1.1 Speed and velocity in
Motion in One and Two
Dimensions
kinematics: the study of motion, the
study of moving objects
1-D Motion
 One dimensional motion or linear motion
occurs when motion can only occur in two
directions which are opposite to each
other. (One set of directions)

Example: A train on a straight track
can only move forwards or backwards.
2-D Motion
 Two dimensional motion is motion in a
plane. (Two sets of directions)

Example: A person running all over a
level field; the person can go east, west,
south, and north.
Scalar Quantities
 Scalar quantities have a number (also
called magnitude or size) and unit, but
no direction.

Examples: distance, speed, time,
energy, and power
Vector Quantities
 Vector quantities have number (also
called magnitude or size), unit, and
direction.

Examples: position, displacement,
velocity, acceleration, and force
Defining Direction Mathematically
 Direction does not fit into an equation well
 We must describe the direction of motion using mathematical
language
 To do this we define direction as either + or –
 This should be done at the beginning of each problem
N+
W-
E+
S-
If you always refer to the
horizontal…
 Vertical component
 Magnitude sin(angle)
 Horizontal component
 Magnitude cos (angle)
 Add the sign and then you are done!
 Distance describes the length of the path
travelled (scalar quantity)
 Displacement describes a change in
position (includes direction)
→
 Its symbol is Δd and its SI unit is metres,
m.
→ → →
→ → →
Δd = d2 – d1 or Δd = df – di
Note that d represents position
Displacement and velocity in Two
Dimensions:
 The total distance is simply the addition
of the magnitude of each displacement
vector.
 The best way to solve for 2D resultant
displacement is to break each vector
into its components.
1.1 Speed and velocity Note
 The best problem solving method is D-GRASS.
 D: Diagram, draw the situation if possible or if
you can’t visualize it in your head
G: Given information, the values in the problem
R: Required, what you are asked to find
1.1 Speed and velocity Note
A: Analysis, determine what information
can be used with equations
S: Solve, substitute into equations and
calculate the desired value
S: Sentence, Write a statement including
the required value as a proper sentence
Sample Problems – Right Angle
Christine walks 15 m [E] and 10 m [N]. What
is her displacement?
Sample Problem – Non-Right Angle
Sifat walks 15 m [E] and then 10 m [E10S].
What is his displacement?
Sample Problem - Components
What is the resultant displacement of an
ATV that goes 4.0 km [N], 5.0 km [E],
7.5 km [SE], 11.0 km [20º N of E], and
finally 8.8 km [30º E from N].
D-Grass outline for Example #2
1. Make a diagram (Include individual
displacements)
2. Required; resultant displacement = ?
3. Analysis…Break into component, total
each type of component (x and y), and
combine them.
D-Grass outline for Example #2
4. Solve to find the resulting total
displacement for each component (x and y).
And then calculate the magnitude of the
resultant displacement with the Pythagorean
Theorem. Use tan to get the direction.
5. Write a proper statement sentence.
1.1 Speed and velocity Note
 Look at diagram on blackboard.
x-comp:
(Let E be “+” and W be “-”)
→
→
→
→
→
→
Δdx = Δd1x+ Δd2x +Δd3x +Δd4x +Δd5x
= 0 + 5.0km +7.5km(sin45º)
+11.0km(sin70º) +8.8km(sin30º)
1.1 Speed and velocity Note
→
Δdx = +5.0km + 5.303km
+10.337km +4.4km
= +25.040 km
= 25.040 km [E]
1.1 Speed and velocity Note
y-comp:
 (Let N be “+” and S be “-”)
→
→
→
→
→
→
Δdy = Δd1y+ Δd2y +Δd3y +Δd4y +Δd5y
= 4.0km + 0 -7.5km(cos45º)
+11.0km(cos70º) +8.8km(cos30º)
1.1 Speed and velocity Note
→
Δdy = 4.0km -5.303km
+3.7622km +7.621km
= +10.080 km
= 10.080 km [N]
1.1 Speed and velocity Note
Draw resultant vector diagram of the
x-comp and the y-comp.
2
2
2
c = a +b
2
2
2
Δd = Δdx + Δdy (SCALARS)
1.1 Speed and velocity Note
Δd2 = Δdx2 + Δdy2 (SCALARS)
Δd = (Δdx2 + Δdy2)0.5
= [(25.040km)2 +(10.080km)2]0.5
= (627.00km2+101.61km2)0.5
= (728.61km)0.5
= 27.0km
1.1 Speed and velocity Note
tan θ = Δdx
Δdy
= 25.040 km
10.080 km
= 2.4841
1.1 Speed and velocity Note
tan θ = 2.4841
θ = tan-1(2.4841)
= 68.1º
Therefore the resultant displacement of
the ATV is 27.0km [N 68.1º E].
Speed
 Note: v = d/t means nothing!!!!
Which v? Δv, v1, v2, vav, or vinst?
vav = d/Δt
where vav = average speed
Δd = distance
Δt = time
Velocity
→
→
vav = Δd/Δt
→
where vav = average velocity
→
Δd
= displacement
Δt
= time
Velocity
 If the velocity is constant then at every
point the velocity equals the average
velocity.
→
→
→
→
vav = v1 = v2 = vinst
also Δv = 0
→
1.1 Speed and velocity Note
Example # 1: (1-D)
A car drives for half an hour and covers
a displacement of 20 km [W]. Find the
average velocity of the car in m/s.
1.1 Speed and velocity Note
Soln
→
vav = ?
→
Δd = 20 km [W] = 20 000 m [W]
Δt = 0.5hX60minX 60s = 1800s
1h 1 min
1.1 Speed and velocity Note
→
→
vav = Δd
Δt
= 20 000 m [W]
1800 s
= 11 m/s [W]
Therefore the car had an average velocity of
11m/s [W].
1.1 Speed and velocity Note
Example #3: (2-D) A car travels at 50.0 km/h
[E] for 10.0 minutes, 80.0 km/h [N] for 25.0
minutes, and finally 100.0 km/h [40º S from
W] for 35.0 minutes.
a) Calculate the average speed in m/s.
b) Calculate the average velocity in km/h and in
m/s.
1.1
Speed
and
velocity
Note
n
Sol
a) Calculate the displacement for each
velocity.
→
→
v = Δd/Δt
→
Therefore
→
Δd = v  Δt
 since velocity is constant for this interval we can use the above
equation.
1.1 Speed and velocity Note
→
→
Δd1 = v1  Δt
= (50.0 km/h [E])( 10.0/60h)
= 8.3333km [E] or 8333.3m [E]
→ →
Δd2 = v2  Δt
= (80.0 km/h [N])(25.0/60h)
= 33.333 km [N] or 33333 m [N]
1.1 Speed and velocity Note
→
→
Δd3 = v3  Δt
= (100.0 km/h [40º S from W])
X (35.0/60h)
= 58.333 km [S 50º W]
or 58333 m [S 50º W]
1.1 Speed and velocity Note
Δd = Δd1 + Δd2 + Δd3
= 8333.3m +33333m +58333m
= 99999 m
Δt = Δt1 + Δt2 + Δt3
= 10.0min + 25.0min + 35.0min
= 70.0min
= 70.0min X (60 s/1min)
= 4200 s
1.1 Speed and velocity Note
vav = Δd/ Δt
= (99999 m)/(4200 s)
= 23.8 m/s
Therefore the average speed is 23.8 m/s.
1.1 Speed and velocity Note
b) Draw a diagram of the displacements
X-component: (Let E be “+” and
let W be “–”)
→
→
→
→
Δdx = Δd1x + Δd2x + Δd3x
= + 8333.3m – 58333m(sin50º)
= + 8333.3 m – 44686 m
= – 36353 m
= 36353 m [W]
1.1 Speed and velocity Note
Y-component: (Let N be “+” and let S be “–”)
→
→
→
→
Δdy = Δd1y + Δd2y + Δd3y
= +33333m –58333m(cos50º)
= + 33333 m – 37495.7 m
= – 4162.7 m
= 4162.7 m [S]
1.1 Speed and velocity Note
 Draw resultant vector diagram of
the x-comp and the y-comp.
Δdr2 = Δdx2 + Δdy2 (SCALARS)
Δdr = (Δdx2 + Δdy2)0.5
Δdr = [(36353m)2 + (4162.7m)2]0.5
Δdr = (1.3215 X 109 m2
+1.7328 X 107 m2)0.5
Δdr = (1.3388 X 109 m2)0.5
Δdr = 36590 m
1.1 Speed and velocity Note
tan θ = Δdx
Δdy
= 363532 m
4163 m
= 8.7896
θ = tan-1(8.7896)
= 83.5º
1.1 Speed and velocity Note
→
→
vav = Δdr/Δt
= (36590m [S 83.5º W])/(4200s)
= 8.7121 m/s [S 83.5º W]
or 31.4 km/h [S 83.5º W]
Therefore the average velocity is 8.71 m/s [S 83.5º
W] or
31.4 km/h [S 83.5º W].
Extra Practice Questions
 Practice #1: What is the resultant
displacement of a jogger that goes 3.0 km
[W], 1.0 km [E],
2.3 km [18º E from S],
1.5 km [63º S of E], and finally 2.2 km
[N]?
Extra Practice Questions
Practice #2: A boat travels at
20.0 km/h [NE] for 23.0 minutes, 40.0 km/h [N]
for 35.0 minutes, and finally 50.0 km/h [40º N of
W] for 55.0 minutes.
a) Calculate the average speed in m/s.
b) Calculate the average velocity in km/h and in
m/s.
Extra Practice Questions
Practice #3: A swimmer travels at 4.0 km/h [13º S
of W] for 18.0 minutes, 3.5 km/h [70º W from S]
for 21.0 minutes, and finally
3.0 km/h [25º N of W] for 11.0 minutes.
a) Calculate the average speed in m/s.
b) Calculate the average velocity in km/h and in m/s.
Extra Practice Questions Answers
1. 1.5 km [S 24.7º W]
2a) 11.3 m/s
b) 34.7 km/h [N 27.0º W]
or 9.64 m/s [N 27.0º W]
3 a) 0.99 m/s
b) 3.4 km/h [S 80.8º W] or 0.95 m/s [S
80.8º W]