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

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

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

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

From

, but

Substitute for the value of v

Thus,

2s = 2ut +at²

Divide through by 2 thus,

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

S = 15 x15 x

S =225+225, = 450m

SIMPLIFIED 10 A body moving in a straight line with an acceleration of 12ms^-2has a velocity of 80ms¹ 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 = S₉ – S₈

S₉ =

S₉ = 20(9) +

x12 x 9², = 180 + 486 =666m

S8 = 20(8) +

x 12 x 8², = 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 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^-2or 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

²=u² +2gh

Vertically upward

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

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

= 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 =

2H = g

= 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

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

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