The concept of Kinematics is the study of motion without the force causing them and the mass of the body. Suppose a body of mass 2kg is pulled a distance of 10m by a force of 5N; kinematics studies only the movement in a giving direction without concern on the 5N force or the 2kg mass of the body.
The motion of a body can either be in any of these forms :
1. Random: Example is gaseous particles,
2. Translational: Example is a moving car,
3. Rotational: Example is a ceiling fan,
4. Oscillatory: Example is swinging pendulum.
STRAIGHT LINE MOTION (SLM).
This is the motion of a body in a straight line. There are four parameters involved in the study of this motion of a body in a straight line. They are Displacement or Distance, Velocity, Acceleration and Time.
This is the distance travelled in a specified direction. It is the shortest distance between two points. In kinematics it it represented with (s) and is mathematically related to velocity thus:
S = vt
Where: s = displacement
v = velocity
t = time
The unit of displacement is metre (m).
Example I. A car with a uniform velocity of 10 m/s left Enugu at 10:00 am and arrived Nsukka at 10:30 am. Calculate its displacement.
First lets write down the terms given to us. We have:
V = 10 m/s
T = (10:30 – 10:00) = 30 mins N:B we need to further change this time to be in seconds. So 30 mins would be:
(30 x 60) sec = 1800 s.
So from the formula, we have that
s = vt
= 10 x 1800
= 18000 m.
Velocity is speed measured in a specified direction. Many misunderstand Speed to be Velocity. In Physics, the two mean different things. Speed is the rate of change of distance with time. Velocity is a vector quantity (i.e it has both magnitude and direction) while speed is a scalar quantity ( it has magnitude but no direction) => these we shall treat soon still, always right here @ www.toscanyacademy.com. Don’t log out.
The unit of velocity is metre per second (m/s). Mathematically:
Velocity (v) = s/t
Where s = displacement = distance.
t = time.
Uniform velocity: A body is said to be in a uniform velocity when the ration (s/t) is constant. I.e unchanging.
Average Velocity (aV): Average velocity is the total distance travelled (Stotal ) divided by the total time taken (Ttotal).
aV = Stotal /Ttotal
Example I. Calculate the velocity of a car that covered a distance of 250 metres in 25 seconds.
We write down the given terms as always:
S = 250m
T = 25s
So, .’. V = s/t
= 10 m/s.
Acceleration is the rate of change of velocity with time. The unit of acceleration is metre per second squared (m/s2). Acceleration like velocity is a vector quantity, that is why it is wrong to think that acceleration means the rate of change of speed with time. Mathematically Acceleration (a) is:
a = v/t
Where v = velocity
t = time.
Uniform acceleration: If the rate of change of velocity is constant, the acceleration is said to be uniform. This implies that:
a = v/t = constant.
If the velocity of a body is increasing with time, it is said to be accelerating, but if the velocity of a body is decreasing with time, it is said to be retarding or decelerating or experiencing retardation or deceleration. And then in such a case,it is important we notice the acceleration is said to be negative.
Example I. A body rolled from point X to point Y in 2 seconds with a velocity of 5m/s. Calculate the acceleration of the car.
First, we note down our terms,
t = 2 s
v = 5 m/s
from the formula a = v/t
= 2.5 m/s2
THE EQUATIONS OF MOTION.
Deriving the equations of motion, it was noticed that if a body starts with initial velocity ‘u’ accelerates uniformly along a straight line with acceleration ‘a’ and covers a distance ‘s’ in a time ‘t’ when its velocity reaches a final value ‘v’, then the distance ‘s’ covered is given by s = average velocity X time.
s = u – v x t ………(equ I)
Also by definition, acceleration ‘a’ = rate of change of velocity and since ‘a’ is constant we will have
a = (v-u)/t = v = u + at ……………(equ II)
Eliminating ‘t’ from (equ I) and (equ II), we fined
v 2 = u 2 + 2as …………(equ III)
Eliminating ‘v’ from (equ I) and (equ II), we find
s = ut + 1/2 at 2 ……….(equ IV)
These four equations of motion are used in solving problems associated with uniformly accelerated motion. When using them, the following points should be noted.
I. Ensure that all the units match. i.e, ‘v’ in m/s , ‘s’ in m, ‘a’ in m/s2 and ‘t’ in s OR ‘v’ in km/h, ‘s’ in km, ‘a’ in km/h2 , ‘t’ in hours.
II. Each of the equations contains four of the five variables u, v, s, a, t. The value of three are normally given and the value of one or both unknowns are required to be found.
III. To determine which equation to use in solving a particular problem,
(a) note down the given variables,
(b) note down the required variable,
(c) note down the un-given variable
(d) then the equation to use is the one that does NOT contain the un-given variable.
‘equ I does not contain ‘a’
‘equ II does not contain ‘s’
‘equ III does not contain ‘t’
‘equ IV does not contain ‘v’
IV. Note the conversion formulae: 1km/h = 1000/(60 x 60)m/s OR 36km/h = 10m/s OR 1m/s = 3.6km/h.
V. Do not confuse ‘s’ for distance with s the unit of time.
Example I. A car moves from rest with an acceleration of 0.2m/s2 . Find its velocity when it has moved a distance of 50m.
We note down the given variables, which are:
a = 0.2m/s2
s = 50m
The required variable is ‘v’ and
The un-given variable is ‘t’ and the equation that doesn’t contain ‘t’ is equ III. So we use equ III.
v2 = u2 + 2as
We should also bear in mind that the initial velocity ‘u’ of a moving body which started at rest is zero.
So with that now, our ‘u’ here will be 0m/s.
.’. v2 = 02 + 2 x 0.2 x 50
v = squr(0 + 2 x 0.2 x 50)
A car is uniformly retarded and brought to rest from a velocity of 36km/h in 5sec. Find:
(a) it’s retardation
(b) the distance covered during this period.
We like before, write down the variebles. Then we have
V = 36km/h
t = 5s
s = ?
(a) First, we should notice that by retardation, we are required to find the accelaration. Then the formula for accelaration is
a = v/t
and here we have v = 36km/h, we will need to change it to m/s before we proceed. So 36km/h = (36 x 1000) / (60 x 60) = 36000/3600 = 10m/s
therefore we now have
a = 10/5
(b) Now we have the acceleration, we need to find the distance covered, so we use (equ IV). And I need somebody to tell us why we should use equ IV.
So from equ IV
s = ut + 1/2at2
N:B initial velocity (u) in this case is Zero, then
with that s = 0 + 1/2 x 2 x 25
Now lets have one or two problem set for you guys to solve.
A train slows from 108km/h with a uniform retardation of 5m/s2 . How long will it take to reach 18km/h?
Question II. What is the distance covered in question I?
Try and solve this questions and prove your answers in the comment box. Later I will provide the solutions to the questions when more than 5 students has solved it.
See you next time.
Hope we all learnt something useful today from this lesson. I will be happy to hear from you about today’s class. Your suggestion and contribution can help us keep improving in our future classes to come.
The following physics topics will be covered in all the classes
Lesson 3: Forces
Lesson 4: Momentum & Energy
Lesson 5: Circular & Satellite Motion
Lesson 6: Energy
Lesson 7: Wave Phenomena/Sound
Lesson 8: Optics
Lesson 9: Electricity
Lesson 10: Magnetism
Lesson 11: Thermodynamics
Lesson 12: Nuclear Energy
Lesson 13: Modern Physics
If there is something you feel we did not cover and it is important for us to do so, kindly let us know. The coming physics classes will be detailed with exercise questions and quiz test.
Keep in touch with www.toscanyacademy.com for more lessons. See you soon. Keep “physicsing!”