Lethbridge College
Applied Physics (PHY143)
Unit 3: Vectors
Scalars vs. Vectors
A scalar measurement has a magnitude only. In comparison, a vector has both a
magnitude and a direction.
In the last unit, we introduced the terms ‘distance’ and ‘displacement’. Distance is a
scalar measurement: the unit is meters. Displacement is a vector: the unit is meters in a
certain direction from point one to point two.
Similarly, speed is a scalar measurement, and velocity is a vector. Speed was defined as
the distance over time, and since distance is a scalar measurement so is speed. When we
say we drove from A to B and back with an average speed, we can see that the direction
we travelled did not matter. Velocity, on the other hand, is defined as displacement over
time. Displacement is a vector from the initial position to the final position, and so is
velocity.
Time is a scalar measurement, since time has no direction.
There are a number of ways to describe the direction of a vector. We will use the number
of degrees from the positive x-axis in a counter-clockwise direction (as shown). There are
360 degrees in a full circle. (Note: when using your calculator, be sure you are set on
degrees rather than radians).
Components of Vectors
In physics, it is often easier to break a vector into an x-component and y-component. The
components take the place of the original vector.
For example, we may be working with a vector described as a displacement of 25 m at an
angle of 65 degrees. We may replace this vector with an x-component and a ycomponent, as shown below.
It is important that you know that the two components exactly replace the original vector.
Trigonometry can be used to determine the magnitude of each component.
x-component = magnitude x cos(angle from zero)
y-component = magnitude x sin(angle from zero)
If you always use the angle from zero (as shown above) the signs of your components
will always be correct.
Example:
A car is travelling at a velocity of 18.0 m/s at an angle of 40 degrees. What are the x- and
y-components?
Vx = 18.0 cos(40) = +13.8 m/s
Vy = 18.0 sin(40) = +11.6 m/s
The vector can be broken into its components. These components replace the vector. For
example, if I were to travel from the origin to the end of the vector, I would travel at 18.0
m/s. I could achieve the same average velocity by travelling in the x-direction at 13.8 m/s
and in the y-direction at 11.6 m/s (I would have covered the same displacement in the
same amount of time).
Note: it is good practice to show the positive directions you are using in your problem (as
shown in the diagram).
Example:
A person has a displacement of 44.0 meters at an angle of 130 degrees (from the positive
x-axis). What are the components of this vector?
Xx = 44.0 cos(130) = -28.3 m
Xy = 44.0 sin(130) = +33.7 m
If the person has travelled 44 meters at an angle of 130 degrees, that person could have
arrived at the same point (the same displacement) by travelling 28.3 m in the negative xdirection, and then 33.7 meters in a positive y-direction. Note how the trigonometric
functions gave you the correct signs.
Combining Vectors
If you are given the components of a vector, you may recombine them into a vector using
Pythagorean theory.
V = sqrt(Vx² + Vy²)
The angle of the vector may also be found using the tangent function.
Angle = tan-1(Vy/Vx)
Note: this will always give you the angle measured from the closest x-axis. You must
know which quadrant you are in to find the actual angle measured from zero degrees
(positive x-axis).
The quadrant can be determined by the signs of Vx and Vy, and then the angle can be
‘corrected’ to the reference point.
Quadrant I
Quadrant II
Quadrant III
Quadrant IV
Vx +
Vx –
Vx –
Vx +
Vy+
Vy+
Vy –
Vy –
Example:
Given the components of Vx = -25.0 m/s and Vy = -15.0 m/s, what is the resultant
vector?
V = sqrt [(-25)² + (-15)²] = 29.2 m/s
Angle = tan-1[(-15)/(-25)] = 31 degrees.
Note that this angle is based on the negative x-axis. The vector is in Quadrant III so the
corrected angle is 180 + 31 = 211 degrees from our reference point (positive x-axis).
Adding Vectors (Graphical)
There are two methods to add vectors. One is graphical (tip-to-tail) method, and the other
is analytical.
Tip-to-Tail:
When you are given a number of vectors (displacement, velocity, acceleration, force, and
so on) you can combine them to create a single, resultant vector. One method is to draw
the vectors to scale, starting the first vector at the origin (reference point), and drawing
each vector beginning where the last one finished.
Example:
You are given five displacement vectors:
X1 = 5 m at 70 degrees (from the positive x-axis which is always our reference point)
X2 = 7 m at 115 degrees
X3 = 3 m at 270 degrees
X4 = 9 m at 330 degrees
X5 = 5 m at 180 degrees
Starting with X1, draw each vector (to scale) from where the last ends (as shown).
It does not matter which order I draw these vectors, the resultant will be the same. If you
draw these vectors to scale, you can determine the magnitude and direction of the
resultant vector directly from the diagram. In this example, we are approximately 4 m at
60 degrees from out reference axis.
Adding Vectors (Analytical)
To add vectors analytically, you must break the vectors into their x- and x-components.
All of the x-components are added together, and all of the y-components are added
together. The sum of the x- and y-components can be recombined into a resultant vector
with a magnitude and direction as shown above.
Using the same example as above:
X1x = 5 cos(70) = 1.71
X2x = 7 cos(115) = -2.96
X3x = 3 cos(270) = 0
X4x = 9 cos(330) = 7.79
X5x = 5 cos(180) = -5.00
Sum of the x-components = 1.71 + (-2.96) + 0 + 7.79 + (-5.00) = +1.54
(Note: if you use the angle from our zero reference point, the cosine will give you the
correct sign).
X1y = 5 sin(70) = 4.70
X2y = 7 sin(115) = 6.34
X3y = 3 sin(270) = -3.00
X4y = 9 sin(330) = -4.50
X5y = 5 sin(180) = 0
Sum of the y-components = 4.70 + 6.34 + (-3.00) + (-4.50) + 0 = +3.54
(Note: if you use the angle from our zero reference point, the sine will give you the
correct sign).
The resultant vector can now be calculated:
R = sqrt [(1.54)² + (3.54)²] = 3.986 m
Angle = tan-1(3.54/1.54) = 66 degrees.
Check which quadrant you are in: The graphical method shows that the resultant vector is
in Quadrant I; and since both the sum of x-components and the sum of y-components are
positive signs, we know it is in Quadrant I. The angle of 66 degrees does not need any
correction to our reference axis.
Subtracting Vectors
If you remember the two kinematics formulae introduced in the last unit, you will
recognize a few vectors.
(Vf + Vi) /2 = (Xf – Xi) / t
a = (Vf – Vi) / t
Velocity, displacement and acceleration are vectors (and time is scalar as it has no
direction).
In the first formula, the Xf – Xi indicates the final displacement vector minus the initial
displacement vector. Subtracting vectors can be confusing.
Supposing we rewrite this difference in displacement as Xf + (-Xi). You can seen that
nothing has changed, except we are now adding the negative of the initial displacement
vector.
To make a vector negative we add 180 degrees to the direction component.
For example, if Xi = 7m at an angle of 35 degrees from our reference axis (positive xaxis). The negative vector (-Xi) is equal to 7 m at (180 + 35) = 215 degrees. This is
illustrated below:
Example:
A ladybug has an initial velocity of 7 m/s at an angle of 30 degrees. Five seconds later,
the ladybug has a velocity of 14 m/s at an angle of 300 degrees. What is the acceleration
of the ladybug?
Vi = 7 m/s at 30 degrees
Vf = 14 m/s at 300 degrees
a = (Vf – Vi) / t
or to be consistent with our discussion, a = [Vf + (-Vi)] / t
To make Vi negative, we must add 180 degrees to the direction.
-Vi = 7 m/s at 210 degrees
Vf = 14 m/s at 300 degrees
Breaking the vectors into components so we can add them:
-Vix = 7 cos(210) = -6.06 m/s
-Viy = 7 sin(210) = -3.50 m/s
Vfx = 14 cos(300) = 7.00 m/s
Vfy = 14 sin(300) = -12.12 m/s
Add the components:
Sum of x-components = -6.06 + 7.00 = 0.94 m/s
Sum of y-components = -3.50 + -12.12 = -15.62 m/s
The resultant vector is:
V = sqrt [(0.94)² + (-15.62)²] = 12.16 m/s
Angle = tan-1(-15.62/0.94) = 86.6 degrees.
What quadrant are we in? With a positive x and negative y, we are in Quadrant IV.
The corrected angle is = 360 – 86.6 = 273.4 degrees (from our reference axis).
Vf – Vi = 12.16 m/s at 273.4 degrees.
To complete the problem, we have to find acceleration:
a = (Vf – Vi) / t = 12.16 (at 273.4 degrees) / 5 seconds
a = 2.4 m/s² at 273.4 degrees
Note: only divide the magnitude by the time (the direction of the acceleration vector is
the same as the change in velocity vector)
Self Test
1. What are the components for the acceleration vector a = 5.0 m/s² at 130 degrees?
2. Given three velocities. V1 = 6.0 m/s at 15 degrees; V2 = 15 m/s at 110 degrees;
and V3 = 9.0 m/s at 270 degrees. Find the resultant vector using both graphical
and analytical methods.
3. Three young men are pulling on life preserver on a sinking ship. One is pulling at
300 N at an angle of 0 degrees; the second is pulling at 450 N at an angle of 165
degrees; and the third is pulling at 600 N at an angle of 260 degrees. What is the
resultant force and direction on the life preserver?
4. You have a displacement vs. time curve that has the relationship of x = 4t – 2t².
Graph the curve for times between 0 and 3 seconds. What is the average velocity
between the times of 1 and 1.5 seconds? Remember that Average Velocity = (Xf Xi) / t. Hint: Find the vector from the origin (t=0) to Xi and the vector from the
origin to Xf. Subtract the vector Xi from Xf and then divide by 0.5 seconds.
5. You are travelling on a bicycle in a circle with a diameter of 15 m at a constant
velocity of 3 m/s. If your initial position is at 0 degrees (travelling counterclockwise), and you reach your final position at 90 degrees in 7 seconds, what is
your acceleration (magnitude and direction)?
Self Test Solutions
1. ax = 5 cos(130) = -3.2 m/s²
ay = 5 sin(130) = 3.8 m/s²
2. V1x = 6.0 cos(15) = 5.8 m/s
V2x = 15 cos(110) = -5.1 m/s
V3x = 9.0 cos(270) = 0
Sum of x-components = +0.7 m/s
V1y = 6.0 sin(15) = 1.6 m/s
V2y = 15 sin(110) = 14.1 m/s
V3y = 9.0 sin(270) = -9.0 m/s
Sum of y-components = +6.7 m/s
Resultant vector = sqrt[(0.7)² + (6.7)²] = 6.7 m/s
Angle = tan-1(6.7/0.7)] = 84 degrees
Check quadrant: both are positive signs (Quadrant I) so angle is correct as it is.
3. F1x = 300 cos(0) = 300 N
F2x = 450 cos(165) = -434.7 N
F3x = 600 cos(260) = -104.2 N
Sum of x-components = -238.9 N
F1x = 300 sin(0) = 0 N
F2x = 450 sin(165) = 116.5 N
F3x = 600 sin(260) = -590.9 N
Sum of y-components = -474.4 N
Resultant vector = sqrt[(-238.9)² + (-474.4)²] = 531.2 N
Angle = tan-1(474.4/238.9)] = 63.3 degrees
Check quadrant: both are negative signs (Quadrant III) so the correct angle is (180
+ 63.3) = 243.3 degrees from our reference point (positive x-axis).
4.
The vector X1 can be written as 2.24 at 63.4 degrees.
The vector X2 can be written as 2.12 at 45 degrees.
We want to subtract the vectors [X2 + (-X1)], so we have to make X1 into a
negative vector (by adding 180 degrees)
X1 becomes 2.24 at 243.4 degrees.
X1x = 2.24 cos(243.4) = -1.0 m
X2x = 2.12 cos(45) = 1.5 m
Sum of x-components = +0.5 m
X1y = 2.24 sin(243.4) = -2.0 m
X2y = 2.12 sin(45) = 1.5 m
Sum of y-components = -.50 m
The resultant vector = sqrt[(0.5)² + (-0.5)²] = 0.71 m
The resultant angle = tan-1(-0.5/0.5) = 45 degrees
The vector is in Quadrant IV, so the corrected angle is (360 – 45) = 315 degrees.
(Note: this looks like the expected velocity as shown in the diagram)
The average velocity = (Xf – Xi) / t
Vave = (0.71 m at 315 degrees) / 0.5 seconds = 1.41 m/s at 315 degrees.
(Note: only divide the magnitude by the time)
5. Acceleration = (Vf – Vi) / t
The vector at the initial point is V1 = 3 m/s at 90 degrees.
The vector at the final point is V2 = 3 m/s at 180 degrees.
We want to make V1 into a negative vector, by adding 180 degrees
-V1 = 3 m/s at 270 degrees.
V1x = 3 cos(270) = 0 m/s
V2x = 3 cos(180) = -3 m/s
Sum of x-components = -3 m/s
V1y = 3 sin(270) = -3 m/s
V2y = 3 sin(180) = 0
Sum of y-components = -3 m/s
Resultant vector = sqrt[(-3)² + (-3)²] = 4.24 m/s
Resultant angle = tan-1(-3/-3) = 45 degrees.
The vector is in Quadrant III, so the corrected angle is (180 + 45) = 225 degrees.
Acceleration = 4.24 m/s at 225 degrees / 7 seconds = 0.6 m/s² at 225 degrees.
You will notice that when travelling in a circle with a constant speed, the
acceleration always points towards the center of the circle.