# PHYSICS CALL: MOTION

Motion is one of the fundamental concepts of physics. Motion
is defined as the change in position of a body, more specifically an object is
said to be in motion when it’s linear or angular displacement changes with time
examples include a car moving on the road, a bird flying in the air etc.

**TYPES OF MOTION**

1.
Translational Motion: this is when a body moves
from one point in space to another along a straight part. E.g. a car moving in a constant direction.

2.
Random motion: this a disorderly motion or an
irregular motion or a zigzag motion e.g. the motion of gas particles

3.
Circular or rotational motion: this is motion in
a circle about a center or axis e.g. rotating fan, the spinning wheel of a car.

4.
Oscillatory or vibration motion: This is a tro-and-
fro or periodic type of motion e.g. a swinging pendulum

**NOTE:**In real life situation, a number of objects undergo more than one type of motion at the same time e.g. A moving football rotates as it moves from one point in the field to another (a combination of rotational and translational motion). Also the earth rotates about its own axis, as its whole body moves from one position to another around the sun (translational and rotational).

**CAUSES OF MOTION**

Motion is caused by
the effect of force.

A force is that which causes a stationary object to move or
tend to move, it also changes the direction of a moving object and stops a
moving object.

**TYPES OF FORCE**

Contact force

Force field (non-contact force)

Examples of force field are

1 Gravitational force

2 Electrical force

3 Magnetic force

**DISTANCE**

Distance is the length between any two point mathematically
distance = speed x time. It is a scalar quantity.

**DISPLACEMENT**

This is the distance moved in a specified direction. It is a
vector quantity.

**DIFFERENCE BETWEEN DISTANCE AND DISPLACEMENT**

1
Distance is determined by magnitude only; direction is not involved while
displacement is by both magnitude
and direction.

2
Distance is a scalar quantity while displacement is a vector

Distance > he walked 45km

Displacement > He walked 45km north east.

**VELOCITY**

Velocity can be defined the rate of change of displacement
with time.

Average velocity =

__Distance Travelled__
Time taken

The SI unit is meter per seconds

= (M/S or MS

^{-1})
The dimension is LT

^{-1}it is a vector quantity**ACCELERATION**

This is the rate of change of velocity with time. It is a
vector quantity

Acceleration

__=__
Where a = acceleration

V = final velocity

U = initial velocity

T = time

**NOTE:**Acceleration =

Retardation =

Another name for retardation is deceleration.

The SI unit of acceleration is MS

^{-2}and the dimension is LT^{-2}**VELOCITY**

**SIMPLIFIED 3**A car moving round a circular racing track takes 120secs to do a lap of 8km. What is the velocity in km/hr?

Solution

Velocity = distance /time

S = 8km, t = 120 seconds, since we are asked to find the
velocity in km/h, we convert 120 seconds to hour.

=

**SIMPLIFIED 4**: A bus leaves Owerri at 10:30am and arrives at a town which is 600km from Owerri at 4:30 find the average velocity for the journey in

(A)
Km/h (B)
M/S

*Solution*

Distance covered = 600km

Time left = 10:30am

Time arrived = 4:30pm

Time taken = Time left – Time arrived

= 10:30am – 4:30pm = 6hrs

Convert 100km/h to M/S

**SIMPLIFIED 5**A cars has a velocity of 84km/hr, how far does it travel in ¼ mins.

Solution

How far means distance, make distance the subject formula

V = 84km/h but the SI unit of velocity is M/S thus we
convert 84km/h to M/S

= 84 x 1000 = 840
=23.33m/s, also convert time from mins. to seconds

60 x 60 36

t = x 60 = 15

S = 23.33 x 15

= 350m

ACCELERATION

**SIMPLIFIED 6**A car with an initial velocity of 20m/s travelled for 2 seconds and finally come to rest with a velocity f 40m/s what is the acceleration.

**Solution**

a = 10m/s

^{2}**SIMPLIFIED 7**A car starts from rest and reaches a velocity of 72km/h in 5 seconds

1 What is its acceleration?

2 Determine the car’s velocity after 12
seconds. If it maintained steady acceleration.

Solution

1.

**whenver a body starts from rest, the initial velocity u = 0**

*Note:*
U = 0 m/s, v = 72km/h

Convert 72km/h to m/s = 20m/s

T= 5 seconds

^{2}

2.

**EQUATION OF MOTION WITH UNIFORM ACCELERATION IN A STRAIGHT LINE**

A particle starts with initial velocity “U” accelerates
along a straight line with acceleration “a” and covers a distance “S” in time
“T”. When its velocity reaches a final velocity “V” we derive the equations of
motion thus.

**DERIVING EQUATIONS OF MOTION**

1.

*Derivation*

From acceleration,

Make v the subject
formula

Thus

2. V

^{2}= u^{2}+ 2as*Derivation*

From, make t the subject
formula

…………………………(1)

From, make t the subject
formula

…………………………(2)

Equating equation (1) and (2)

2as = (v-u) (v+u)

2as = v

^{2}-vu-vu-u^{2}
2as = v

^{2}-u^{2}**v**

^{2}=u^{2 }**+**

^{ }2as
3.

^{2 }
From , but

^{}

Substitute for the value of v

Thus,

2s = 2ut +at

^{2 }
Divide through by 2 thus,

^{2}
Thus the equations of motion in a straight line are

^{1. }

^{}^{2. }

^{2}**=u**

^{2}+2as^{3. }

^{}
These four equations
are used in solving problems associated with uniformly accelerated motion.
Using them, the following point should be noted.

1.
Ensure that all unit match i.e. v = m/s, s =m, a
= m/s

^{2}and t in seconds
2.
Each of
the equation contains four of the five variables u, v, s and t, you are
normally given the value of three of them and required to find one or both
of the unknowns

3.
The best
way to select which equation to use is to look at the problem or question to find out
which of the five variable is not given in addition to the one required. Then
find the equation which does not contain the variable that is not given. This
gives you the required equation.

**Simplified 8**A body starts from rest with an acceleration of 0.4ms

^{-2}find its velocity when it has moved a distance of 80m.

*Solution*

U =0ms

^{-1}(start from rest), V =? , S = 80m, a=0.4ms^{-2}
Using

^{2}=u^{2}+2as
v

^{2}= 0^{2}+ 2x0.4x80
v

^{2}= 0+ 0.8 x 80
v

^{2}= 64
= 8ms

^{-1}**SIMPLIFIED 9**if a train travels at 54km/h and then accelerates at 2ms

^{-2}for ¼ min, find the distance covered.

^{}

*Solution*

a = 2ms

^{-2}, = 15 seconds
u = 54km/h convert to m/s

=

^{-1}^{2}

S = 15 x15 x x2

S =225+225, = 450m

**SIMPLIFIED 10**A body moving in a straight line with an acceleration of 12ms

^{-}2

^{ }has a velocity of 80ms

^{1}after 5seconds find

1.
The initial velocity of the body

2.
The distance covered in the 9

^{th}seconds*Solution*

From

U = v-at,
u = 80-12 x10, 80-60

U = 20ms

^{-1}
Note this does not mean distance in 9 seconds rather 9

^{th}seconds thus distance in 9 seconds – distance in 8 seconds
= S

_{9th}= S_{9}– S_{8}
S

_{9}=^{2}
S

_{9}= 20(9) + x12 x 9^{2}, = 180 + 486 =666m
S8 = 20(8) + x 12 x 8

^{2}, = 160 + 384 = 544m
= 160 + 384 = 544m

Distance in 9

^{th}seconds = 666m – 544m = 122m**Deceleration**or

**Retardation**is a negative acceleration in problem solving, we change the “+” sign to “-“in the equation of motion

**SIMPLIFIED 11**A train slows or retards uniformly from 98m51 to 38m5

^{1}in 10 seconds find the declaration.

*Solution*

38 = 98 – 10a

= - 6ms

^{-2}**SIMPLIFIED I2**A car travelling at 20ms

^{-1}Undergoes retardation of 2ms

^{-2}when brakes are applied calculate the time taken to come to rest and the distance travelled from the place where the brakes where applied.

*Solution*

U = 20ms

^{-2}, 2ms^{-1}, v=0
Using, 0 =20 - 2t

2t = 20, t = 10seconds

(b) Distance,

S=20 x 10 - x 2 x 10

^{2}, S = 200 – 100
S =100m

**MOTION UNDER GRAVITY**

A body falling under the influence of gravity has uniform
accelerated motion .Acceleration due to gravity is defined as force of
attraction of the earth on a unit mass. It is a vector quantity, the valve of g
is constant and the valve is 9.8ms

^{-2 }or approximately 10ms^{-2}. The valve of g varies on the earth surface because the earth is not a perfect square. When a body is moving upwards, it’s acceleration due to gravity is (-g), but when the body is moving downwards, the acceleration is (+g).when a body is released from a height above the ground, the initial velocity U = 0, when it is thrown up its final velocity V = 0
Thus the equations of motion under gravity are given as follows.

**Vertically downward**

1.

2.

^{2}
3.

^{2}=u^{2}+2gh**Vertically upward**

1.

2.

^{2}
3 .V

^{2}= u^{2}– 2gh
g = 10ms

^{-2}or 9.8ms^{-2}**SIMPLIFIED 13**A palm fruit dropped to the ground from the op of a tree 45m tall. How long does it take to reach the ground? Take g = 10ms

^{-2}

*Solution*

U = 0ms

^{-1}, a = 10ms^{-2}
Using

^{2}
45 = 0 x t + x 10 x t

^{2}, 45 = 5t^{2},
T

^{2}= 9, = 3seconds**SIMPLIFIED 14**A body is thrown up vertically with a velocity of 20ms

^{-1}calculate

1.The maximum
height reached

2.The time to
reach the maximum height

*Solution*

U = 20ms

^{-2},v = 0, g = 10ms^{-2}
Using the equation v

^{2}= u^{2}- 2gh
0

^{2}= 20^{2}– 2x10h
0=400 – 20h

20h = 400

h = 20m

(b) v = u – gt, 0 = 20 – 10t,

-20 = - 10t, t = 2seconds

**SIMPLIFIED 14**A stone is thrown vertically upward from the ground with a velocity of 15ms

^{-1}calculate

1.
Maximum height
reached

2.
Time to reach
maximum height

3.Time to
reach the ground again after the stone is thrown up

4.Velocity
reached way to the maximum height (g = 10ms

^{-2})
Solution

U = 15ms

^{-1}, v = 0ms^{-1},g = 10ms^{-2}
Using the equation

V

^{2}= u2 – 2gh
0

^{2}= 15^{2}– 2x10h
0

^{2}= 15^{2}– 20h
0

^{2}= 225 – 20h
20h = 225

H =

(b) V = u - gt

0 = 15 – 10t

10t = 15, t =1.5seconds

(c) T = 2t, = 2 x 1.5

= 3 seconds

(d) ¾ of the maximum height is

¾ x 11.25 = 8.438m

New high = 8.438

Using the equation

V

^{2}= u^{2 }– 2gh
V

^{2}= 15^{2}– 2 x 10 x 5.438, V^{2}= 225 – 168.75
V2
= 56.25

v = 56.25 = 7.3m5

^{1}**SIMPLIFIED**15 A stone is thrown horizontally from the top of a building with a velocity of 15m5

^{1}if the height of the building is 60m calculate

Time of flight

Final velocity of the stone

Range

Solution

This is a horizontal motion thus

(a) H = u

H =

^{2}
2H = g

^{2}
= 12

V

^{2}= u2 + 2gh
V

^{2}= 02 +2x10x60
V

^{2}= 1200
V = 1,200
V = 34.64m5

^{1}
Range
is the horizontal distance

Range
= velocity x time

=
15x3.46 = 51.9m

**VELOCITY TIME GRAPH**

Velocity
time graph provides useful information about a moving object, such as
acceleration, retardation, distance covered and average speed of the object.

The
motion of the object can form shapes such as square, triangle, trapezium
rectangle, e.t.c or even a combination of 6100 or more of these shapes. Thus
the sum of areas of these shapes formed corresponds to the total distance
covered by the object. Hence we conclude that the total distance covered by the
object is = Area of the shape.

Simplified
16. A car starts from rest and acceleration uniformly at 0.6m5

^{2}for 10secs and then continues at a steady speed for a further 15seconds.
Draw
a velocity time graph of the case motion

Determine
the total distance travelled.

V = u + a

0+0.6x10 = 6m5

^{1}
(b) Total distance travelled = Area of trapezium

=

=

Simplified
16 A train starts from rest and accelerates uniformly a

1.5m3

^{2}for 5 seconds, before maintaining a steady speed for a further 15 seconds, after which it retards to rest to rest in 10seconds.
(a)
Draw the velocity time graph of the train’s motion

(b)
Find the distance travelled

(c)
What is the average speed of the train in m5

^{1}
Solution

V = u + a

V
= 0x1.5x5

V
= 7.5m/s

(B)

Distance
travelled = area of trapezium

=

=
½ (AB + 0D) x h

=

(C)

Average speed = Total distance

Total time

= 168.75 = 5.63

30

**FRICTION:**

Friction
is the fore which opposes relative motion when two solid bodies are in contact

F
& R

F
= NN, N =

Where
F = Static or limiting frictional force

R
= Normal reaction, contact, form each body exert on the other

N
= co-efficient of static friction which is detained as the ratio o the limiting
friction force and normal reaction (R)

Note
R = mg

Coefficient
of static friction has no 51 unit

For
a body to move, the applied force must be greater that the force of friction.

F - R = ma

**LAWS OF FRICTION**

Friction
depends on the nature of the surface

Friction
opposes motion

Friction
is independent of the area in contact

Frictional
force is proportional to the normal reaction (R)

**ADVANTAGES OF FRICTION**

Friction
between the ground and on feet makes walking possible

The
friction between the road and tyres of a car causes the car to stop when its
break is applied.

Without
friction we could not hold a pen with our hand

The
belt that drives the wheel in a machine depends on friction to make the wheel
to turn without slipping.

**DISADVANTAGES OF FRICTION**

Friction
reduce the efficiency of machine

It
leads to wear and tear on the money parts of machine

It
causes the heating of engine

**REDUCING FRICTION**

The
use of lubricants like oil, grease, air etc

The
use of balls and roller-bearings lubricants prevents direct contacts of the two
metals. Roller bearing replaces sliding friction with rolling friction which is
much less than sliding friction

Simplified 17 A black of wood of mass 6kg rest on a
horizontal table, a horizontal force of 25 is just enough to causes it to slide find

The
co-efficient of friction

The
angle of friction

Solution

F = 25

R = 6x10 = 60

Âµ = 25/60 = 0.42

Angle of friction = TanÎ¸ = Âµ

Î¸
= tan

^{-1}Âµ = Tan^{-1}0.42 = 22.8^{0}
Simplified 18 a body o mass 40kg is given an acceleration of
10m5

^{2}on a horizontal ground for which the co-efficient of friction is 0.5 calculate the force required to accelerate the body.
Solution

The equation of motion of the body is given by F- NR = MÎ¸

Where F is the applied force

R = mg = 40x10 = 400N

F = mÎ¸ + ÂµR

= 40x10+0.5 x 400

= 400 + 200 = 600Âµ

Simplified 19 A block of wood resting on a horizontal table
is just moved by a horizontal force of 15Âµ. If the cp-efficient of friction
between the table and block is 0.35, find the mass of the block?

Solution

Âµ = f/R but R = mg

Âµ = f/mg make m subject formula

M = f/mg = 15 =
4.3g

0.35x10

**MOTION ON AN INCLINED PLANE**

For motion on an inclined plane F(limiting friction force) =
mgsinÎ¸

R (Normal Reaction) = mgcosÎ¸

Thus Âµ = f/R = mgsinÎ¸ = tanÎ¸

MgcosÎ¸

Âµ= tanÎ¸

At equilibrium

Required force p = mgsinÎ¸ + F

But F = ÂµR = Âµ MgcosÎ¸

P = mgsinÎ¸ + ÂµMgcosÎ¸

Simplified 20 an object of mass 8kg rest on a wooden plane
inclined at 35

^{0}to the horizontal, it is found that the least force parallel to the plane which causes the object to slide up is 10SN. What is the co-efficient of sliding friction between the object and the wood (g = 10M5^{2})
Solution

Required force p = mgsinÎ¸ + F

105 = 8x10 sin 35 + f

105 = (80x 0.5735) +F

105 = 45.856 + f

F = 105 – 45.886

F = 59.114Âµ

But Âµ = f/R = 59.114 = 59.114

8x10cos 35 65.532

Âµ
= 0.90

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