3506FETQR
Introductory Foundation Physics
Lecture 5
Kinematics 1
By
Dr. Vida Zadeh
Kinematics
Speed
Average Velocity
Instantaneous Velocity
Velocity – Time Graphs
Acceleration
Instantaneous Acceleration
Kinematics
We are familiar with speed and its units of kilometres per hour, metre per
second etc.
The speed is the ratio of the distance travelled, to the time required for the
travel.
Average speed is the total distance, s, divided by time taken, t, to travel
that distance:𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑆𝑝𝑒𝑒𝑑 =
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑙𝑒𝑑
𝑡𝑖𝑚𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙
=
𝑠
𝑡
If the average speed is the same for all parts of a trip, the speed is
constant.
“The speed of a body is constant, when it travels equal distance in equal
periods of time”
Edexcel Physics 1 – pp 15-16
Kinematics
Example:
How far will the tiger travel, if its speed is 31 m/s and
it travels for 3.2 seconds?
By multiplying both sides by of the definition of average
speed by time, we get the distance travelled
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 × 𝑡𝑖𝑚𝑒
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑆𝑝𝑒𝑒𝑑 × 𝑡𝑖𝑚𝑒 =
𝑡𝑖𝑚𝑒
= 31 𝑚 𝑠 × 3.2 𝑠 = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 99.2 𝑚 = 99 𝑚
Edexcel Physics 1 – pp 16-17
Average Velocity
Suppose a car is located at point 𝑥1 at a time 𝑡1 and another point 𝑥2
at a later time 𝑡2
The average velocity 𝑣 , over the time interval is:
𝑣=
𝑓𝑖𝑛𝑎𝑙 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 −𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛
𝑓𝑖𝑛𝑎𝑙 𝑡𝑖𝑚𝑒−𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑡𝑖𝑚𝑒
=
𝑥2 − 𝑥1
𝑡2 − 𝑡1
In the previous slide, we used the terms
initial position and final position,
measured from a zero point, in our
reference system of coordinates
Velocity
In determining velocity, we refer to a starting point
(initial position) and the destination point (final Position)
We set up a system of coordinates
Within this coordinate system, is a zero point , from
which all positions are measured. Frequently, the zero
point is chosen as the starting point for any motion.
Any change in position, including the sign of the change,
is called the displacement.
Kinematics
Most often, we have chosen the zero to be the starting point for the
motion. Note, we can have both positive and negative positions
measured from the origin.
In one dimension, if we define position measured
to the right of the origin being positive, then positions measured to
the left of the origin are negative. In this case, positive
velocity indicates motion to the right, while a
negative velocity indicates motion to the left.
-ve
+ve
Kinematics
Example: A pigeon is released from a roof in Rayyan street
(a) what is its average velocity ???
(i) If it travels 50 km due East from Rayan St. in one hour?
(ii) If it travels 50 km West from Rayan St. in one hour?
(iii) If it starts 10 km East of Rayan St., travels to 20 km East and
then travels a distance of 50 km West in one hour?
(b) What would the speed be in each of these cases?
final position − initial position
v=
elapsed time
Kinematics
Starts from here
0 ref Point
Solution
(a)From the statement of the problem, we see that all of the motion is confined along a
line in an East and West direction
(1) Make a diagram of the situation
(2) Choose a coordinate system, the positive and negative directions. In this case, East is
positive, West negative
(3) Write down the formula for average velocity
Notice that the duration is given by the difference between initial and final times, this is
the elapsed time
𝒗=
𝒇𝒊𝒏𝒂𝒍 𝒑𝒐𝒔𝒊𝒕𝒊𝒐𝒏 −𝒊𝒏𝒊𝒕𝒊𝒂𝒍 𝒑𝒐𝒔𝒊𝒕𝒊𝒐𝒏
𝒇𝒊𝒏𝒂𝒍 𝒕𝒊𝒎𝒆−𝒊𝒏𝒊𝒕𝒊𝒂𝒍 𝒕𝒊𝒎𝒆
=
𝒙𝟐 − 𝒙𝟏
𝒕𝟐 − 𝒕𝟐
Kinematics
The difference between speed and velocity is more than just an algebraic sign; it involves
the difference between the total DISTANCE travelled (for speed) and the net change in position
(for velocity) = DISPLACEMENT.
a) So for the 3 cases
i) 𝑣 =
50 𝑘𝑚 − 0 𝑘𝑚
= +50 𝑘𝑚/ℎ
1ℎ
ii) 𝑣 =
−50 𝑘𝑚 − 0 𝑘𝑚
= −50 𝑘𝑚/ℎ
1ℎ
−30 𝑘𝑚 − 10 𝑘𝑚
= −40 𝑘𝑚/ℎ
1ℎ
Notice the average speed is always a
positive number
iii) 𝑣 =
b) The average speed is found from
50 𝑘𝑚
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑆𝑝𝑒𝑒𝑑 =
= 50 𝑘𝑚/ℎ
1ℎ
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑆𝑝𝑒𝑒𝑑 =
50 𝑘𝑚
= 50 𝑘𝑚/ℎ
1ℎ
For the third case, the distance travelled is 10 km East, plus
50 km West, so a total distance of 60 km. Thus
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑆𝑝𝑒𝑒𝑑 =
60 𝑘𝑚
= 60 𝑘𝑚/ℎ
1ℎ
In situation (iii) the magnitude of the displacement and the total distance travelled are not the same.
Consequently, the numerical value of the average speed is not the same as that of the average velocity
𝑨𝒗𝒆𝒓𝒂𝒈𝒆 𝑺𝒑𝒆𝒆𝒅 =
𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒕𝒓𝒂𝒗𝒆𝒍𝒍𝒆𝒅
𝒆𝒍𝒂𝒑𝒔𝒆𝒅 𝒕𝒊𝒎𝒆
A Graphical Interpretation of Velocity
Consider two people:- one running, A, and one walking, B,
with constant velocity.
We plot the elapsed time along the horizontal axis
and the velocity
along the vertical axis.
In this case, the horizontal line 𝑣𝐴 represents the particular
constant velocity of the runner. A slower moving person, but one with constant velocity, also gives
rise to the horizontal line 𝑣𝐵
We can also plot the distance travelled against the elapsed time for the two cases (see below)
Our definitions of both velocity and speed,
show that the distance at constant velocity
or speed, is directly proportional to the time.
When one variable is directly proportional
to the other, e.g. as distance is to time, a straight line results, i.e. :-
a linear relationship
Edexcel Physics 1 – pp 18-19
A Graphical Interpretation of Velocity
In the below graph, we plot distance against time for a case in which the velocity is not
constant.
Over the portion of the trip between times 𝑡1 and 𝑡2 , the line is straight and the velocity
is constant.
Between times 𝑡2 and 𝑡3 , the velocity remains the same and is still constant.
Between times 𝑡4 and 𝑡5 , the velocity is constant, but is not the same, as
between times 𝑡1 and 𝑡2 , or between 𝑡2 and 𝑡3
The velocity is not constant, for the time interval
between 𝑡3 and 𝑡4 , neither
is it constant between 𝑡1 and 𝑡5
So, we can define the velocity for any time interval,
𝑡1 to 𝑡2 , or 𝑡2 to 𝑡3 or any other interval.
A Graphical Interpretation of Velocity
To use our graphical method further, we need a
more formal definition for the slope of a line.
Suppose an object moves with constant velocity,
so that its position-time graph, is like the graph
on the right.
If the object was at point 𝑥1 at time 𝑡1 and at another point 𝑥2
at a later time, 𝑡2 , then the net change in position is given by
∆𝑥 = 𝑥2 − 𝑥1
(∆ 𝑖𝑠 𝑡ℎ𝑒 𝐺𝑟𝑒𝑒𝑘 𝑐𝑎𝑝𝑖𝑡𝑎𝑙 𝑙𝑒𝑡𝑡𝑒𝑟, 𝑑𝑒𝑙𝑡𝑎 , used to indicate a change in something)
Similarly, the change in time is given as ∆𝑡 = 𝑡2 − 𝑡1 the average velocity of the object during the
time interval from 𝑡2 − 𝑡1 , is then written:
𝑣=
𝑥2 − 𝑥1
∆𝑥
=
𝑡2 − 𝑡1
∆𝑡
Instantaneous Velocity
Suppose that a runner’s velocity is not constant, but changes as illustrated in the displacement-time
graph
Now, that the line is curved rather straight, so how do we find the velocity at a particular instant of
time?
To determine the velocity at point 𝐴 for instance, we draw a tangent to the curve at that point.
The slope of that tangent, which is a straight line, is given by 𝑣𝐴 = ∆𝑥𝐴 ∆𝑡𝐴 . The subscripts refer to
the line that is tangent to the curve at point 𝐴.
The velocity measured at any given moment, for example point 𝐴, is called the instantaneous
velocity.
Instantaneous Velocity
Example:
At the start of a 100 m race a sprinter is poised for action. When the starter fires his gun, the
sprinter pushes off the starting block and quickly reaches maximum velocity. The graph shows the
displacement of a sprinter as a function of the time elapsed, after the starting signal. Determine the
instantaneous velocity of the sprinter at 0.5 s, 1.5 s and 2.5 s.
Instantaneous Velocity
Example:
At the start of a 100 m race a sprinter is poised for action. When the starter fires his gun, the
sprinter pushes off the starting block and quickly reaches maximum velocity. The graph shows the
displacement of a sprinter as a function of the time elapsed, after the starting signal. Determine the
instantaneous velocity of the sprinter at 0.5 s, 1.5 s and 2.5 s.
Solution
The results can be read directly from the graph. We have drawn
tangent lines through the points corresponding
to t = 0.5 s, 1.5 s and 2.5 s
t = 0.5 s , we have
𝑣 𝑡 = 0.5 𝑠 =
∆𝑥
2.0 𝑚
=
= 4.0 𝑚/𝑠
∆𝑡
0.50 𝑠
t = 1.5 s
𝑣 𝑡 = 1.5 𝑠 =
4.0 𝑚
= 8.0 𝑚/𝑠
0.50 𝑠
t = 2.5 s
𝑣 𝑡 = 2.5 𝑠 =
4.8 𝑚
= 9.6 𝑚/𝑠
0.50 𝑠
Velocity–Time Graph
With reference to the runner or walker, moving at constant speed, but different
velocities, we can use a velocity-time graph to analysis their motion.
The height of the vertical line from the
horizontal axis to the line 𝑣𝑎 is
equal to the velocity. Since v = distance/time, so the
net distance travelled = v x t
Looking at this graph, you can see that v x t is actually
equal to the area of the rectangle under the line v.
This quantity is the net distance travelled 𝑠,
which is the displacement
We can extend this observation to state a
general principle:The area under any velocity-time curve between
two times, is equivalent to the displacement during
that time interval.
Velocity–Time Graph
Since the area under the velocity-time curve represents the displacement of the moving body,
we can deal with non-uniform velocity.
Figure (a) represents a case, in which the velocity decreases
from some initial value 𝑣0 to zero in a linear way.
The area under the curve represents the distance to stop.
This area is half times the height times the base, so the distance
1
travelled is = 𝑣0 𝑡
2
In (b), we are starting from rest and smoothly bringing,
say a car ,up to a final velocity 𝑣𝑓
The distance travelled during the time the velocity
1
increasing is given by = 𝑣𝑓 𝑡
2
Average Velocity and Average Acceleration
We define the average velocity of an object as its change in position divided by time
elapsed
∆𝒙
𝒗=
∆𝒕
This tells us how the object position changes with time
It is reasonable to define a quantity that indicates how the object’s velocity changes
with time.
We define the average acceleration 𝑎 as the change in velocity divided by the time
required for the change
The average acceleration can be written as:-
𝑎=
𝑣2 − 𝑣1
∆𝑣
=
𝑡2 − 𝑡1
∆𝑡
Acceleration
Example:
Average acceleration
A cyclist starts from rest and increases her velocity
at a constant rate, until she reaches a speed
of 2.0 𝑚 𝑠 in 5.0 s. What is her average acceleration?
Acceleration
Example:
Average acceleration
A cyclist starts from rest and increases her velocity
at a constant rate, until she reaches a speed
of 2.0 𝑚 𝑠 in 5.0 s. What is her average acceleration?
Solution:
We are given a change in velocity over a time interval, so we employ 𝑎 =
2.0 𝑚/𝑠 − 0𝑚/𝑠
𝑚/𝑠
𝑎=
= 0.40
5.0 𝑠
𝑠
𝑎 = 0.40 𝑚/𝑠 2
𝑣2 − 𝑣1
𝑡2 − 𝑡1
=
∆𝑣
∆𝑡
Acceleration
Example:
A motorcyclist starts from rest and accelerates in one direction to a constant acceleration of
𝟒. 𝟎 𝒎/𝒔𝟐 for 10 s. What is the velocity of the rider at the end of the 10 s?
Edexcel Physics 1 – pp 17
Acceleration
Example:
A motorcyclist starts from rest and accelerates in one direction to a constant acceleration of
𝟒. 𝟎 𝒎/𝒔𝟐 for 10 s. What is the velocity of the rider at the end of the 10 s?
Solution:
Here, we are given a constant acceleration and a time interval and need to find the change in
velocity.
We rearrange the formula as:𝒂=
𝒗𝟐 − 𝒗𝟏
𝒕𝟐 − 𝒕𝟏
𝒗𝟐 = 𝒂 𝒕𝟐 − 𝒕𝟏 + 𝒗𝟏
Inserting the values, we get:𝒗𝟐 = (𝟒. 𝟎 𝒎 𝒔𝟐 × 𝟏𝟎 𝒔) + 𝟎 = 𝟒𝟎 𝒎/𝒔
Edexcel Physics 1 – pp 17
Instantaneous Acceleration
A graphical approach offers further insight into the relationships
between acceleration, velocity, time, and distance.
From the definition of average acceleration, we see that acceleration
may also be described as the slope of the curve of velocity versus time.
In Figure (a), the horizontal line (zero slope) corresponds to an object
moving with constant velocity and therefore zero acceleration.
In Figure (b), the slope of the line is everywhere the same indicating
that the velocity is increasing at a constant rate. Therefore, the
acceleration is positive and constant.
In Figure (c):- How can we determine the acceleration at any
particular instant of time?
We define the instantaneous acceleration in a manner similar
to the way we define the instantaneous velocity.
The instantaneous acceleration is the limiting value of ∆𝑣 ∆𝑡
as the time interval becomes vanishingly small.
𝒂=
∆𝒗
∆𝒕
Instantaneous Acceleration
Example: Acceleration of a sprinter.
A sprinter at rest (𝑣 = 0) at the state of a race, quickly accelerates to maximum
velocity. Determine the acceleration of the sprinter at 0.5 s, 1.5 s and 2.5 s.
At t = 0.5 s
𝑎 𝑡 = 0.5𝑠 =
∆𝑣
∆𝑡
=
3.0 𝑚 𝑠
0.5 𝑠
= 6.0 𝑚 𝑠 2
At t = 1.5 s, we have:∆𝑣 1.25 𝑚 𝑠
𝑎 𝑡 = 1.5𝑠 =
=
= 2.5 𝑚 𝑠 2
∆𝑡
0.50 𝑠
At t = 2.5 s, we have:𝑎 𝑡 = 2.5𝑠 =
∆𝑣 0.40 𝑚 𝑠
=
= 0.80 𝑚 𝑠 2
∆𝑡
0.50 𝑠
Summary
• Distance-time graph: its gradient or slope, gives
you information about the speed and velocity.
• Velocity-time graph: we can use it to get two
pieces of information:1. Its gradient or slope, gives you the acceleration
or deceleration.
2. The area under the velocity-time graph gives the
total distance moved.
What did we learn today?
Speed
Average Velocity
Instantaneous Velocity
Velocity – Time Graphs
Acceleration
Instantaneous Acceleration