MOTION IN A STRAIGHT LINE
MOTION IN A STRAIGHT LINE |
Motion is the change in position and orientation of an object. The motion of a rigid object (one which does not change shape) is made up translational motion or translation. i.e. movement of the centre of mass from one place to another and rotational motion or rotation i.e. movement around its mass. The study of the motion of points is called kinematics.
Linear motion or motion in a straight line
Linear or rectilinear motion is movement in a straight line and is the simplest form of translational motion. The linear motion of a rigid object is described as the motion of its centre of gravity.
Distance and displacement
Distance is the measurement of length between two points. It is a scalar quantity. The S I unit of distance is a metre. Other units include millimetres, centimetres, hectometres kilometres etc.
Displacement is defined as the distance from a fixed point to another point in a specified or given direction. It is also defined as the distance and direction of an object from a fixed reference point. It is a vector quantity. The position of an object can be expressed by its displacement from a specified point. The S I unit of distance is a metre. Other units include millimetres, centimetres, hectometres kilometres etc.
Speed and velocity
Speed is defined as the distance travelled per unit time. It is also defined as the distance an object travels in a certain length of time. It is a scalar quantity. If the speed of an object is constant, it is said to be moving with uniform speed. The average speed of an object over a time interval is the distance travelled by the object divide by the time interval. The instantaneous speed is the speed at any moment. The S I unit of speed is metre per second (m/s). Other units include kilometre per hour (km/h).
Thus, speed = distance travelled
Time taken
Velocity is defined as the displacement per unit time or the distance travelled in specified direction per unit time.
Velocity is the speed and direction of an object (i.e. its displacement in a given time). It is a vector quantity. If the speed of an object is constant, it is said to be moving with uniform velocity. The average velocity of an object over a time interval is the displacement or distance travelled by the object in specified direction divide by the time interval. The instantaneous velocity is the velocity at any moment. The S I unit of velocity is metre per second (m/s). Other units include kilometre per hour (km/h).
That is: velocity = displacement
Time taken
Velocity = distance travelled in a straight line
Time taken
Relative velocity is the velocity which an object appears to have when seen by an observer who may be moving. This is known as the velocity of an object relative to the observer.
Acceleration
Acceleration is the rate of change of velocity. That is the change of velocity of an object in a certain time. It is a vector quantity.
Thus, acceleration = change in velocity
Time
Or acceleration = final velocity - initial velocity
Time
Initial velocity is the starting velocity and final velocity is the finishing velocity. The S I unit of acceleration is metre per second squared m/s2).
Deceleration or retardation in one direction is acceleration in the opposite direction to the motion (negative acceleration).
An object whose velocity is changing the same amount in equal amount of time is moving with uniform acceleration.
In a situation where acceleration is not constant or acceleration of the object changes from time to time, the actual acceleration of the object will be the average acceleration.
Examples
- Change 20m/s into kilometre per hour
20 x 60 x 60 100072 km/h
- Change 72 km/h in m/s.
72 x 1000
60 x 60
20 m/s
Velocity – time – graphs
Imagine that a body moves with uniform velocity u for a time t.
From the formula of velocity:
Velocity = displacement
Time
It follows that:
Displacement = velocity x time
Therefore:
Distance = velocity x time
The symbol for distance is s, then
s = u x t
s = ut
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Graph 
u A B O C t
The area under the graph AB is the area of the rectangle OABC which is ut. It follows that the area under the velocity – time graph is the distance s.
Velocity - time graph of linear motion
Consider the diagram:
v
Graph  |
A B C t
O
From the diagram, OA represents a uniform acceleration of an object. It means that from O to A the object accelerated uniformly to a velocity v. From AB the object travelled with uniform velocity. This means the object maintained the same velocity v up to B, and BC represents a uniform deceleration or retardation.
In the above figure, the area under the graph OABC is the distance covered by the object.
Equations for uniformly accelerated motion
It should be noted that the velocity – time graph for a body which is uniformly accelerated is a straight line.
Consider the diagrams below:
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V B | | | | | |  | ||||||||
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v - u |
u A D
t | | ||
O C time T
In the above u is the initial velocity and v is the final velocity. If we are to find acceleration, by definition:
Acceleration = change in velocity
Time
Thus a = v – u
t
Now if we are to make velocity v the subject of the formula we obtain:
v = u + at this is the first equation |
We know that the area under the graph is distance travelled by an object. Consider the same figure above, this is also shown below:
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Graph V B | | | | | | | ||||||||
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v - u |
u A D
t |
O C time t
The area under the graph AB within time t is given by the area of triangle ABD plus the area of the rectangle OACD. Distance is always denoted by s.
Thus s = area of triangle ABD + area of rectangle OACD.
= 1/2 (v – u) t + ut
From the first equation of motion, which is v = u + at, it we replace v it follows that:
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s = 1/2 u + at - u t +ut
= 1/2 at x t +ut
= 1/2 at2 + ut
Thus s = ut + 1/2 at2
s = ut + 1/2 at2 this the second equation of motion |
The first and the second equations of motion can be combined to give the third equation of motion.
i.e. v = u + at
We square both sides of the equation we obtain:
v2 = (u + at) 2
Or v2 = u2 +2uat + a 2t 2
Or v2 = u2 + 2a (ut + ½ at 2)
But s = u t + 1/2 a t 2,
Thus v2 = u2 +2 a s
v2 = u2 +2 a s this is the second equation of motion |
Examples
- A body moving with a velocity of 30m/s is accelerated uniformly to a velocity of 50m/s in 5s. Calculate the acceleration of the body and the distance travelled by the body. Draw the velocity time graph represented by the body.
Given that: u = 30m/s
v = 50m/s
t = 5s
We can use the first equation of motion to find acceleration a.
i.e. v = u + at
50 = 30 + a x 5
50 – 30 = 5a
5a = 20
a = 4m/s2
We can use the third equation of motion to find distance s.
v2 = u2 +2as
50 x 50 = 30 x 30 + 2 x 4 x s
2500 = 900 + 8 s
8s = 2500 – 900
8s = 1600
s = 400 m
v
50m/s | |
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30m/s |
time 5s
ACCELERATION DUE TO GRAVITY
During the 16th century, Galileo Galilei disapproved the theory that the acceleration of a falling body depends on the mass of that body.
Light bodies are observed to fall down slowly because they are very much affected by air resistance. If a feather and a steel ball are released from the same height at the same time, the steel ball strikes the ground first followed by the feather which takes take to reach the ground. This is because the feather is very much affected by the air resistance. But if the feather and the steel ball are made to fall in a vacuum, they fall down at the same time. This is because there is no air resistance to affect the feather. Heavy objects are considered to fall freely because the air resistance on them is too small to be taken into consideration.
Equation of motion of free fall
If a body falls freely from rest, its acceleration a is equal to gravity g, and initial velocity u is equal to 0.
a = g, u = 0.
From the first equation v = u + at, it follows that the velocity of the body falling freely after t seconds is;
v = 0 + gt
v = gt acting as the first equation of motionThe distance travelled by the body is given by the equation;
s = u t + 1/2 a t 2 where a = g and u = 0.
s = 0 + 1/2 g t 2
s = 1/2 g t 2 acting the second equation of motionA body which is thrown vertically upwards has its initial velocity u and acceleration -g because the acceleration is against the gravity. The velocity of the body v and the distance s can be calculated using the formula;
v = u + at, where a = -g
Therefore, v = u – gt,
We know that;
s = u t + 1/2 a t2, where a = -g
Thus s = u t - 1/2 g t 2
The equation v2 = u2 +2as is applicable also to the bodies falling under gravity.
Since u = 0 and a = g
It follows that:
v2 = 0 +2gs
v2 = 2gs
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v = √ 2gs
The simple pendulum
This is a small heavy body suspended by a light inextensible string. This was first discovered by Galileo Galilei while he was in church. He discovered that the period of the swing of a lamp suspended from a roof was constant even if the oscillation was small. | |||||||||||||||||||||||||
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String  |
A B
Pendulum bob O
When the bob is displaced through a small angle, it swings through an oscillation A to B and back to A. the journey from point A to B and back to A is what is known as the oscillation. The time taken to complete the one oscillation (i.e. from A to B and back to A) is known as the period. The distance between the point of suspension of the string to the centre of gravity of the pendulum bob is called the length of the pendulum. When the pendulum swings the maximum displacement is called the amplitude and the angle between the string and the vertical axis is called the angular amplitude.
Given that the angular amplitude is small, the period of one oscillation depends of the length of the pendulum.
The formula for finding the period is: 
T = 2Π _l_ g
Where T is the period, l is the length of the pendulum and g is the acceleration due to gravity.
If we square both sides of the formula we obtain;
T2 = 4 Π2 l
g
Measuring g using a simple pendulum
Arrange the pendulum as shown in the figure above. Using a stop watch, record the time for 50 oscillations. Use the table below for recording.
Length(l) | Time for 50 oscillations | Period T(s) | T2(s)2 | ||
t1 | t2 | Average t(s) | |||
Plotting the graph of T2 against the length of pendulum l, a straight line is obtained.
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C 
A B |
The slope of the straight line formed is equal to;
4 Π2
g
therefore, g = 4 Π2 where slope = BC
slope AB
DYNAMICS |
Dynamics is the study of relationship between the motion of an object and the forces acting on it. A single force on an object causes it to change speed and and/or direction (i.e. accelerate). If two or more forces act and there is no resultant force, the object does not accelerate, but changes shape.  | ||
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Inertia
The tendency of an object to resist a change of velocity (i.e. to resist a force trying to accelerate it) is called inertia. It is measured as mass.
This is observed when you are in a bus. The bus tends to throw you behind because you resist motion when it has just started moving. It also throws you forward if it is stopped suddenly.
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Mass
A measurement of the inertia of an object is mass. The force needed to accelerate an object by a given amount depends on its mass – a larger mass needs a larger force.
Momentum
The mass of an object multiplied by its velocity is called momentum. Since velocity is a vector quantity, so is momentum.
Momentum = mass x velocity = mv
Impulse
Impulse is the force acting on an object multiplied by the time for which the force acts. From Newton’s second law of motion, impulse is equal to the change in momentum of an object. An equal change in momentum can be achieved by a small force for a long time or a large force for a short time.
Impulse = force x time = Ft
Since force is the rate of change of momentum, then
Impulse = change in momentum
Collision
An occurrence which results in two or more objects exerting forces on each other is called collision. This is not the everyday idea of a collision, because the objects do not necessarily have to be in contact.
Law of conservation on linear momentum
This law states that when two or more objects exert forces on each other (are in collision), their total momentum remains constant, provided no external forces act. If the time for the collision is very small and the system is considered just before and just after the collision, forces such as friction can be ignored.
Newton’s laws of Motion
There are three laws of motion formulated by Newton in the late 1770’s which relate force and motion.
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Object at rest | |||||||||
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Newton’s First Law of Motion
This law states that if an object is at rest or if its speed and direction are constant, then the resultant force on it is zero.
Newton’s Second Law of Motion
The second law of motion states that momentum of an object change, i.e. if it accelerates, then there must be a resultant force acting on it. Normally, the mass of the object is constant, and the force is thus proportional to the acceleration of the object. The direction of the acceleration is the same as the direction of the force.
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If mass remains constant, then:
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Newton’s Third law of Motion
This law states that if an action force exerts a force on the object, the object exerts an equal but opposite force to the action called the reaction. Forces always occur in equal and opposite pairs called the action and reaction. Thus if object A exerts a force on object B, object B exerts an equal but opposite force on A. these forces do not cancel each other out, as they act on different objects.
Self-check Exercise
20 Velocity m/s | |
Time (s)5 10
The velocity time graph in figure 1 above is for a car travelling along a straight road.
a) What is the acceleration of the car during the first 5 seconds?
b) How far has the car travelled after 5 seconds?
c) How far has the car travelled after 10 seconds?
- A car travels at a uniform velocity of 20m/s for 5 seconds. The brakes are then applied and the car comes to rest with uniform retardation in a further 8 seconds. Draw a sketch graph of velocity against time. How far does the car travel after the brakes are applied?
- A motor car moves from rest and attains a velocity of 20m/s in 12 seconds. It then maintained this velocity for 8 seconds. Then brakes are then applied and it stops in 15 seconds.
i. Draw the velocity time graph for the motion.
ii. What was the acceleration when the car started moving from rest?
iii. Find retardation of the car
iv. What was the total distance travelled by the car?
v. Find the total time taken for the journey from rest up to the time it was made to stop.
- A body at rest is allowed to fall freely under gravity. If the body strikes the ground with a speed of 50 m/s, calculate;
i. The time taken by the body to reach the ground
ii. The distance travelled by the body before it strikes the ground (Acceleration due to gravity = 10m/s2)
- For a body moving in a straight line with a uniform acceleration, which of the following will be a suitable graph to represent this motion?
A. Distance against time graph
B. Acceleration against time graph
C. Velocity time graph
D. Displacement against time graph
- A car moving uniformly at a velocity of 60 km/h decelerates uniformly and stopped after 5 seconds. The acceleration is:
A. 12 m/s2
B. -0.183 m/s2
C. 3.3 m/s2
D. – 3.3 m/s2
a) What is a free fall motion?
b) An iron ball is dropped from a tower of a certain building near a beach and takes 3.5 seconds to reach the ground. Find:
i. The velocity with which it strikes the ground.
ii. The maximum height of the tower.
a) Distinguish between speed and velocity
b) A car starts from rest and moves with a constant acceleration of 3 m/s2 until it reaches a velocity of 24 m/s. it maintains this velocity for ten seconds and the brakes to rest with retardation of 4 m/s2.
iii. Draw the velocity – time graph for this motion
iv. Find the total distance travelled
v. Find the average velocity for the journey
Labels: MOTION
posted by Michael Malika @ 2:41 AM
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