Motion

1. Kinematics:

• Kinematics is the study of motion without considering forces. It deals with the description of motion, including displacement, instantaneous speed, average speed, velocity, and acceleration.

• a) Displacement, instantaneous speed, average speed, velocity and acceleration:

• ⇒ Displacement:
• Displacement is the distance between an object’s initial and final positions (the shortest distance between two points). It’s a vector quantity, meaning it has both magnitude (amount of displacement) and direction.
• 
• Figure 1 The shortest distance between two points
• Formula:
• [math]Displacement (d) = final position (x_2) – initial position (x_1)[/math]
• [math]d = x_2 – x_1[/math]
• Units:
• meters (m)
• Displacement is different from distance traveled. Distance traveled is the total length of the path an object follows, while displacement is the straight-line distance between the starting and ending points.
• Example:
• An object moves from point A ( [math]x_1[/math] = 2 m) to point B ([math]x_2[/math]  = 6 m). Find the displacement.
• Given data:
• Point A = [math]x_1[/math] =  2m
• Point B= [math]x_2[/math] =  6m
• Find data:
• Displacement = d =?
• Formula:
• [math]Displacement (d) = final position (x_2) – initial position (x_1)[/math]
• [math]d = x_2 – x_1[/math]
• Solution:
• [math]d = x_2 – x_1[/math]
• [math]d=6-2[/math]
• [math]d=4m[/math]
• The displacement is [math]4m[/math](meters) in the direction from point A to point B.
• ⇒ Speed and velocity:
• Speed and distance are both scalar quantities because they have only a size (magnitude); velocity and displacement are vector quantities because they have both a size and a direction.
• [math]\text{Average speed} = \frac{\text{total distance traveled}}{\text{time taken}}
  [/math]
• [math]\text{Average velocity} = \frac{\text{total displacement}}{\text{time taken}}
  [/math]
• Example:
• “A car travels from city A to city B and returns to city A in 4 hours, covering a total distance of 240 km.”
• Given data:
• Distance = d = 240 km
• Displacement = d = 0 (due to car return its initial position)
• Time taken = t = 4h
• Find data:
• Average speed = s =?
• Average velocity = v =?
• Formula:
• [math]\text{Average speed} = \frac{\text{total distance traveled}}{\text{time taken}}
  [/math]
• [math]\text{Average velocity} = \frac{\text{total displacement}}{\text{time taken}}
  [/math]
• Solution:
• a) Average speed
• [math]\text{Average speed} = \frac{\text{total distance traveled}}{\text{time taken}}
  [/math]
• [math]\text{Average speed} = \frac{240}{4}
  [/math]
• [math]Average speed=60 km/h[/math]
• b) Average velocity
• [math]\text{Average velocity} = \frac{\text{total displacement}}{\text{time taken}}
  [/math]
• [math]\text{Average velocity} = \frac{0}{4}
  [/math]
• [math]Average velocity = 0[/math]
• ⇒ Acceleration:
• Acceleration is the rate of change of velocity, its unit meter per second per second [math]\frac{m}{s^2}
  [/math] and it’s a vector quantity.
• Formula:
• [math]\text{Acceleration} = \frac{\text{Change in velocity}}{\text{time taken}}
  [/math]
• Example:
• A car may accelerate from 2.0 ms^-1 to 14.0 ms^-1 over a period of 4 seconds and find its acceleration.
• Given data:
• Change in velocity =[math]V_f – V_i = 14 – 2 = 12 \, \text{ms}^{-1}
  [/math]
• Time taken = t = 4 s
• Find data:
• Acceleration = a =?
• Formula:
• [math]\text{Acceleration} = \frac{\text{Change in velocity}}{\text{time taken}}
  [/math]
• Solution:
• [math]\text{Acceleration} = \frac{\text{Change in velocity}}{\text{time taken}}
  [/math]
• [math]\text{Acceleration} = \frac{12}{4}
  [/math]
• [math]Acceleration=3 ms^(-2)[/math]

b) Graphical representations of displacement, speed, velocity and acceleration

• ⇒ Graphical representations of displacement-time:
• Look at Figure 2, which represents the motion of a long-distance runner during a training run.
• 
• Figure 2 Displacement – time graph
• The total distance covered was 24 km – that is, 12 km outwards and 12 km back home.
• The total displacement was zero – he is a distance of 0 km from his starting point at the end of the run.
• The average velocity of the runner was zero because the displacement is zero and
• [math]\text{average velocity} = \frac{\text{total displacement}}{\text{time}}
  [/math]
• The average speed was 6 kmh⁻¹ because 24 km was covered in a time of 4 hours.
• ⇒ Velocity-time graph:
• As the graph shows, the velocity is constant throughout the interval.

  
  Figure 3 Velocity-time graph

• No particles of matter how much the time changes, the velocity will be c at every instant. In this case, we have taken the initial velocity to be positive.
• The graph will be different if the initial velocity is negative.
• Another to illustrate the velocity -time graph (figure 4)
• The gradient of a velocity–time graph represents acceleration. A horizontal line represents a constant velocity. A line along the x -axis at a value of zero on the y -axis represents zero velocity.
• An acceleration is calculated using:
• [math]\text{Acceleration} = \frac{\text{change in velocity}}{\text{time taken for the change}}
  [/math]
• A diagonal line with a positive gradient represents a constant acceleration.

  
  Figure 4 Velocity-time graph

• A diagonal line with a negative gradient represents a constant negative acceleration.
• A curve indicates the velocity change is not uniform – the gradient of the tangent to the curve at any point gives the rate of change of velocity at that instant, i.e., the acceleration.

2. Linear motion:

• When a resultant force acts on a body, the body will accelerate. The force may cause the body to speed up or to slow down.
• If the change in velocity of the object is the same for each second of its motion, then we say that the body has a constant acceleration.
• If drag forces are ignored and we drop a rock from the top of a cliff, its velocity will increase by the same amount each second as it falls.
• The force is due to its weight, and the typical acceleration is equal to 9.81 ms^-2 although this value does vary slightly over the Earth.
• a) Constant acceleration in a straight line:
• For a body moving with constant acceleration, the distance travelled from a starting point plotted against time will be a parabola (Figure 5).
• 
• Figure 5 Graph of displacement against time for constant acceleration
• Its velocity will vary in a linear fashion with time (Figure 6)
• 
• Figure 6 Graph of velocity against time for constant acceleration
• From figure 5 and 6 the acceleration will be constant (Figure 7).
• 
• Figure 7 Graph of acceleration against time for constant acceleration
• The calculations involved in interpreting such events requires the ability to work from algebraic equations.
• ⇒ Equations of motion:
• When dealing with constant acceleration over a given time, we always use the standard algebraic symbols shown in Table 1
• Table 1 Some quantities which are use in equations of motion

• The equations of motion that use these symbols are often referred to as the ‘suvat’ equations. However, their correct name is ‘equations of motion’ or ‘kinematic equations’. You need to be able to apply these four equations to any kinematics problem.
• [math]v = u + at \qquad s = \frac{1}{2} (u + v)t
  [/math]
• [math]s = ut + \frac{1}{2}at^2  \qquad v^2 = u^2 + 2as [/math]
• ⇒ Techniques and procedures used to investigate the motion and collisions of objects
• some techniques and procedures used to investigate the motion and collisions of objects according to kinematics equations:
• – Identify the known quantities: List the given information, such as initial velocity, final velocity, displacement, time, and acceleration.
• – Choose the relevant kinematics equation: Select the appropriate equation based on the known quantities, such as:
• d = vt(distance = velocity x time)
• v = u + at(final velocity = initial velocity + acceleration x time)
• [math]v^2 = u^2 + 2as [/math](Final  = initial  + 2 x acceleration x displacement)
• [math]s = ut + \frac{1}{2}at^2[/math](displacement = initial velocity x time + 0.5 x acceleration x )
• – Rearrange the equation: Rearrange the equation to solve for the unknown quantity.
• – Plug in the values: Substitute the known values into the equation.
• – Solve for the unknown: Calculate the unknown quantity.
• – Verify the units: Ensure the units of the answer are consistent with the problem.
• – Use graphs and diagrams: Visualize the motion using graphs and diagrams to aid in understanding and solving the problem.
• – Consider collision types: Identify the type of collision (elastic, inelastic, or perfectly inelastic) and apply the appropriate conservation laws (momentum, energy, or both).
• – Apply conservation laws: Use the laws of conservation of momentum and energy to analyze collisions.
• – Check the answer: Verify the answer is reasonable and consistent with the problem’s context.
• b) i) Acceleration g of free fall:
• The acceleration of free fall, commonly referred to as g, is approximately:
• – g = 9.8 meters per second squared ([math]\frac{m}{s^2}[/math]) on Earth’s surface. This is the acceleration due to gravity that an object experiences when it is in free fall, meaning it is falling under the sole influence of gravity, without any other forces acting on it.
• Note that the value of g can vary slightly depending on factors such as:
• – Altitude (g decreases with increasing altitude)
• – Latitude (g varies slightly with latitude due to Earth’s slightly ellipsoidal shape)
• – Local geology (g can vary slightly depending on the density of the surrounding terrain)
• However, for most practical purposes, the standard value of g = 9.8[math]\frac{m}{s^2}[/math] is used.
• ii) Techniques and procedures used to determine the acceleration of free fall using trapdoor and electromagnet arrangement or light gates and timer:
• To determine the acceleration of free fall using a trapdoor and electromagnet arrangement or light gates and timer, follow these techniques and procedures:
• Trapdoor and Electromagnet Arrangement:
• – Set up the trapdoor and electromagnet arrangement, ensuring the trapdoor is level and the electromagnet is calibrated.
• – Place an object on the trapdoor, connected to a spring or a release mechanism.
• – Release the object, and the trapdoor will open, allowing the object to fall.
• – The electromagnet will measure the object’s velocity at specific points during its fall.
• – Use the velocity data to calculate the acceleration due to gravity (g).
• Light Gates and Timer:
• – Set up two light gates, spaced apart, and a timer.
• – Place an object on a track or a drop mechanism, ensuring it will pass through both light gates.
• – Release the object, and the timer will start when the object passes through the first light gate.
• – The timer will stop when the object passes through the second light gate.
• – Measure the time (t) and distance (d) between the light gates.
• – Calculate the acceleration due to gravity (g) using the equation: g = 2[math]\frac{d}{t^2}[/math] .
• Common Procedures:
• – Ensure accurate measurements and calibration of equipment.
• – Repeat the experiment multiple times to ensure reliable data.
• – Use data analysis software to calculate the acceleration due to gravity (g).
• – Compare results with the accepted value of g (9.8[math]\frac{m}{s^2}[/math] ) to verify accuracy.
• By following these techniques and procedures, you can accurately determine the acceleration of free fall using either the trapdoor and electromagnet arrangement or light gates and timer.

3. Projectile motion:

• ‘Free fall’ is the acceleration of a body under the action of a gravitational fi eld, with air resistance and buoyancy being ignored. Objects of different masses fall at the same rate under the influence of gravity.
• Projectile motion is a type of motion where an object is thrown or launched into the air and follows a curved trajectory under the influence of gravity.
• 
• Figure 8 Projectile motion (a) The trajectory of a kicked football  (b) Ball cover the maximum displacement at 45 degree angle
• a) Independence of the vertical and horizontal motion of a projectile:
• The independence of the vertical and horizontal motion of a projectile is a fundamental concept in physics. It states that:
• The vertical motion of a projectile is independent of its horizontal motion.
• In other words, the vertical component of the projectile’s motion (its rise and fall) does not affect its horizontal motion (its left-right motion), and vice versa.
• – The horizontal velocity of a projectile remains constant.
• – The vertical velocity of a projectile changes due to gravity.
• This independence is due to the fact that the gravitational force acting on the projectile is perpendicular to its horizontal motion, and therefore does not affect its horizontal velocity.
• As a result, we can analyze the vertical and horizontal motions separately, using the following equations:
• – Vertical motion:
• [math]v_y = v_{i_y} – gt[/math]
• – Horizontal motion:
• [math]v_x = v_i x[/math]
• Where and are the initial horizontal and vertical velocities, respectively, and g is the acceleration due to gravity.
• b) Two-dimensional motion of a projectile with constant velocity in one direction and constant acceleration in a perpendicular direction.
• Two-dimensional motion of a projectile:
• – Motion in two dimensions (x and y axes)
• – Projectile moves with:
  • – Constant velocity ([math]v_x[/math]) in the horizontal direction (x-axis)
  • – Constant acceleration ([math]a_y[/math]) in the vertical direction (y-axis)
• Initial Conditions:
• – Initial position: ([math]x_i, y_i[/math] )
• – Initial velocity: ( [math]v_x, v_y[/math] )
• – Initial acceleration: (0, [math]a_y[/math])
• Motion Equations:
• – Horizontal motion (x-axis):
• [math]x = x_i + v_x t[/math]
• [math]v_x= constant[/math]
• – Vertical motion (y-axis):
• [math]y = y_i + v_y t – \frac{1}{2} a_y t^2[/math]
• – [math] v_y[/math]  = initial vertical velocity
• – [math] a_y[/math] = constant acceleration (due to gravity)
• Solving for the trajectory:
• – Combine the horizontal and vertical motions to get the parametric equations:
• [math]x = x_i + v_x t[/math]
• [math]y = y_i + v_y t – \frac{1}{2} a_y t^2[/math]
• – Eliminate t to get the equation of the trajectory (a parabola):
• [math]y = \frac{v_y}{v_x} x – \frac{a_y}{2 v_x^2} x^2 + y_i[/math]
• This describes the two-dimensional motion of a projectile with constant velocity in one direction (horizontal) and constant acceleration in a perpendicular direction (vertical). The resulting trajectory is a parabola.
• 
• Figure 9 Two-dimensional motion of a projectile with constant velocity