1) The document discusses key concepts in physics including position, displacement, velocity, acceleration, and how to interpret graphs related to these concepts. It provides definitions, formulas, example problems and explanations. 2) Position refers to an object's location relative to a reference point, while displacement describes the change in position. Velocity is the rate of change of position (or displacement divided by time) and acceleration is the rate of change of velocity. 3) Graphs are presented that relate position, velocity, displacement and acceleration to time, with the slope of certain graphs representing velocity or acceleration. Example problems are worked through to illustrate these concepts.

1. Physics 30S
Page 1 of 7
Name:______________
Position, Displacement, Velocity & Acceleration
A review
An object exists at a particular location. To describe this location you need a reference
point, a unit of distance and a direction grid.
Position
• is the way we describe this location.
• symbol usually used is “x”
• usually measured in metres.
Do you get it.
• Bob is located at x = +5 m. What does this mean? (in words)
5 m away from the refrence point, in a positive direction
• Bob is at x = -22 m. What does this mean? (in words)
22 m away from the reference point, In a negative direction
Displacement
• describes the change of position an object undergoes.
• measured in metres.
• has a direction (indicated by (+/-) or an angle.)
• formula is d = Δx or d = x2 – x1
• consecutive displacements add up to form a net displacement.
Dnet = d1 + d2 + d3 + … (vector sum)
Do you get it?
• Bob moves from x = +5 m to x = +12 m. What is his displacement? +7m
• Bob moves from x = +15 m to x = +2 m. What is his displacement? -13m
• Bob moves from x = +5 m to x = -22 m. What is his displacement?-27m
• Bob starts at +5 m then makes a displacement of +12 m. What is his final position?
+17m
• Bob starts at +15 m then makes a displacement of –25 m. What is his final position?
-10m
• Bob starts at -22 m then makes a displacement of -12 m. What is his final position?
-34m
• Bob makes consecutive displacements of +12 m then +18 m then – 35 m. What is his
net displacement? -41 m

2. Physics 30S
Page 2 of 7
10
5
0
-5
-10
1 2 3 4 5 6
displacement-Time Graphs
A visual representation of an objects location as time elapses.
• displacement is dependent variable.
• Displacement (Δx) represents the “rise” on the graph (x2 – x1)
Do you get it?
• What is the position of the object at t = 1 s, t = 2 s, t = 3 s?
5m, 0m, 0m
• What is the object’s displacement between t = 0 to t = 2 sec, between t = 1 to t = 3
sec, between t = 0 to t = 3 sec?
-10m, -5m, -10m
• What is the object’s position at t = 2, at t = 6, at t = 5.5 sec?
0m, +10m, +5m
• What displacement does the object make between t = 2 and t = 6 s, between t = 3
and t = 5 s, between t = 1 and t = 4 s?
+10m, 0m, -15m
• Assuming this is a one-dimensional motion, what is the total displacement over the
whole trip? What is the total distance traveled by the object?
Disaplcement was 0 m, distance was 40m
Velocity
As an object moves the rate at which it changes its location is called its velocity.
• Measured in m/s
• Has a magnitude and a direction (i.e. it is a vector)
• formula is v= Δx/Δt or v = d/t or d = (x2 – x1)/ Δt
• average velocity vav is measured over some interval of time (how far did you go –
displacement and how long did it take)
• instantaneous velocity vinst is measured at some moment (actually it is an average
but over a very tiny interval of time since even a moment has some duration.)
d
(m)
t (s)

3. Physics 30S
Page 3 of 7
40
20
0
-20
-40
5 10 15 20
Do you get it?
• Bob moves from x = +3 m to x = +13 m in 5 s. What is his velocity? Is that an
average or an instantaneous velocity?
+2m/s that was an average velocity
• Bob moves from x = +5 m to x = -25 m in 4 seconds. What is his average velocity
over that interval.
-7.25 m/s
Velocity-Time Graph
• A velocity time graph is a visual representation of an objects movement over
some time period.
Do you get it?
• Bob’s velocity changes as shown on the graph.
Describe in words Bob’s motion.
Desribe using words
What is his instantaneous velocity at t = 5, t = 10,
t = 12, t = 20 seconds.
+20 m/s
-20m/s
0m/s
0m/s
What is Bob’s change in velocity (Δv) between
t= 0 and t=10 s, between t=5 and t=15 s and
Between t=7.5 and t=20 seconds.
-60 m/s, +20 m/s, 0m/s
• The average velocity during an interval (the middle velocity in that interval) times
the length of time in the interval represents the change of position or displacement
made during that interval. Note that this also corresponds to the area under the
graph during that interval. (Δx = vav Δt or Δx = .5(vf + vi) Δt )
v
(m/s)
t (s)

4. Physics 30S
Page 4 of 7
10
5
0
-5
-10
1 2 3 4 5 6
Do you get it?
• Bob’s velocity changes uniformly as shown on the graph. What displacement does he
make during interval #1, during interval #2, during interval #3, during the whole
trip?
Back to position-time graphs. The slope of a position time graph represents the velocity
during that interval. This is because v= Δx/Δt = slope.
Do you get it?
Bob’s position varies with time as shown on the graph.
• What is Bob’s average velocity in each of the four
intervals shown?
-5 m/s
0m/s
-10 m/s
+10m/s
• What is Bob’s instantaneous velocity at t=1, at
t=3.5 and at t=5 seconds?
-5 m/s
-10 m/s
+10 m/s
5 10 15 17 20 25 30 t (s)
10
5
v(m/s)
0
-4
v (m/s)
p
(m) t (s)

5. Physics 30S
Page 5 of 7
Acceleration
As an object moves the rate at which it increases or decreases (i.e. changes) its velocity is
called its rate of acceleration.
• Measured in m/s/s or m/s2
• Has a magnitude and a direction (i.e. it is a vector)
• formula is a= Δv/Δt or a = (v2 – v1)/ Δt
• average acceleration aav is measured over some interval of time (what did your
velocity change by and how long did it take)
• instantaneous acceleration ainst is measured at some moment (actually it is an
average but over a very tiny interval of time since even a moment has some
duration.) When a car accelerates from zero to 100, it’s (instantaneous)
acceleration is not as great at the end as it was when it first started.
• Note: an object at rest can be accelerating (it just hasn’t got any speed at that
moment)
• In high school we deal almost exclusively with accelerations that are constant (i.e
the object is gaining/losing speed at a constant rate)
Do you get it?
• Bob’s velocity changes from +4 m/s to +20 m/s in a time of 8 seconds. What was his
rate of acceleration? Is that average or instantaneous?
That is average and is + 2 m/s2
• Bob is moving at 30 m/s then slows to 10 m/s in 4 s. What is his rate of
acceleration?
-5 m/s2
• Bob is moving North at 12 m/s and 6 s later he is moving South at 18 m/s. What is
his rate of acceleration.
-5 m/s2
• Bob is moving at +26 m/s and accelerates at –2 m/s2 for 5 s. What is his velocity
now?
Final velocity is +16 m/s

6. Physics 30S
Page 6 of 7
4
2
0
-2
-4
5 10 15 20
Acceleration-Time graphs
An acceleration-time graph is a visual representation of an objects movement over some
time period. It shows the rate at which an objects speed changes with time.
• Since Δv = a Δt (the area under a velocity time graph represents the change in
velocity which an object has undergone during that time interval. If the
acceleration is constant (which it always is in grade 11) then the velocity time
section that corresponds will be linear (i.e. the velocity increases at a constant
rate).
Do you get it?
Bob’s undergoes three uniform accelerations as shown on the graph.
• Assuming Bob started out at rest, what was his
change in velocity during interval #1, during
interval #2, during interval #3?
+40 m/s
-10 m/s
-20 m/s
• If Bob started at rest, what was his final velocity
at the end of interval #3?
+10 m/s
a
t
v
t
Δv
Area=Δv
a
(m/s
2
)
t (s)

7. Physics 30S
Page 7 of 7
Back to velocity-time graphs. The slope of a velocity-time graph represents the
acceleration occurring during that interval. This is because a= Δv /Δt = slope.
Do you get it?
Bob’s velocity changes with time as shown on the graph.
• What is his acceleration in each of the three intervals?
+ .5 m/s2
-2 m/s2
0 m/s2
• Draw the corresponding acceleration-time graph.
5 10 15 17 20 25 30 t (s)
10
5
v(m/s)
0
-4
v (m/s)