two stopovers. The first leg is due east for
620.0 km; the second leg is southeast (45.0) for 440.0 km; and the third leg is at 53.0 south of west
for 550.0 km. What is the plane’s total displacement?
D1y = 0
D1x = 620.0 km
Dx
=?
D1
45.0
D2
D2x: cos 45.0 = D2x / 440.0 km
D2x = (cos 45.0)(440.0 km) = 311 km
D2y: sin 45.0 = D2x / 440.0 km
D2y = (sin 45.0)(-440 km) = -311 km
53.0
D
Dy
D3
D3x : cos 53.0 = D3x / 550.0 km
D3x = (cos 53.0)(-550.0 km) = - 331 km
D3y: sin 53.0 = D3y / 550.0 km
D3y = (sin 53.0)(550.0 km) = -439 km
Dx = Dx1 + Dx2 + D3x = 600. km
Dy = Dy1 + D2y + D3y = - 750. km
D = Dx2 + Dy2 = (600. km)2 + (-750. km)2 = 960. km
tan  = o/a = Dy / Dx = - 750. km / 600. km = - 1.25
 = tan-1 –1.25 = - 51.3
51.3 south of east
Set 3A
1. A car is driven 125 km west and then 65 km southwest. What is the displacement of the car from the point
of origin (magnitude and direction)? Draw a diagram.
2. A delivery truck travels 14 blocks north, 16 blocks east, and 26 blocks south. What is its final displacement
from the origin? Assume the blocks are equal length.
3. If Vx = 18.80 units and Vy = - 16.40 units, determine the magnitude and direction of V.
4. V is a vector 24.3 units in magnitude and points at an angle of 54.8 above the negative x axis. (a) Sketch
this vector. (b) Find Vx and Vy.
5. The figure below shows two vectors, A and B, whose magnitudes are A = 8.31 units and B = 5.55 units.
Determine C if (a) C = A + B, (b) C = A - B (c) C = B - A. Give the magnitude and direction
for each.
A
B
6. Vector V1 is 8.08 units long and points along the negative x axis. Vector V2 is 4.51 units long and points at
+ 45.0 to the positive x axis. (a) What are the x and y components of each vector? (b) Determine the sum
of the two vectors (magnitude and angle).
7. An airplane is traveling 785 km/h in a direction 38.5 west of north. (a) Find the components of the
velocity vector in the northerly and westerly directions. (b) How far north and how far west has the plane
traveled after 3.00 h?
B = 40.0 units
56.0
A = 66.0 units
28.0
C = 46.8 units
8. Three vectors are shown in the figure above. Their magnitudes are given in arbitrary units. Determine the
sum of the three vectors. Give the resultant in terms of (a) components, (b) magnitude and angle with the x
axis.
3-5 PROJECTILE MOTION
8. PROJECTILE MOTION – motion in two dimensions
Examples: hitting a golf ball, a thrown or batted baseball, kicked footballs, speeding bullets
Although air resistance can often be important, we are going to ignore its effect.
Acceleration due to gravity = 9.80 m/s2 .
9. Galileo showed that projectile motion could be understood by: analyzing the horizontal and vertical
components of the motion separately.
Motion begins at t = 0 at the origin of an xy coordinate system. Therefore, x0 = 0 and y0 = 0 .
10. Galileo predicted that: an object projected horizontally will reach the ground in the same time as an object
dropped vertically.
See fig. 3-19
Use the kinematic equations from chapter 2 to solve for the motion in both directions.
3-6 SOLVING PROBLEMS INVOLVING PROJECTILE MOTION
Example 3-3: A movie stunt driver on a motorcycle speeds horizontally off a 50.0 m high cliff. How fast
must the motorcycle leave the cliff-top if it is to land on level ground below, 90.0 m from the base of
the cliff where the cameras are?
horizontally:
x0 = 0
x = 90.0 m
a = 0
v0 = ? Don’t know v or t
Therefore, we need to look at the vertical motion first to find the time.
vertically:
x0 = 0
x = x0 + v0 t + ½ a t2
horizontally:
x0 = 0
x = x0 + v0 t + ½ a t2
a = 9.80 m/s2
v0 = 0
x = 50.0 m
t2 = 2 x / a = 2 (50.0 m) / 9.80 m/s2
x = 90.0 m
a = 0
t = 3.19 s
t =?
t = 3.19 s
v0 = ?
v0 = x / t = 90.0 m / 3.19 s = 28.2 m/s
Example 3-4: A football is kicked at an angle  = 37.0 with a velocity of 20.0 m/s. Calculate
(a) the maximum height, (b) the time of travel before the football hits the ground,
(c) how far away it hits the ground, (d) the velocity vector at the maximum height, and
(e) the acceleration vector at maximum height. Assume the ball leaves the foot at ground level.
v = 20.0 m/s
 = 37.0
(a)
Up:
vy
vo = 12.0 m/s
v = v0 + a t
vy: sin 37.0 = Vy / 20.0 m/s
vy = (sin 37.0)(20.0 m/s) = 12.0 m/s
vx: cos 37.0 = Vx / 20.0 m/s
vx = (cos 37.0)(20.0 m/s) = 16.0 m/s
a = - 9.80 m/s2
v = 0
x0 = 0
x = ?
t = - v0 / a = - 12.0 m/s / - 9.80 m/s2 = 1.22 s
x = x 0 + v0 t + ½ a t 2
=
(12.0 m/s) (1.22 s) + ½ (-9.80 m/s2)(1.22 s)2 = 7.35 m
(b)
Time up = time down
1.22 s + 1.22 s = 2.44 s
(c)
Horizontally: a = 0
x = vt
x = (16.0 m/s)(2.44 s) = 39.0 m
(d)
At the maximum height vy = 0. vx is a constant 16.0 m/s
(e)
There is no horizontal acceleration, only vertical, which is a constant 9.80 m/s2.
Conceptual Example 3-5: A child sits upright in a wagon which is moving to the right at constant speed.
The child extends her hand and throws an apple straight upward (from her own point of view), while
the wagon continues to travel forward at constant speed. Will the apple land (a) behind the wagon,
(b) in the wagon, or (c) in front of the wagon? Explain your answer.
(b) In the wagon. As it moves up and then back down it is also moving forward at the same
constant velocity as the child.
Conceptual Example 3-6: A boy on a small hill aims his water-balloon slingshot horizontally, straight at a
second boy hanging from a tree branch a distance d away. At the instant the water balloon is
released, the second boy lets go and falls from the tree, hoping to avoid being hit. Show that he
made the wrong move. (He hasn’t studied physics yet.)
Both the balloon and the boy are falling the same distance at the same time due to the
acceleration due to gravity. The balloon will hit the boy. (See fig 3-19, pg. 59)
Set 3B
1. A tiger leaps horizontally from a 7.5 m high rock with a speed of 4.5 m/s. How far from the base of the
rock will she land?
2. A diver running 1.6 m/s dives out horizontally from the edge of a vertical cliff and reaches the water below
3.0 s later. How high was the cliff and how far from its base did the diver hit the water?
3. A ball is thrown horizontally from the roof of a building 56 m tall and lands 45 m from the base. What was
the ball’s initial speed?
4. A football is kicked at ground level with a speed of 20.0 m/s at an angle of 37.0 to the horizontal. How much later
does it hit the ground?
5. A ball is thrown horizontally at 22.2 m/s from the roof of a building lands 36.0 m from the base of the
building. How high is the building?
6. A shot-putter throws the shot with an initial speed of 14 m/s at a 40 angle to the horizontal. Calculate the
horizontal distance traveled by the shot if it leaves the athlete’s hand at a height of 2.2 m above the ground.
7. An athlete executing a long jump leaves the ground at a 30 angle and travels 7.80 m. (a) What was the
takeoff speed?
8. The pilot of an airplane traveling 160 km/h wants to drop supplies to flood victims isolated on a patch of
land 160m below. The supplies should be dropped how many seconds before the plane is directly
overhead?
9. A projectile is fired with an initial speed of 75.2 m/s at an angle of 34.5 above the horizontal on a long flat
firing range. Determine (a) the maximum height reached by the projectile, (b) the total time in the air, (c)
the total horizontal distance covered (that is, the range), and (d) the velocity of the projectile 1.50 s after
firing.