Kinematics of Linear and Projectile Motion

Learning Outcomes and Introduction to Kinematics

• Kinematics Definition: The description of the motion of objects without consideration of what causes the motion (specifically, mass or force are not considered).

• Categories of motion covered:

  • 1-Dimension (1D): Linear or straight-line motion, specifically horizontal uniform accelerated motion.

  • 2-Dimension (2D): Projectile motion, involving objects launched horizontally ($\theta = 0^{\circ}$) or at an angle $\theta$.

• Core Learning Objectives: Students must define instantaneous, average, and uniform velocity/acceleration; interpret and determine physical quantities from displacement-time ($s-t$), velocity-time ($v-t$), and acceleration-time ($a-t$) graphs; and solve problems using kinematics equations and projectile motion principles.

Linear Motion Fundamentals

• Scalar Quantities vs. Vector Quantities:

  • Distance ($s$): Defined as the total path length traversed in moving from one location to another. It is a scalar quantity and is always positive. SI Unit: metre ($m$).

  • Displacement ($s$): Defined as the shortest distance (straight line) between the initial and final points. It is a vector quantity and can be positive, negative, or zero depending on direction. SI Unit: metre ($m$).

  • Speed ($v$): Distance travelled per unit time interval ($v = \frac{s}{t}$). It is a scalar quantity.

  • Velocity ($v$): Speed in a particular direction; defined as the time rate of change of displacement ($v = \frac{s}{t}$). It is a vector quantity. SI unit: $m\,s^{-1}$.

• Specific Velocity Definitions:

  • Instantaneous Velocity: The velocity at a specific instant of time along the path of motion. It corresponds to the slope of the tangent line on an $s-t$ graph at that specific point. Formula: $v = \lim_{\Delta t \to 0} \frac{\Delta s}{\Delta t} = \frac{ds}{dt}$.

  • Average Velocity: The rate of change of displacement over a finite interval of time. Formula: $v = \frac{\Delta s}{\Delta t} = \frac{s_2 - s_1}{t_2 - t_1}$.

  • Uniform Velocity: Constant velocity where displacement increases by equal amounts in equal times. For an object moving with uniform velocity, its instantaneous velocity equals its average velocity at any time, and acceleration ($a$) is $0\,m\,s^{-2}$.

Acceleration

• Definition: The rate of change of velocity. Since velocity is a vector, acceleration results from a change in speed (magnitude), direction, or both.

• Type: Vector quantity. SI unit: $m\,s^{-2}$.

• Acceleration vs. Deceleration:

  • Acceleration: Moving in the forward direction or speeding up. Occurs when velocity and acceleration have the same signs (e.g., $+v$ and $+a$ or $-v$ and $-a$).

  • Deceleration: Slowing down. Occurs when velocity and acceleration have opposite signs (e.g., $+v$ and $-a$ or $-v$ and $+a$).

• Specific Acceleration Definitions:

  • Average Acceleration ($a_{av}$): Defined as the rate of change in velocity: $a_{av} = \frac{\Delta v}{\Delta t} = \frac{v_2 - v_1}{t_2 - t_1}$. The direction is the same as the motion if accelerating, and opposite to motion if decelerating.

  • Instantaneous Acceleration ($a$): The acceleration at a particular instant of time: $a = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} = \frac{dv}{dt}$. It is represented by the gradient of the tangent line on a $v-t$ graph.

  • Uniform Acceleration: Constant acceleration that does not change over time. The object's velocity changes at a constant rate along a fixed direction.

Graphical Representation of Motion

• Displacement-Time Graph ($s-t$):

  • Gradient: Represents velocity ($v = \frac{s_2 - s_1}{t_2 - t_1}$).

  • Horizontal line: Object is stationary / at rest ($v = 0$).

  • Straight line with constant slope: Uniform (constant) velocity.

  • Curved line: Changing velocity (acceleration or deceleration). A steeper slope indicates higher speed; a gentler slope indicates lower speed.

  • Negative slope: Uniform velocity returning to the starting point.

• Velocity-Time Graph ($v-t$):

  • Gradient: Represents acceleration ($a = \frac{v_2 - v_1}{t_2 - t_1}$).

  • Area under the graph: Represents displacement.

  • Horizontal line ($v \neq 0$): Uniform velocity ($a = 0$).

  • Positive slope: Speeding up (acceleration).

  • Negative slope: Slowing down (deceleration) or speeding up in the opposite direction.

• Acceleration-Time Graph ($a-t$):

  • Area under the graph: Represents the change in velocity ($\Delta v$).

Physical Meanings of Linear Graph Scenarios

• Scenario 1 (Horizontal line at $s=0, v=0, a=0$): Object is at rest/static/stationary.

• Scenario 2 (Straight upward diagonal on $s-t$): Moving in positive direction, constant velocity, $a = 0$.

• Scenario 3 (Concave upward curve on $s-t$): Moving in positive direction, velocity increases, constant positive acceleration.

• Scenario 4 (Concave downward curve on $s-t$): Moving in positive direction, velocity decreases, constant negative deceleration.

• Scenario 5 (Straight downward diagonal on $s-t$ starting from positive displacement): Moving in negative/opposite direction, constant velocity, $a = 0$.

• Scenario 6 (Curve toward negative displacement on $s-t$): Moving in negative direction, velocity increases (negatively), constant negative acceleration.

• Scenario 7 (Decurving toward horizontal on $s-t$ in negative direction): Moving in negative direction, velocity decreases, constant positive deceleration.

Uniformly Accelerated Motion

• Kinematics Equations (Equations of Motion):

  1. $v = u + at$

  2. $s = \frac{1}{2}(u + v)t$

  3. $s = ut + \frac{1}{2}at^2$

  4. $v^2 = u^2 + 2as$

• Implicit Data for Problem Solving:

  • Initially at rest: $u = 0\,m\,s^{-1}$.

  • Finally stops: $v = 0\,m\,s^{-1}$.

  • Decelerates or brakes: Acceleration $a$ is negative.

  • Uniform velocity: $a = 0\,m\,s^{-2}$ (Equations simplify to $s = vt$).

• Derivation Principles:

  • Velocity is the gradient of the $s-t$ graph.

  • Acceleration is the gradient of the $v-t$ graph.

  • Displacement is the area under the $v-t$ graph (Area of trapezium).

Projectile Motion

• Definition: A form of motion experienced by a projectile thrown near Earth's surface moving along a curved parabolic path under the action of gravity only (air resistance negligible).

• Key Principle: Horizontal and vertical motions are independent. They do not influence each other and are analyzed separately along perpendicular axes.

• Motion Components:

  • Horizontal ($x$-direction): No acceleration ($a_x = 0$). Velocity is constant ($u_x = v_x$).

  • Vertical ($y$-direction): Acceleration is constant and directed downward due to gravity ($a_y = -g = -9.81\,m\,s^{-2}$. Velocity changes by $9.81\,m\,s^{-1}$ every second.

• Equations for Projectile Motion:

  • Horizontal:

    • $u_x = u \cos(\theta)$

    • $v_x = u_x$ (constant)

    • $s_x = u_x t$

  • Vertical:

    • $u_y = u \sin(\theta)$

    • $v_y = u_y - gt$

    • $s_y = u_y t - \frac{1}{2}gt^{2}$

    • $v_y^2 = u_y^2 - 2gs_y$

• Resultant Velocity at time $t$:

  • Magnitude: $v = \sqrt{v_x^2 + v_y^2}$

  • Direction: $\theta = \tan^{-1}(\frac{v_y}{v_x})$ (measured from the horizontal axis).

• Specific Cases:

  • Case 1 (Full Projectile): Returns to the same vertical level. Total flight time $T = 2 t_H$ (where $t_H$ is time to max height). At max height, $v_y = 0$. Range ($R$) is max when $\theta = 45^{\circ}$.

  • Case 2 (Launched from height at an angle): Final vertical displacement $s_y = -h$ because the landing point is lower than the starting point.

  • Case 3 (Launched horizontally, $\theta = 0^{\circ}$): Initial vertical velocity $u_y = 0$. Displacement $s_y = -h$. Horizontal velocity $v_x = u$.

Problem Solving Examples

• Example 2.1.1 (Toy Train): Displacement-time graph analysis shows intervals of constant velocity ($0.67\,cm\,s^{-1}$, then stopped, then $1.50\,cm\,s^{-1}$). Instantaneous velocity calculation at $t=12\,s$ gives $1.50\,cm\,s^{-1}$.

• Example 2.1.2 (Lift Motion): Analysis of $v-t$ graph. Total distance found by adding total areas ($115\,m$). Total displacement found by taking signed sum of areas ($15\,m$). Average acceleration between $20\,s$ and $40\,s$ is $-0.4\,m\,s^{-2}$.

• Example 2.2.1 (Plane landing): Given $u = 50\,m\,s^{-1}$, $s = 1\,km$, $v = 0$. Found deceleration $a = -1.25\,m\,s^{-2}$ and time to stop $t = 40\,s$.

• Example 2.2.3 (Bus vs Car): A bus ($30\,m\,s^{-1}$) passes a stationary car. The car starts after $5\,s$ with $a = 2\,m\,s^{-2}$. The car takes $15\,s$ to reach the bus's velocity. Distance travelled when level (overlapping displacement) is $1183\,m$ at $t = 39.4\,s$.

• Example 2.3.1 (Superman): Initial speed $200\,m\,s^{-1}$ at $60^{\circ}$. At $t = 2\,s$, position is $(200\,m, 326\,m)$. Time to reach max height is $17.6\,s$. Max height $H = 1525\,m$. Horizontal range $R = 3520\,m$.

• Example 2.3.4 (Basketball Shot): Player ($2.00\,m$ tall) shoots from $10.0\,m$ distance to a hoop ($3.05\,m$ high) at a $40.0^{\circ}$ angle. Relative height $s_y = 1.05\,m$. Initial speed required calculated as $u = 10.67\,m\,s^{-1}$.