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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/s2
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.      
Note: whenver a body starts from rest, the initial velocity u = 0
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. V2 = u2 + 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 = v2-vu-vu-u2
2as = v2-u2
  v2=u2 + 2as
3.2
From , but
Substitute for the value of v
Thus,
2s = 2ut +at2
Divide through by 2 thus,
2
Thus the equations of motion in a straight line are
1.               
2.                2=u2+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/s2 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=u2+2as
v2 = 02 + 2x0.4x80
v2 = 0+ 0.8 x 80
v2 = 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 80ms1 after 5seconds find
1.       The initial velocity of the body
2.       The distance covered in the 9th 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 9th seconds thus distance in 9 seconds – distance in 8 seconds
= S9th = S9 – S8
S9 = 2
S9 = 20(9) + x12 x 92, = 180 + 486 =666m
S8 = 20(8) + x 12 x 82, = 160 + 384 = 544m
= 160 + 384 = 544m
 Distance in 9th 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 38m51 in 10 seconds find the declaration.
Solution
38 = 98 – 10a
= - 6ms-2
              SIMPLIFIED I2 A car travelling at 20ms-1Undergoes 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 102, 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=u2 +2gh
Vertically upward
1.
2. 2
3 .V2 = u2 – 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 t2, 45 = 5t2,
T2 = 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 v2 = u2 - 2gh
02 = 202 – 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
V2 = u2 – 2gh
02 = 152 – 2x10h
02 = 152 – 20h
02 = 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
V2 = u2 – 2gh
V2 = 152 – 2 x 10 x 5.438, V2 = 225 – 168.75
V2 = 56.25
 v = 56.25 = 7.3m51
SIMPLIFIED 15 A stone is thrown horizontally from the top of a building with a velocity of 15m51 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 = g2
 

=  12               

V2 = u2 + 2gh
V2 = 02 +2x10x60
V2 = 1200
V = 1,200    V = 34.64m51

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 0.6m52 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 = 6m51

(b) Total distance travelled = Area of trapezium
=
=

Simplified 16 A train starts from rest and accelerates uniformly a
1.5m32 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 m51


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
µ = 25/60 = 0.42
Angle of friction = Tanθ = µ
  θ = tan-1 µ = Tan-1 0.42 = 22.80
Simplified 18 a body o mass 40kg is given an acceleration of 10m52 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 = Mθ
Where F is the applied force
R = mg = 40x10 = 400N
F = mθ + µR
= 40x10+0.5 x 400
= 400 + 200 = 600µ
Simplified 19 A block of wood resting on a horizontal table is just moved by a horizontal force of 15µ. If the cp-efficient of friction between the table and block is 0.35, find the mass of the block?

Solution
µ = f/R but R = mg
µ = 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) = mgsinθ
R (Normal Reaction) = mgcosθ
Thus µ = f/R =  mgsinθ = tanθ
                         Mgcosθ
 µ= tanθ
At equilibrium
Required force p = mgsinθ + F
But F = µR = µ Mgcosθ
 P = mgsinθ + µMgcosθ
Simplified 20 an object of mass 8kg rest on a wooden plane inclined at 350 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 = 10M52)
Solution 
Required force p = mgsinθ + F
105 = 8x10 sin 35 + f
105 = (80x 0.5735) +F
105 = 45.856 + f
F = 105 – 45.886
F = 59.114µ
But µ = f/R = 59.114         =         59.114
                     8x10cos 35            65.532
 µ = 0.90















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