Kinematics - (describing how things move)
Scalar (no direction)
Vector (w/ direction)
Distance (d)
Displacement (d)
Change in position
(How far you travel in a given
direction)
How far you travel
Speed (s)
How fast you travel
Velocity (v)
How fast you travel (in a given
direction)
Acceleration (a)
Rate of change of velocity
Describing Motion
There are lots of different ways to describe motion….
1. Words
2. Sketches
3. Time elapsed photographs
4. Physical Expressions (Equations)
5. Graphical Representation
Kinematics Equations that Make Sense!
Average speed:
sav = d / change in t
SI unit: m/s
sav = d / ∆t = d / tf - ti
Average velocity:
vav = ∆d / ∆t
SI unit: m/s
vav = (df - di) / t
df = di + vav t
Average acceleration: aav = ∆v / ∆t
SI unit: m/s/s = m/s2
aav = (vf - vi) / t
vf = vi + aav t
Note: if the time intervals are very small we call these quantities instantaneous
Using Split Times!
Position (m)
0-5
5-10
10-15
15-20
20-25
Split Time (s)
1.6
2.4
3.0
3.5
4.0
Av. Velocity (m/s)
3.1
2.1
1.7
1.4
1.2(5)
Determine the average velocity for each distance interval
Determine average velocity of the object over the time recorded
vav = df - di / t
= 25m - 0m / 14.5 s
= 1.7 m / s
Determine the average acceleration over the time recorded
a = vf - vi / t
= (1.25 m/s - 3.1m/s) / 14.5 s
= - 0.13 m / s2
Note: the a is negative because the change in v is negative!!
Reference Frames
All measurements are made relative to a frame of reference..
Sitting at your desk right now you don’t seem to be moving but you
are infact revolving at 30 km/s (67,000 mi/h) around the sun
The earth also has a rotational speed about its axis. People on
earth experience different tangential speeds depending on their
latitude. The space shuttle at Kennedy Space Center is already
traveling at 410 m/s (917 mi/h) before it even gets off the ground. A
geosynchronous satellite that is in a stationary orbit over the earth
is traveling at over 3080 m/s (6890 mi/h)
Someone moving south at 5 km/h on a train which itself is moving
south at 80 km/h will be moving at 85 km/h relative to the earth. If
the person had been moving north on the train their velocity would
have been 75 km/h relative to the earth.
Solving Kinematics Problems
1.
Assign a coordinate system – Define which directions are positive and
negative.
2.
Write down your known variables and show unknowns with a question
mark.
3.
Write down the kinematics expression that will allow you to solve for
one variable. All the others in your expression should be known.
Rearrange if necessary.
4.
Substitute numbers and units into your physical expression.
5.
Solve the equation for your unknown and include the correct units.
6.
Check your answer.
Does the magnitude of your answer make sense?
Do the units come out right?
Can you use another expression to check your answer?
Example 1 Distance Run by a Jogger
How far does a jogger run in 1.5 hours (5400 s) if his
average speed is 2.22 m/s?
t = 5400s
sav = 2.22 m/s
d=?
sav = d / t
So……
d = sav t
= (2.22 m/s)(5400s)
d = 12000 m
Check: Units come out right (m) when multiplied
Example 2 The World’s Fastest Jet-Engine Car
Andy Green in the car ThrustSSC set a world record of 341.1 m/s in
1997. To establish such a record, the driver makes two runs through
the course, one in each direction,to nullify wind effects. From the data,
determine the averagevelocity for each run.
a) t = 4.740 s x = +1609m vav = ?
vav = x / t = (+1609m) / (4.740 s)
vav = + 339.5 m/s
b) t = 4.695 s x = -1609m vav = ?
vav = x / t = (-1609m) / (4.695 s)
vav = - 342.7 m/s
Example 3 Acceleration and Increasing Velocity
Determine the average acceleration of the plane.
vi = 0 km/h
vf = 260 km/h
ti = 0s
tf = 29s
aav = (vf - vi) / (tf – ti)
aav = (+260 km/h – 0 km/h) / (29s – 0s)
aav = + 9.0 km/h /s
Graphical Representation of Motion
Kinematics Relationships Through Graphing:
1.
The slope of a d-t graph at any time tells you the av. velocity of
the object.
2. The slope of a v-t graph at any time tells you the av.
acceleration of the object.
3. The area under a v-t graph tells you the displacement of the
object during that time.
4. The area under a a-t graph tells you the change in velocity
of the object during that time
Constant Motion
60
20
10
18
40
30
20
8
acceleration (m/s)
16
velocity (m/s)
position (m)
50
14
12
10
8
6
4
10
6
4
2
2
0
0
0
1
2
3
4
5
6
0
0
1
2
time (s)
On the d-t graph at any point in
time …
vav = ∆d / ∆t
vav = (50 - 0)m / (5 - 0)s
vav = 10 m/s
The slope is constant on this graph
so the velocity is constant
3
4
5
6
0
1
2
time (s)
On the v-t graph at any point in
time… aav = vf - vi / t
aav = (10 - 10)m/s / (5 - 0)s
aav = 0 m/s2
Looking at the area between the
line and the x-axis….
Area of rectangle = b x h
Area = 5s x 10 m/s = 50 m
Which is of course displacement
3
4
5
6
time (s)
On the a-t graph the area between
the line and the x-axis is….
Area of rectangle = b x h
Area = 5s x 0 m/s2 = 0 m/s
The area thus represents….
∆v = aav ∆t
Change in velocity
Changing Motion
60
60
25
25
7
acceleration (m/s/s)
20
20
40
40
15
15
30
30
10
10
20
20
10
55
10
0
0
0
-100
8
velocity (m/s)
velocity (m/s)
position
position(m)
(m)
50
50
5
4
3
2
1
1
1
2
3
4
5
6
2
3
4
5
6
00
00
0
11
22
time
time (s)
(s)
On the d-t graph at any point in
time …
6
vav = ∆d / ∆t
The slope is constantly increasing
on this graph so the velocity is
increasing at a constant rate
The slope of a tangent line drawn
at a point on the curve will tell you
the instantaneous velocity at this
position
33
44
55
time
time(s)
(s)
On the v-t graph at any point in
time… aav = vf - vi / t
aav = (20 - 0)m/s / (5 - 0)s
aav = 4 m/s2
Looking at the area between the
line and the x-axis….
Area of triangle = 1/2 (b x h)
Area = 1/2 (5s x 20 m/s) = 50 m
Which is of course displacement
66
0
1
2
3
4
5
time (s)
On the a-t graph the area between
the line and the x-axis is….
Area of rectangle = b x h
Area = 5s x 4 m/s2 = 20 m/s
The area thus represents….
Change in velocity
6
To determine the velocity at any point in time you need to find the slope of the distance-time graph. This means that
you need to find the slope of the tangent line drawn at the point of interest. By selecting two points spaced evenly
on either side of the point of interest, a line can be drawn between them that has the same slope as the tangent.
(shown below).
Slope between 1s and 3s shows the
velocity at 2s
The velocity at 2s is p/t = (18m - 2m)/ (3s - 1s) = 8m/s
The acceleration is given by the
slope of the velocity-time graph.
Therefore:
a = v / t = 20m/s / 5s = 4m/s2
Example Problem
A student is late for the school bus. She runs east
down the road at 3 m/s for 30s, then thinks that
she has dropped her calculator so stops for 10s
to check. She jogs back west at 2 m/s for 10s,
stops for 5 s then accelerates uniformly from
rest to 4 m/s east over a 10 second period.
a) Sketch the velocity-time graph of the student’s
motion
b) Determine the total distance and displacement of
the student during this time
c) Determine the student’s average velocity during
this time
Velocity-Time Graph of the Student’s Motion
V-t Graph for Student Going To School
5
velocity (m/s)
4
3
2
1
0
-1 0
10
20
30
40
-2
-3
time (s)
50
60
70
Total distance traveled by the student is….
dtotal = d1 + d2 + d3 + d4 + d5
dtotal = s1t1 + s2  t2+ s3 t3 + s4 t4 + s5 t5
dtotal = (3m/s)(30s) + (0m/s)(10s) + (2m/s)(10s) +
………(0m/s)(5s) + (1/2(4m/s)(10s)
dtotal = 130 m
Total displacement by the student is….
dtotal =  d1 +  d2 +  d3 +  d4 +  d5
+ east - west
dtotal = v1t1 + v2  t2+ v3 t3 + v4 t4 + v5 t5
dtotal = (3m/s)(30s) + (0m/s)(10s) + (-2m/s)(10s) +
………(0m/s)(5s) + (1/2(4m/s)(10s)
dtotal = + 90 m (East)
Average velocity of the student is…..
vav = dtotal / ttotal
= + 90m East / 65s
vav = 1.4 m/s East
V-t Graph for Student Going To School
5
velocity (m/s)
4
3
2
1
+ 90 m
+ 20 m
0
-1 0
10
20
- 20 m50
30
40
-2
-3
time (s)
60
70
More Kinematics Equations that Make
Sense!
df = di + vav t
but
vav = (vi + vf) / 2
df = di + (vi + vf) /2 t
but
vf = vi + aav t
df = di + (vi + (vi + aav t) /2 t
df = di + vi t + 1/2 aavt2
df = di + (vi + vf) /2 t
or…
but
d = vi t + 1/2 aavt2
t = (vf - vi ) / aav
df = di + (vi + vf) /2 (vf - vi) /aav
d = (vi + vf) /2 (vf - vi) /aav
and…
vf2 = vi2 + 2aav d
So…
d = (vf2 - vi2 ) /2aav
Free Fall
1.
Free fall describes the motion of an object which is only under the influence of
gravity. I.e. a ball thrown upwards or dropped
2.
An object in free fall experiences a constant uniform acceleration of 9.8 m/s2 in the
downwards direction
3.
Kinematics equations can be used for solving free fall problems by replacing aav in
the expressions with g where g is 9.8 m/s2 downwards
vf = vi + g t
4.
vf2 = vi2 + 2g dv
dv = vi t + 1/2 g t2
Air resistance limits the time of free fall. Eventually a falling object will reach a
constant velocity downwards known as its terminal velocity
+
vi is negative
vi is positive
g is positive
g is negative
If you define up as the positive direction, g must
be negative because the velocity gets less
positive over time
+
If you define up as the negative direction, g
must be positive because the velocity gets less
negative over time
Freely Falling Bodies
Example 1
A Falling Stone
A stone is dropped from the top of a tall building. After 3.00s
of free fall, what is the displacement y of the stone?
Freely Falling Bodies
dv
a
?
- 9.8 m/s2
vf
vi
t
0 m/s
3.00 s
Freely Falling Bodies
dv
a
?
-9.80 m/s2
vf
vi
t
0 m/s
3.00 s
dv = vi t + ½ g t2
= (0 m/s)(3.00s) + ½ (-9.8 m/s2)(3.00s)2
= - 44.1 m
Freely Falling Bodies
Example 2 How High Does it Go?
The referee tosses the coin up
with an initial speed of 5.00m/s.
In the absence if air resistance,
how high does the coin go above
its point of release?
Freely Falling Bodies
dv
a
vf
vi
?
- 9.80 m/s2
0 m/s
+ 5.00 m/s
t
Freely Falling Bodies
2
dv
a
vf
vi
?
-9.80 m/s2
0 m/s
+5.00
m/s
vf =
vi 2
+ 2 g dv
dv 
0 m s    5.00 m s 



2 9.80 m s
vf  v
2
2
dv
t
2

2g
2
= 1.28 m
2
i