Topic 2: Kinematics of Linear and Projectile Motion

Overview of Kinematics of Linear Motion

• Definition of Kinematics: Kinematics is the description of the motion of objects without consideration of the causes behind that motion, such as mass or force.
• Dimensionality and Classification:
  • 1 Dimension (1D): Linear or straight-line motion.
  • 2 Dimension (2D): Projectile motion.
• Subtopics covered:
  • 2.1 Linear motion.
  • 2.2 Uniformly accelerated motion.
  • 2.3 Projectile motions.

Linear Motion Fundamentals

• Distance:
  • Defined as the total path length traversed in moving from one location to another.
  • Scalar quantity.
  • Always positive.
  • S.I. Unit: meter ($m$).
• Displacement ($s$):
  • Defined as the shortest distance (straight line) between the initial and final points.
  • Vector quantity.
  • Can be positive, negative, or zero.
  • S.I. Unit: meter ($m$).
• Speed:
  • Defined as distance traveled per unit time interval.
  • Scalar quantity.
  • S.I. Unit: $m\,s^{-1}$.
• Velocity ($v$):
  • Defined as the time rate of change of displacement (speed in a particular direction).
  • Vector quantity.
  • S.I. Unit: $m\,s^{-1}$.

Types of Velocity

• Average Velocity: The rate of change of displacement over a finite interval of time. It is defined by the change between two points.
• Instantaneous Velocity: Velocity at a specified position or instant of time along the path of motion. Graphically, it is the slope of the tangent line at a point on a displacement-time graph.
• Uniform Velocity: Velocity that remains constant. For an object moving with uniform velocity, its instantaneous velocity equals the average velocity at any time.
• Differentiating Average and Instantaneous: If a body speeds up or slows down during displacement, the average velocity is not the same as the velocity at a given instant of time.

Acceleration

• Definition: Acceleration is the rate of change of velocity.
• Variables: It is a vector quantity with an S.I. unit of $m\,s^{-2}$.
• Nature of Change: Since velocity is a vector, acceleration can result from a change in:
  1. Speed (magnitude).
  2. Direction.
  3. Both speed and direction.
• Average Acceleration: Defined as the change in velocity divided by the time taken to make that change ($\Delta v / \Delta t$).
• Instantaneous Acceleration: The acceleration at a particular instant of time or position. Graphically, it is the gradient of the tangent line at point $Q$ on a velocity-time ($v-t$) graph.
• Uniform Acceleration: When an object moves with uniform acceleration, its instantaneous acceleration is equal to the average acceleration at any point.
• Directional Context:
  • Accelerate in forward direction.
  • Decelerate in forward direction.
  • Decelerate in opposite direction.
  • Accelerate in opposite direction.
  • Note: Acceleration (increasing speed) and deceleration (decreasing speed) should not be confused with the direction of velocity.

Kinematic Graphs and Interpretation

Graph Gradients and Areas

• Displacement-Time Graph ($s-t$):
  • Gradient = Velocity.
• Velocity-Time Graph ($v-t$):
  • Gradient = Acceleration.
  • Area under the graph = Displacement.
• Acceleration-Time Graph ($a-t$):
  • Area under the graph = Velocity change.

Characterizing Motion from Graphs

• Scenario A:
  • $s$ increases exponentially with time in the positive direction; rapidly increases nearing final position.
  • $v$ increases directly with time in the positive direction.
  • $a$ is constant in the positive direction.
• Scenario B:
  • $s$ increases exponentially with time in the positive direction; slowly increases nearing final position.
  • $v$ decreases directly with time in the positive direction.
  • Deceleration is constant in the negative direction.
• Scenario C:
  • $s$ decreases directly with time in the positive direction.
  • $v$ is constant in the negative direction.
  • Acceleration is zero ($a = 0\,m\,s^{-2}$).
• Scenario D:
  • $s$ decreases exponentially with time in the positive direction; rapidly decreases nearing final position.
  • $v$ increases directly with time in the negative direction.
  • Acceleration is constant in the negative direction.

Uniformly Accelerated Motion Equations

The Four Kinematic Equations

For objects moving with a constant (uniform) rate of velocity change:

1. $v = u + at$
2. $s = ut + \frac{1}{2}at^2$
3. $v^2 = u^2 + 2as$
4. $s = \frac{1}{2}(u + v)t$

Derivation Logic

• Equation 4 ($s = \frac{1}{2}(u+v)t$): Obtained by calculating the shaded area under a velocity-time graph (area of a trapezium).
• Equation 1 ($v = u + at$): Derived from the definition of acceleration as the slope of the $v-t$ graph: $a = \frac{v - u}{t}$.
• Equation 2 ($s = ut + \frac{1}{2}at^2$): Derived by substituting Equation 1 into Equation 4.
• Equation 3 ($v^2 = u^2 + 2as$): Derived by substituting time $t = \frac{v - u}{a}$ from Equation 1 into Equation 4.

Projectile Motion

• Definition: A form of motion experienced by an object (projectile) thrown near the Earth's surface that moves along a curved parabolic path under the action of gravity only. Air resistance is assumed to be negligible.
• Principle of Independence: Horizontal and vertical motions are independent of each other.

Components of Motion

• Horizontal ($x$ component):
  • Forces: No external force (ignoring air resistance).
  • Acceleration ($a_x$): Always $0\,m\,s^{-2}$.
  • Velocity ($v_x$): Constant throughout the flight.
  • Initial Velocity: $u_x = u\cos(\theta)$.
  • Displacement: $s_x = u_x t$.
• Vertical ($y$ component):
  • Forces: Gravity (downward).
  • Acceleration ($a_y$): Always $-g$ (specifically $-9.81\,m\,s^{-2}$) regardless of whether the object is moving up, down, or is at the highest point.
  • Velocity ($v_y$): Changes by $9.81\,m\,s^{-1}$ every second.
  • Initial Velocity: $u_y = u\sin(\theta)$.
  • Displacement: $s_y = u_y t - \frac{1}{2}gt^2$.
  • Final Velocity at time $t$: $v_y = u_y - gt$.
  • Velocity-Displacement relation: $v_y^2 = u_y^2 - 2gs_y$.

Key Terms

• Launch angle ($\theta$): Measured from the horizontal axis.
• Range ($R$): Total horizontal displacement ($s_x$).
• Maximum Height ($H$): The peak vertical displacement ($s_y$) where $v_y = 0$.
• Time of Flight ($T$): Total time the projectile is in the air.
• Velocity Magnitude and Direction:
  • Magnitude $v = \sqrt{v_x^2 + v_y^2}$
  • Direction $\phi = \tan^{-1}\left(\frac{v_y}{v_x}\right)$

Examples and Problems

Example 1: Vector Displacement

• Scenario: Object P moves $30\,m$ East, then $15\,m$ South, and finally $40\,m$ West.
• Solution:
  • Horizontal net displacement: $30 - 40 = -10\,m$ (West).
  • Vertical net displacement: $-15\,m$ (South).
  • Magnitude: $OP = \sqrt{15^2 + 10^2} = 18\,m$.
  • Direction: $\theta = \tan^{-1}\left(\frac{15}{10}\right) = 56^{\circ}$ from west to south.

Example 4: Toy Train Motion Analysis

• Scenario: Toy train moves on a straight track; graph showing $s$ vs $t$.
• Average Velocity Calculation: $v_{av} = \frac{s_2 - s_1}{t_2 - t_1} = \frac{10 - 4}{14 - 0} = 0.429\,cm\,s^{-1}$.

Example 5: Motorcycle Motion Analysis

• Intervals:
  • $0\,s$ to $2\,s$: Constant acceleration.
  • $2\,s$ to $5\,s$: Constant velocity (zero acceleration).
  • $5\,s$ to $6\,s$: Constant deceleration.
• Calculations: Deceleration is the gradient; total displacement is the area under the $v-t$ graph.

Example 6: Plane Landing

• Parameters: $u = 50\,m\,s^{-1}$, $s = 1.0\,km = 1000\,m$, $v = 0\,m\,s^{-1}$.
• Deceleration: $v^2 = u^2 + 2as \Rightarrow 0 = 50^2 + 2a(1000) \Rightarrow a = -1.25\,m\,s^{-2}$.
• Time: $v = u + at \Rightarrow 0 = 50 + (-1.25)t \Rightarrow t = 40\,s$.

Example 7: Bus and Car

• Scenario: Bus travels at steady $30\,m\,s^{-1}$. A car starts $5\,s$ later from rest with $a = 2\,m\,s^{-2}$.
• Car travel time ($t_{car}$): When velocities match, $v_{car} = 30 = 0 + 2(t_{car}) \Rightarrow t_{car} = 15\,s$.

Example 8: Projectile Launch at $\theta = 0^{\circ}$

• Scenario: Plane moving at $115\,m\,s^{-1}$ horizontally at altitude $1050\,m$ releases a package.
• Vertical Calculation: $s_y = u_y t - \frac{1}{2}gt^2 \Rightarrow -1050 = 0(t) - \frac{1}{2}(9.81)t^2 \Rightarrow t = 14.63\,s$.

Example 9: Baseball Home Run

• Scenario: Ball lands $7.5\,m$ above hit point with $v_{final} = 36\,m\,s^{-1}$ at $28^{\circ}$ below horizontal.
• Horizontal velocity component: $v_x = 36\cos(28^{\circ}) = 31.78\,m\,s^{-1}$ (constant).
• Vertical velocity component: $v_y = -36\sin(28^{\circ}) = -16.90\,m\,s^{-1}$.
• Initial vertical velocity: $(-16.90)^2 = u_y^2 - 2(9.81)(7.5) \Rightarrow u_y = 20.80\,m\,s^{-1}$.

Questions & Discussion

• Question 1: Average velocity is defined as?
  • Answer: A. Rate of change of displacement.
• Question 2: Which is correct about uniform acceleration?
  • Answer: Velocity changes with a constant rate (implies Magnitude of acceleration is constant).
• Question 3: What results in a straight line graph for uniform acceleration?
  • Answer: D. (Graph of $s/t$ vs $t$ or $v$ vs $t$).
• Question 4: An object moving with uniform acceleration of $5\,m\,s^{-2}$. A displacement-time graph gradient will…
  • Answer: A. Increase with time (since gradient is velocity and velocity is increasing).
• Question 5: If velocity vs time is a straight line at $30^{\circ}$ with the time axis, the object is…
  • Answer: B. Moving with constant non-zero acceleration.
• Question 6 (Projectile 1): Which angle gives maximum range?
  • Answer: C. $45^{\circ}$.
• Question 7 (Projectile 2): Projectile motion is free fall if launched at what angle from horizontal?
  • Answer: D. $90^{\circ}$.
• Question 8 (Projectile 3): True statement about projectile motion?
  • Answer: B. The vertical component of velocity is zero at the highest point of its flight.
• Question 9 (Projectile 4): An object in projectile motion has a constant…
  • Answer: B. Vertical acceleration.