Kinematics

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Aslevels Physics Notes

Topic 3:  Kinematics

Distance the distance travelled by a body is how far a body travels on a path during motion. Its SI unit is metre (m).

Displacement: is the change in position of an object. Displacement takes into account only where the motion starts and finishes; whether the motion was directly between these points or took a complex route has no effect on its value. The sign of the displacement indicates the direction in which the position has changed.
Displacement is measured in meters (m).

Speed and velocity

Speed and velocity are both quantities that give an indication of how fast an object moves or, more precisely, of how quickly the position of an object is changing. Both terms are in common use and are often assumed to have the same meaning. In physics, however, these terms are defined differently.
  • Speed is defined in terms of the distance travelled per unit time and so, like distance, speed is a scalar. Thus, a direction is not required when describing the speed of an object. 
     
  •  Velocity is defined in terms of the displacement per unit time (rate of change of displacement is velocity) and so is a vector quantity. The SI unit for speed and velocity is meters per second (ms-1 ); kilometers per hour (kmh-1 ) is also commonly used.
    The speed of a body is also magnitude of its velocity. Speed is a scalar quantity and velocity is a vector quantity.

Instantaneous speed and velocity

Instantaneous speed and instantaneous velocity give a measure of how fast something is moving at a particular moment or instant in time

Average speed and velocity

Average speed and average velocity both give an indication of how fast an object is moving over a time interval. The average speed of a car that takes 1 hour to travel 30 km is 30kmh-1 . However, this does not mean that the car travelled the whole distance at this speed. In fact, it is more likely that the car was moving at 60 kmh-1 for a significant amount of time, but some time was also spent not moving at all.  

Displacement – time graphs, (position – time graph)

We can represent the changing position of a moving object by drawing a displacement-time graph. The gradient (slope) of the graph is equal to its velocity. The steeper the slope greater the velocity. A graph like this can also tell us if an object is moving forwards or backwards. If the gradient is negative, the object’s velocity is negative – it is moving backwards.
Combining displacements: the process of adding two displacements together is obtained as vector addition. When two or more vectors are added together, their combined effect is known as the resultant of the vectors .

A position-time (displacement-time) graph indicates the position of an object at any time for motion that occurs over an extended time interval. However, the can also provide additional information.

Consider a swimmer laps of a 50 m pool. Her position-time data are shown in the table. The starting point is treated as the origin for this motion.

 An analysis of the table reveals several features of the swim. For the first 25 s, the swimmer swims at a constant rate. Every 5 s she travels 10 m in a positive direction, i.e. her velocity is +2 ms-1 . Then from 25 s to 35 s, her position does not change; she seems to be resting, i.e. is stationary, for this 10 s interval. Finally, from 35 s to 60 s she swims back towards the starting point, i.e. in a negative direction. On this return lap, she maintains a more leisurely rate of 5 metres every 5 seconds, i.e. ends 25 m from the start. These data are shown on a position-time graph.
The displacement of the swimmer can be determined by comparing the initial and final positions. Her displacement between 20 s and 60 s is for example:
 π‘₯ = π‘“π‘–π‘›π‘Žπ‘™ π‘π‘œπ‘ π‘–π‘‘π‘–π‘œπ‘› − π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™ π‘π‘œπ‘ π‘–π‘‘π‘–π‘œπ‘› = 25 − 40 = −15 m

By further examining the graph, it can be seen that during the first 25 s, the swimmer has a displacement of +50 m. thus her average velocity is +2 ms-1. This value can be obtained from this section of the gradient of the graph. 

Combining velocities: velocity is a vector quantity ad so two velocities can be combined by vector addition in the same way as for two or more displacements (vectors) 

Example:
An aircraft is flying due north with velocity of 200 ms-¹ . A side wind of velocity 50 ms-¹ is blowing due east, the resultant velocity of the aircraft can be obtained by vector addition of two velocities. The resultant vector will give the magnitude and direction of velocity.  

Acceleration

Acceleration is a measure of how quickly velocity changes. Acceleration is defined as the rate of change of velocity. Acceleration is a vector quantity whose direction is that of the velocity change.

A negative acceleration can mean that the object is slowing down in the direction of travel. 


A velocity-time graph can also be used to find the displacement of the body under consideration.
In the graph, the object is moving with a constant velocity of +3ms-1 for 4 s, then slows from 3 ms-1 to zero in the next 2 s. her displacement during this time can be determined from the v-t graph:

Equations of motion

here is set of equation which allows us to calculate the quantities involved when an object is moving with a constant acceleration. The quantities we are concerned with are: 
The five equations of motion relating the various quantities (mentioned above) which define the motion of a particle in a straight line with uniform acceleration are

These equations can only be used  when ;
  • For the motion is a straight line 
  • For an object moving with constant acceleration
When solving problems using these equations, it is important that you think about the problem and try to visualize what is happening.

The following steps are advisable.
1. Draw a simple diagram of the situation. 
2. Neatly write down the information that has been given in the question, using positive and negative values indicating directions. Convert all units to SI form. 
3. Select the equation that matches your data. 
4. Use the appropriate number of significant figures in your answer. 
5. Include units with the answer and specify a direction if the quantity is a vector 

Derivation of equations of motion

Equation 1:

The graph in the figure is a straight line, means that the object is moving with a constant acceleration. If the velocity of the object changes from u ms-1 to v ms-1 in a time interval of t, the acceleration is defined as



Here a is the gradient of the line. Rearranging this gives us the first equation of motion

Equation 2:

As we know that displacement is area under a velocity-time graph. Figure shows that the average velocity of an object is half way between u and v. so the object’s average velocity calculated by averaging its initial and final velocities, is given by: 
The object’s displacement is the shaded area in the figure. This is a rectangle, and so we have:
π‘‘π‘–π‘ π‘π‘™π‘Žπ‘π‘’π‘šπ‘’π‘›π‘‘ = π‘Žπ‘£π‘’π‘Ÿπ‘Žπ‘”π‘’ π‘£π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦ × π‘‘π‘–π‘šπ‘’ π‘‘π‘Žπ‘˜π‘’π‘›

Equation 3:

Look at the figure, displacement is area under the graph which consists of two parts i.e. one is a rectangle and the other is a triangle 
  • Area of rectangle is give as the product of base and height i.e.
    π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘Ÿπ‘’π‘π‘‘π‘Žπ‘›π‘”π‘™π‘’ = β„Žπ‘’π‘–π‘”β„Žπ‘‘ × π‘π‘Žπ‘ π‘’ = 𝑒𝑑 − − − − − −(1)
  • Area of triangle is given as
     π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ ∆= 1 2 × β„Žπ‘’π‘–π‘”β„Žπ‘‘ × π‘π‘Žπ‘ π‘’
    Where height of the triangle is (v-u) and base is t
    ∴ π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ ∆= 1/2 (𝑣 − 𝑒)𝑑
     π‘π‘’𝑑 𝑣 − 𝑒 = π‘Žπ‘‘ (π‘’π‘ž 1)
    ∴ π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ ∆= 1/2 (π‘Žπ‘‘)𝑑 = 1/2 π‘Žπ‘‘² − − − − − (2)
Now the total area under the graph is given by adding eq. 1 and 2 which represents the displacement S.

𝑆 = π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘Ÿπ‘’π‘π‘‘π‘Žπ‘›π‘”π‘™π‘’ + π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘‘π‘Ÿπ‘–π‘Žπ‘›π‘”π‘™e

Equation 4

This equation can also be derived from the same graph of eq. 3 i.e. Area under graph = area of bigger rectangle – area of triangle.

Equation 5:


Uniform and non-uniform acceleration

It is important to note that the equations of motion only apply to an object which is moving with a constant acceleration. If the acceleration was changing, you would not know what to put in the equations.
The velocity-time graph in the figure shows non-uniform acceleration. It is not a straight line; its gradient is changing (in this case, decreasing).
The acceleration at any instant in time is given by the gradient of the velocity-time graph. The triangle in the figure shows how to find acceleration at t=20 seconds. 
  •  At the time of interest, mark a point on the graph. 
  • Draw a tangent to the curve at that point. 
  • Make a large right-angled triangle, and use it to find the gradient. 

Displacement; (area under velocity-time-graph)

You can find the change in displacement of a body as it accelerates by determining the area under the velocity-time graph.
To find the displacement of the object in the above figure between t = 0 and 20 s, the most straightforward, but lengthy, method is just to count the number of small squares.  

Acceleration due to gravity

Even today, many people think that heavy objects fall faster than light ones. The cause of confusion is usually related to the effects of air resistance. Some objects are greatly affected by air resistance, for example a feather and a balloon. This is why these objects do not speed up as they fall. However if air resistance is ignored, all free-falling bodies near the surface of Earth will move with an equal downwards acceleration. In other words, mass of the object does not matter.


If you drop a ball or stone, it falls to the ground. The figure, based on a multiflash photograph, shows the ball at equal intervals of time. You can see that the ball’s velocity increases as it falls because the spaces between the images of the ball increase steadily. The ball is accelerating.

If we measure the acceleration of a freely falling object on the surface of the Earth, we find a value of about 9.81 ms-² . This is known as the acceleration of free fall, and is given the symbol g: 

The value of g depends on where you are on the Earth’s surface, but for examination purposes we take g = 9.81 ms-² . 

Motion in two dimensions – Projectile

When an object is thrown into the air at an angle to the vertical, and is moving freely under the gravity, it follows a parabolic path. Such type of two dimensional motion under gravity is called projectile motion. The path followed by the projectile is called its trajectory.
 Following are the characteristics of projectile motion:
  • Horizontal component of velocity of projectile is constant. i.e. horizontal acceleration is zero. 
  • Vertical component of velocity changes. 
  • Vertical component of velocity is zero at the highest point. 
  • Vertical component of acceleration is constant i.e. –g (-9.81ms-² ) throughout the motion. 
  • Magnitude of vertical component of velocity decreases from the point of projection till the highest point and then increases with constant acceleration.

Maximum height of projectile

If an object is projected in the air with initial velocity u at an angle πœƒ to the vertical then vertical component of

Time taken to reach the maximum height

Let 𝑑₁ be the time taken to reach the maximum height, then

Height 𝒉 of projectile at any time 𝒕. 

𝑆 = 𝑒𝑑 + 1/2 π‘Žπ‘‘²

β„Ž = (π‘’π‘†π‘–π‘›πœƒ)𝑑 − 1/2 𝑔𝑑²

Total time 𝑇 of flight

When the object lands on the other side of the ground, β„Ž = 0


 Range 𝑅 of a projectile

Horizontal component of velocity 𝑣𝐻 of a projectile is constant
                         π‘£π» = π‘’πΆπ‘œπ‘ πœƒ = π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘ 
β„Žπ‘œπ‘Ÿπ‘–π‘§π‘œπ‘›π‘‘π‘Žπ‘™ π‘‘π‘–π‘ π‘π‘™π‘Žπ‘π‘’π‘šπ‘’π‘›π‘‘ = π‘£π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦ × π‘‘π‘–π‘še
π‘₯ = (π‘’πΆπ‘œπ‘ πœƒ)t

Maximum horizontal displacement covered by a projectile is called its range, when time is 𝑇.
𝑅 = (π‘’πΆπ‘œπ‘ πœƒ)T

 Maximum range of a projectile

Maximum range of a projectile depends on the maximum value of 𝑆𝑖𝑛2πœƒ i.e. 𝑆𝑖𝑛2πœƒ = 1, π‘€β„Žπ‘’π‘› 2πœƒ = 90°
πœƒ = 45°

The result shows that the range of a projectile is maximum when the angle of projection is 45°.


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