Describing Motion Around Us

Foundations of Motion and Descriptive Quantities

• Overview of Motion in Nature

  • Everything in nature is in continuous motion, spanning from massive astronomical objects down to subatomic particles.

  • Examples of natural motion include:

  • Flitting butterflies

  • Slithering snakes

  • Hopping hares

  • Galloping horses

  • Tendrils of climbing plants twinning around a support

  • Closing mechanism of flytraps

  • Dancing dust particles suspended in a sunbeam

  • Smoke particles dispersing in air

  • Rising and falling of ocean tides

  • Gathering of atmospheric clouds

• Scientific Methodology and Simplification

  • To analyze complex physical phenomena, scientists first investigate idealized and simplified models.

  • Primary simplified classifications of motion include:

  • Linear motion (motion along a straight path)

  • Circular motion (motion along a curved circular path)

  • Oscillatory motion (repetitive back-and-forth movement about a central position)

Linear Motion, Reference Frames, and Position

• Definition of Linear Motion

  • When an object moves strictly along a straight line, its movement is defined as linear motion or motion in a straight line.

  • Linear motion represents the simplest physical form of motion.

  • Real-world examples of linear motion include:

  • Swimmers competing in a straight swimming race

  • A ball falling vertically downwards under gravity

  • A vehicle traveling along a straight stretch of a highway

  • A train moving along a straight railway track

• Describing Position

  • Describing the motion of an object requires specifying its position at various consecutive instants of time.

  • Reference Point (Origin $O$): A fixed reference point must be chosen to define position. The position of an object at any instant is specified by both its distance and its direction relative to this fixed reference point.

  • Condition for Motion vs. Rest:

  • Motion: An object is in motion if its position relative to the chosen reference point changes as time progresses.

  • Rest: An object is at rest if its position relative to the reference point remains completely unchanged over time.

• Directional Conventions in One Dimension

  • For straight-line motion, an object can move in only two opposite directions: forward and backward.

  • Direction is denoted using algebraic signs:

  • Positions located to the right of the reference point $O$ are conventionally assigned a positive sign ($+$).

  • Positions located to the left of the reference point $O$ are assigned a negative sign ($-$).

• Time Concepts

  • Instant of Time: A single precise reading on a clock corresponding to a specific moment.

  • Time Interval: The elapsed time duration between two distinct instants of time (the difference between two clock readings).

• Scalar and Vector Quantities

  • Scalars: Physical quantities that are completely specified by their numerical value and associated units (magnitude) alone.

  • Vectors: Physical quantities that require both a magnitude (numerical value with units) and a specific spatial direction for a complete description.

Distance Travelled versus Displacement

• Definitions and Distinction

  • Total Distance Travelled: The actual total length of the path traversed by a moving object during a given time interval, regardless of the direction of motion. Distance is a scalar quantity.

  • Displacement: The net change in position of an object between two specific instants of time. Displacement is a vector quantity.

  • Magnitude of Displacement: The shortest straight-line distance measured from the initial position to the final position.

  • Direction of Displacement: Directed straight from the initial position towards the final position.

  • SI Unit: The International System of Units (SI) unit for both distance travelled and displacement is the metre ($m$).

• Detailed Trajectory Example (Athlete on a Straight Track)

  • Consider an athlete running on a straight track relative to an origin $O$ at $0\,m$:

  • At time $t = 0\,s$, the athlete starts at point $O$ ($0\,m$).

  • At time $t = 4\,s$, the athlete reaches point $B$ at position $40\,m$.

  • At time $t = 10\,s$, the athlete reaches point $A$ at position $100\,m$.

  • The athlete then reverses direction and runs back along the same line, reaching point $B$ ($40\,m$) at time $t = 16\,s$

  • Calculations between $t = 0\,s$ and $t = 16\,s$:

  • Total distance travelled = path length $OA + AB = 100\,m + 60\,m = 160\,m$

  • Net displacement = final position $-$ initial position = position of $B -$ position of $O = 40\,m - 0\,m = +40\,m$ (or $40\,m$ in the positive direction).

  • Key Conclusion: The total distance travelled ($160\,m$) and the magnitude of displacement ($40\,m$) are not equal when direction changes.

• Conditions for Equality

  • For linear motion, the total distance travelled is equal to the magnitude of displacement if and only if the object moves continuously in a single direction without turning back.

• Analysis of Vertical Motion (Ball Thrown Upwards)

  • Consider a ball thrown vertically upwards from origin $O$ ($0\,cm$), passing point $A$ ($40\,cm$), reaching maximum height at point $B$ ($100\,cm$), falling back down past point $C$ ($80\,cm$), and returning to origin $O$ ($0\,cm$).

  • Data Table:

  • Position $O$ (Start): Total distance = $0\,cm$; Displacement = $0\,cm$

  • Position $A$ (Ascending): Total distance = $40\,cm$; Displacement = $40\,cm$ upwards

  • Position $B$ (Apex): Total distance = $100\,cm$; Displacement = $100\,cm$ upwards

  • Position $C$ (Descending): Total distance = $100\,cm + 20\,cm = 120\,cm$; Displacement = $80\,cm$ upwards

  • Position $O$ (Returned): Total distance = $100\,cm + 100\,cm = 200\,cm$; Displacement = $0\,cm$

  • General Properties of Displacement:

  • Displacement can be equal to zero even if total distance travelled is non-zero.

  • The magnitude of displacement is always less than or equal to the total distance travelled ($\text{magnitude of displacement} \le \text{total distance}$).

Average Speed and Average Velocity

• Average Speed

  • Average speed measures how fast or slow an object covers distance over a given time interval.

  • Formula:     $\text{average speed} = \frac{\text{total distance travelled}}{\text{time interval}}$

  • Average speed is a scalar quantity and carries no spatial direction.

• Uniform and Non-Uniform Linear Motion

  • Uniform Linear Motion: An object moving in a straight line covers equal distances in equal intervals of time (for all choices of time intervals). The object moves with a constant speed.

  • Non-Uniform Linear Motion: An object covers unequal distances in equal intervals of time. The speed changes over time (increasing, decreasing, or fluctuating).

• Average Velocity

  • Average velocity measures how fast the position of an object changes and in which direction.

  • Formula:     $\text{average velocity} = \frac{\text{change in position}}{\text{time interval}} = \frac{\text{displacement}}{\text{time interval}}$

  • Expressed algebraically:     $v_{av} = \frac{s}{t}$     where $v_{av}$ is average velocity, $s$ is displacement, and $t$ is the time interval.

  • Rate of Change: Average velocity represents the average rate of change of position with respect to time.

  • Direction: The direction of average velocity matches the direction of displacement, indicated by a positive ($+$) or negative ($-$) sign in one-dimensional motion.

  • SI Units: Metres per second ($m\,s^{-1}$ or $m/s$). Another common unit is kilometres per hour ($km\,h^{-1}$).

• Comparative Example (Sarang in a Swimming Pool)

  • Pool length = $25\,m$. Sarang swims from one end to the other and back to the starting point in a time interval of $50\,s$.

  • Total distance travelled = $25\,m + 25\,m = 50\,m$

  • Displacement = $0\,m$

  • Average speed = $\frac{50\,m}{50\,s} = 1\,m\,s^{-1}$

  • Average velocity = $\frac{0\,m}{50\,s} = 0\,m\,s^{-1}$

• Instantaneous Velocity

  • Instantaneous velocity is the velocity of an object at a specific instant of time.

  • As the time interval around an instant approaches an infinitesimally small value, the average velocity approaches the instantaneous velocity.

  • Speedometer Analogy: A vehicle's speedometer reading provides the magnitude of instantaneous velocity, while the direction of the tires indicates the direction of instantaneous velocity.

Historical Mathematical Contributions to Motion

• Ancient and Medieval Indian Scientific Contributions

  • The concept of speed as distance divided by time is documented in classical Indian mathematical literature.

  • Aryabhatiya (5th century CE): Contains foundational references to speed and rates of movement.

  • Ganitakaumudi (14th century CE): Authored comprehensive problems regarding relative linear motion.

• Historical Example Problem from Ganitakaumudi

  • Problem Statement: Two postmen start walking towards each other separated by a total distance of $210\,\text{yojanas}$ (a traditional Indian unit of distance). One postman travels at $9\,\text{yojanas per day}$ and the other covers $5\,\text{yojanas per day}$. Determine how many days pass before they meet.

  • Solution:

  • Distance covered by both postmen combined in one day = $9\,\text{yojanas} + 5\,\text{yojanas} = 14\,\text{yojanas per day}$

  • Total distance needed to meet = $210\,\text{yojanas}$

  • Time taken to meet = $\frac{210\,\text{yojanas}}{14\,\text{yojanas/day}} = 15\,\text{days}$

  • In $15\,\text{days}$, the first postman covers $15 \times 9 = 135\,\text{yojanas}$, and the second postman covers $15 \times 5 = 75\,\text{yojanas}$.

Average Acceleration and Free-Fall Dynamics

• Definition of Average Acceleration

  • Acceleration characterizes the rate at which the velocity of an object changes over time.

  • Formula:     $\text{average acceleration} = \frac{\text{change in velocity}}{\text{time interval}} = \frac{\text{final velocity} - \text{initial velocity}}{\text{time interval}}$

  • Expressed algebraically:     $a = \frac{v - u}{t_2 - t_1}$     where $u$ is the initial velocity at time $t_1$, $v$ is the final velocity at time $t_2$, and $a$ is the average acceleration.

  • SI Unit: Metres per second squared ($m\,s^{-2}$ or $m/s^2$).

• Direction of Acceleration

  • For straight-line motion:

  • If the magnitude of velocity is increasing over time, acceleration acts in the same direction as velocity.

  • If the magnitude of velocity is decreasing over time (deceleration or braking), acceleration acts in the direction opposite to velocity.

  • Acceleration can result from a change in velocity magnitude, a change in velocity direction, or both.

  • High Speed vs Zero Acceleration: An object can move at an extremely high speed and still have zero acceleration if its velocity remains completely constant (e.g., a high-speed bus moving along a straight road at uniform speed).

• Detailed Bus Acceleration Example

  • A bus travels on a straight highway with an initial velocity $u = 36\,km\,h^{-1}$.

  • Unit conversion: $36\,km\,h^{-1} = 36 \times \frac{1000\,m}{3600\,s} = 10\,m\,s^{-1}$

  • Phase 1 (Speeding up): The driver depresses the accelerator for $t = 10\,s$, increasing velocity to $v = 54\,km\,h^{-1} = 15\,m\,s^{-1}$.     $a = \frac{15\,m\,s^{-1} - 10\,m\,s^{-1}}{10\,s} = \frac{5\,m\,s^{-1}}{10\,s} = 0.5\,m\,s^{-2}$     Acceleration is $+0.5\,m\,s^{-2}$ in the direction of motion.

  • Phase 2 (Braking): The driver sees an obstacle and applies brakes, bringing the bus to a full stop ($v = 0\,m\,s^{-1}$) from $u = 15\,m\,s^{-1}$ in $t = 5\,s$     $a = \frac{0\,m\,s^{-1} - 15\,m\,s^{-1}}{5\,s} = \frac{-15\,m\,s^{-1}}{5\,s} = -3\,m\,s^{-2}$     The negative sign indicates that acceleration acts opposite to the direction of velocity.

• Constant Acceleration and Gravitational Free Fall

  • Constant Acceleration Definition: If an object moving in a straight line undergoes changes in velocity by equal amounts in equal intervals of time, its acceleration is constant.

  • Free-Fall Example: An object dropped from rest ($u = 0\,m\,s^{-1}$ at $t = 0\,s$) near Earth's surface falls vertically downward under gravity:

  • At $t = 0\,s$, $v = 0\,m\,s^{-1}$

  • At $t = 1\,s$, $v = 9.8\,m\,s^{-1}$ $\implies a = \frac{9.8 - 0}{1} = 9.8\,m\,s^{-2}$

  • At $t = 2\,s$, $v = 19.6\,m\,s^{-1}$ $\implies a = \frac{19.6 - 9.8}{1} = 9.8\,m\,s^{-2}$

  • At $t = 3\,s$, $v = 29.4\,m\,s^{-1}$ $\implies a = \frac{29.4 - 19.6}{1} = 9.8\,m\,s^{-2}$

  • At $t = 4\,s$, $v = 39.2\,m\,s^{-1}$ $\implies a = \frac{39.2 - 29.4}{1} = 9.8\,m\,s^{-2}$

  • The acceleration is constant across all successive 1-second intervals, equal to $9.8\,m\,s^{-2}$ directed vertically downwards. This is denoted as the acceleration due to gravity, $g$.

Graphical Representation of Linear Motion

• Purpose of Motion Graphs

  • Graphs visually depict how position, velocity, and acceleration change with time.

  • They allow for the extraction of physical quantities, comparison between different objects, and classification of motion as uniform or non-uniform.

• Graph Construction Conventions

  • Horizontal axis ($X$ axis): Represents time ($t$).

  • Vertical axis ($Y$ axis): Represents position ($s$) or velocity ($v$).

  • Origin ($O$): Point of intersection of axes.

• Position-Time Graphs ($s-t$ Graphs)

  • Interpretation of Graph Shapes:

  • Straight line parallel to X-axis: The object's position does not change with time; the object is stationary (at rest).

  • Straight line with positive constant slope: The object moves with a constant velocity (uniform linear motion).

  • Curved line (increasing steepness): The object's displacement per unit time is increasing; velocity is changing (accelerated motion).

  • Calculating Velocity from $s-t$ Graph Slope:

  • Select two points $A(t_1, s_1)$ and $B(t_2, s_2)$ on the line.

  • Construct a right-angled triangle $ABC$ where $BC = s_2 - s_1$ (change in position) and $AC = t_2 - t_1$ (time interval).

  • The slope of the line equals the average velocity:       $\text{Slope} = \frac{BC}{AC} = \frac{s_2 - s_1}{t_2 - t_1} = v_{av}$

  • A steeper slope corresponds to a higher magnitude of velocity.

• Velocity-Time Graphs ($v-t$ Graphs)

  • Interpretation of Graph Shapes:

  • Straight line parallel to X-axis: Velocity is constant over time; acceleration $a = 0\,m\,s^{-2}$.

  • Straight line sloping upwards: Velocity increases by equal amounts in equal time intervals; acceleration is positive and constant.

  • Straight line sloping downwards: Velocity decreases by equal amounts in equal time intervals; acceleration is negative and constant.

  • Calculating Acceleration from $v-t$ Graph Slope:

  • The slope of a velocity-time line gives the acceleration of the object:       $\text{Slope} = \frac{BC}{AC} = \frac{v_2 - v_1}{t_2 - t_1} = a$

  • Calculating Displacement from $v-t$ Graph Area:

  • The area bounded by the velocity-time curve and the time axis for a chosen time interval represents the net displacement ($s$) during that interval.

  • Constant Velocity Case: Area of rectangle = $\text{height} \times \text{width} = v \times t = s$

    • Example: For $v = 20\,m\,s^{-1}$ over $t = 6\,s$, $\text{area} = 20\,m\,s^{-1} \times 6\,s = 120\,m$.

  • Constant Acceleration Case: Area of trapezium = area of rectangle + area of triangle       $\text{Displacement } s = \left(v_1 \times (t_2 - t_1)\right) + \frac{1}{2}\left((t_2 - t_1) \times (v_2 - v_1)\right)$

    • Example: For a car accelerating from $v_1 = 5\,m\,s^{-1}$ at $t_1 = 10\,s$ to $v_2 = 10\,m\,s^{-1}$ at $t_2 = 20\,s$:         $\text{Rectangle Area} = 5\,m\,s^{-1} \times (20 - 10)\,s = 50\,m$         $\text{Triangle Area} = \frac{1}{2} \times (20 - 10)\,s \times (10 - 5)\,m\,s^{-1} = 25\,m$         $\text{Total Displacement } s = 50\,m + 25\,m = 75\,m$

Kinematic Equations for Straight-Line Motion with Constant Acceleration

• Derivation of the First Kinematic Equation ($v = u + at$)

  • By definition of constant acceleration:     $a = \frac{v - u}{t - 0} = \frac{v - u}{t}$

  • Multiplying both sides by $t$ gives $at = v - u$.

  • Rearranging yields:     $v = u + at$

• Derivation of the Second Kinematic Equation ($s = ut + \frac{1}{2}at^2$)

  • Displacement $s$ equals the total area under the $v-t$ graph from time $0$ to $t$, bounded between initial velocity $u$ and final velocity $v$:     $s = \text{Area of rectangle} + \text{Area of triangle}$     $s = (u \times t) + \frac{1}{2} \times t \times (v - u)$

  • Substituting $v - u = at$ from the first kinematic equation:     $s = ut + \frac{1}{2} \times t \times (at)$     $s = ut + \frac{1}{2}at^2$

• Derivation of the Third Kinematic Equation ($v^2 = u^2 + 2as$)

  • From the first equation, express time as $t = \frac{v - u}{a}$.

  • Substitute this expression for $t$ into the second kinematic equation:     $s = u\left(\frac{v - u}{a}\right) + \frac{1}{2}a\left(\frac{v - u}{a}\right)^2$     $s = \frac{uv - u^2}{a} + \frac{a(v^2 - 2uv + u^2)}{2a^2}$     $s = \frac{2(uv - u^2) + (v^2 - 2uv + u^2)}{2a}$     $s = \frac{2uv - 2u^2 + v^2 - 2uv + u^2}{2a}$     $s = \frac{v^2 - u^2}{2a}$

  • Rearranging yields:     $2as = v^2 - u^2$     $v^2 = u^2 + 2as$

• Additional Derived Kinematic Forms

  • Expression without initial velocity $u$:     $s = vt - \frac{1}{2}at^2$

  • Expression using average velocity formula (Trapezium Area method):     $s = \frac{1}{2}(u + v)t$

• Application Conditions and Rules

  • These equations are valid only when acceleration $a$ is constant.

  • Vector directions must be handled via explicit sign conventions ($+$ and $-$) for $u$, $v$, $a$, and $s$.

• Detailed Sample Problem: Car Stopping Distance

  • A car undergoes braking acceleration $a = -4\,m\,s^{-2}$ to come to a stop ($v = 0\,m\,s^{-1}$).

  • Using $v^2 = u^2 + 2as \implies 0^2 = u^2 + 2(-4)s \implies 0 = u^2 - 8s \implies s = \frac{u^2}{8}$.

  • Case (i): Initial Velocity $u = 54\,km\,h^{-1} = 15\,m\,s^{-1}$     $s = \frac{(15)^2}{8} = \frac{225}{8} = 28.125\,m$

  • Case (ii): Initial Velocity $u = 108\,km\,h^{-1} = 30\,m\,s^{-1}$     $s = \frac{(30)^2}{8} = \frac{900}{8} = 112.5\,m$

  • Societal/Safety Takeaway: Doubling the initial speed quadruples the required stopping distance ($112.5\,m$ vs $28.125\,m$).

Two-Dimensional Motion and Uniform Circular Motion

• Motion in Two and Three Dimensions

  • Motion in a Plane (Two Dimensions): Movement constrained to a two-dimensional surface. Examples include vehicles overtaking, trajectory of a kicked ball, or circular motion of a satellite.

  • Motion in Space (Three Dimensions): Movement through three-dimensional space. Examples include a car climbing a winding mountain road, a bird flying, or an aircraft navigating air space.

• Uniform Circular Motion Definitions

  • Circular Motion: Motion of an object along a circular path.

  • Uniform Circular Motion: Motion along a circular path at a constant (uniform) speed.

• Distance, Displacement, and Speed in Uniform Circular Motion

  • For an object moving on a circular path of radius $R$ making one complete revolution in time period $T$:

  • Total distance travelled in one revolution = circumference = $2\pi R$

  • Net displacement in one full revolution = $0\,m$ (returns to starting location).

  • Average speed $v_{av}$ over one revolution:       $v_{av} = \frac{2\pi R}{T}$       Because speed is constant in uniform circular motion, this value represents the exact speed at every point along the circular path.

  • Average velocity over one complete revolution = $\frac{0\,m}{T} = 0\,m\,s^{-1}$.

• Direction of Velocity and Tangential Motion

  • As an object moves along a circular path, the direction of its velocity changes continuously.

  • At any specific point on the circle, the velocity vector is directed strictly along the tangent to the circle at that point in the direction of motion.

  • Ring and Marble Investigation: When a marble spinning inside a circular ring is freed by lifting the ring, it immediately moves off in a straight line tangent to the circle at its point of release.

• Acceleration in Uniform Circular Motion

  • Because velocity is a vector possessing both magnitude and direction, any change in direction constitutes a change in velocity.

  • Even though the magnitude of velocity (speed) is completely constant, uniform circular motion is an accelerated motion due to the continuous change in the direction of velocity.

Practice Questions, Analytical Problems, and Solutions

• Question 1: Multi-trip Distance and Displacement

  • Scenario: A person leaves home, walks $250\,m$ along a straight road to a shop, forgets a bag, returns $250\,m$ home, walks $250\,m$ back to the shop, and finally walks $250\,m$ home.

  • Total Distance Travelled: $250\,m + 250\,m + 250\,m + 250\,m = 1000\,m$

  • Displacement from Home: $0\,m$ (final position equals initial position).

• Question 2: Stair Climbing Analysis

  • Scenario: A student runs from the ground floor to the 4th floor, then returns down to the 2nd floor. Each floor height is $3\,m$

  • Calculations:

  • Distance ground floor to 4th floor = $4 \times 3\,m = 12\,m$

  • Distance 4th floor to 2nd floor = $2 \times 3\,m = 6\,m$

  • Total vertical distance travelled = $12\,m + 6\,m = 18\,m$

  • Final displacement = height of 2nd floor relative to ground floor = $2 \times 3\,m = 6\,m$ vertically upwards.

• Question 3: Acceleration with Constant Speedometer Reading

  • Question: Can a scooter accelerate while its speedometer reading remains constant?

  • Answer: Yes. If the scooter travels along a curved path or takes a turn at constant speed, the direction of its velocity changes continuously, producing non-zero acceleration.

• Question 4: Straight Line Acceleration and Distance Calculation

  • Problem: A car starts from rest ($u = 0\,m\,s^{-1}$) and reaches $v = 24\,m\,s^{-1}$ in $t = 6\,s$.

  • Average Acceleration:     $a = \frac{v - u}{t} = \frac{24\,m\,s^{-1} - 0\,m\,s^{-1}}{6\,s} = 4\,m\,s^{-2}$

  • Distance Travelled:     $s = ut + \frac{1}{2}at^2 = (0 \times 6) + \frac{1}{2}(4)(6^2) = 2 \times 36 = 72\,m$

• Question 5: Motorbike Braking Problem

  • Problem: A motorbike with initial velocity $u = 28\,m\,s^{-1}$ stops ($v = 0\,m\,s^{-1}$) after travelling $s = 98\,m$ under constant acceleration.

  • Acceleration Calculation:     $v^2 = u^2 + 2as \implies 0^2 = (28)^2 + 2a(98)$     $0 = 784 + 196a \implies 196a = -784 \implies a = -4\,m\,s^{-2}$

  • Time Taken Calculation:     $v = u + at \implies 0 = 28 + (-4)t \implies 4t = 28 \implies t = 7\,s$

• Question 6: Velocity Comparison from $s-t$ Graph

  • Question: Do two objects $A$ and $B$ moving along parallel straight tracks ever have equal velocity if their position-time graph displays parallel straight lines?

  • Answer: Yes, if their $s-t$ lines are parallel, their slopes are identical at all times, meaning their velocities are continuously equal.

• Question 7: Analysis of Position-Time Curves

  • Scenario: Two objects $A$ and $B$ start at position $0\,m$ at $t = 0\,s$ and arrive at position $10\,m$ at $t = 10\,s$.

  • True Statement: The average velocity of both over the $10\,s$ interval is equal because both undergo identical net displacement ($10\,m - 0\,m = 10\,m$) over the same time duration ($10\,s$), giving $v_{av} = 1\,m\,s^{-1}$.

• Question 8: Truck Speed Reduction Distance

  • Problem: A truck decelerates from $54\,km\,h^{-1}$ ($15\,m\,s^{-1}$) to $36\,km\,h^{-1}$ ($10\,m\,s^{-1}$) in $t = 36\,s$ under uniform acceleration.

  • Acceleration:     $a = \frac{10 - 15}{36} = -\frac{5}{36}\,m\,s^{-2}$

  • Distance Travelled:     $s = \left(\frac{u + v}{2}\right)t = \left(\frac{15 + 10}{2}\right) \times 36 = 12.5 \times 36 = 450\,m$

• Question 9: Three-Stage Car Motion

  • Stage 1: Accelerates uniformly from rest ($u = 0$) to $20\,m\,s^{-1}$ in $5\,s$     $s_1 = \left(\frac{0 + 20}{2}\right) \times 5 = 50\,m$

  • Stage 2: Travels at constant velocity $20\,m\,s^{-1}$ for $10\,s$     $s_2 = 20 \times 10 = 200\,m$

  • Stage 3: Brakes uniformly to a stop ($v = 0$) in $6\,s$     $s_3 = \left(\frac{20 + 0}{2}\right) \times 6 = 60\,m$

  • Total Distance Travelled:     $s_{total} = 50\,m + 200\,m + 60\,m = 310\,m$

• Question 10: Obstacle Avoidance and Reaction Time

  • Problem: Bus velocity $u = 36\,km\,h^{-1} = 10\,m\,s^{-1}$. Obstacle is $30\,m$ ahead. Driver reaction time = $0.5\,s$. Braking deceleration $a = -2.5\,m\,s^{-2}$.

  • Distance covered during reaction time:     $s_{reaction} = u \times t_{reaction} = 10\,m\,s^{-1} \times 0.5\,s = 5\,m$

  • Distance covered during braking:     $v^2 = u^2 + 2as \implies 0^2 = (10)^2 + 2(-2.5)s_{braking}$     $0 = 100 - 5s_{braking} \implies 5s_{braking} = 100 \implies s_{braking} = 20\,m$

  • Total Stopping Distance:     $s_{stopping} = 5\,m + 20\,m = 25\,m$

  • Conclusion: Since 25\,m < 30\,m, the bus stops safely before colliding with the obstacle.

• Question 11: Earth Reference Frame Analysis

  • Question: Considering Earth moves around the Sun, can an object on Earth be considered at rest?

  • Answer: Rest and motion are relative terms that depend entirely on the frame of reference. Relative to a reference frame attached to the Earth, the object is at rest. Relative to a reference frame attached to the Sun, the object is in motion.

• Question 12: Car Displacement from Specified Graph

  • Scenario: A car moves at constant velocity $6\,m\,s^{-1}$ for $2\,\text{minutes}$ ($120\,s$), then accelerates at $1\,m\,s^{-2}$ for $6\,s$

  • Segment 1 ($0$ to $120\,s$):     $s_1 = 6\,m\,s^{-1} \times 120\,s = 720\,m$

  • Segment 2 ($120$ to $126\,s$):     $s_2 = ut + \frac{1}{2}at^2 = (6 \times 6) + \frac{1}{2}(1)(6^2) = 36 + 18 = 54\,m$

  • Total Displacement:     $s_{total} = 720\,m + 54\,m = 774\,m$

• Question 13: Wall Clock Minute Hand Calculations

  • Scenario: Wall clock minute hand length $r = 7\,cm = 0.07\,m$. Time interval: 6:00 PM to 7:30 PM ($1.5\,\text{hours} = 90\,\text{minutes} = 5400\,s$).

  • Number of Revolutions: $1.5$ complete rotations.

  • (i) Distance Travelled:     $\text{Distance} = 1.5 \times (2\pi r) = 1.5 \times 2 \times \frac{22}{7} \times 7\,cm = 66\,cm = 0.66\,m$

  • (ii) Displacement:

  • At 6:00 PM, minute hand tip is at the 12 mark.

  • At 7:30 PM, minute hand tip is at the 6 mark.

  • Displacement is a straight line pointing from 12 to 6 mark (diameter):       $\text{Displacement} = 2 \times r = 2 \times 7\,cm = 14\,cm = 0.14\,m \text{ (directed vertically downward)}$

  • (iii) Average Speed:     $\text{Speed} = \frac{66\,cm}{90\,\text{min}} = 0.733\,cm\,\text{min}^{-1} = \frac{0.66\,m}{5400\,s} = 1.22 \times 10^{-4}\,m\,s^{-1}$

  • (iv) Average Velocity:     $\text{Velocity} = \frac{14\,cm}{90\,\text{min}} = 0.156\,cm\,\text{min}^{-1} = \frac{0.14\,m}{5400\,s} = 2.59 \times 10^{-5}\,m\,s^{-1} \text{ (downward)}$

Practical Investigations and Advanced Extensions

• Rotating Cardboard Disc Investigation

  • A cardboard disc of radius $8\,cm$ has numbers 1 to 12 written on its outer border ($7\,cm$ from center) and letters 'ABCDEF' on its inner border ($4\,cm$ from center).

  • Observation: When spun rapidly, outer numbers blur and fade faster than inner letters.

  • Explanation: Linear speed in circular motion is proportional to radius ($v = \omega R$, where $\omega$ is angular velocity). The outer track has a larger radius (7\,cm > 4\,cm), so its markings move at a higher linear speed, causing greater motion blur.

• Smartphone Accelerometer Experiments

  • Built-in smartphone accelerometers utilize micro-electromechanical systems (MEMS) to register precise acceleration changes.

  • Applications include detecting minute physiological tremors in medical movement disorder research.

• Braking Distance Safety Factors

  • Factors increasing vehicle stopping distance:

  • Wet or icy road surfaces (reduces friction coefficient)

  • Worn-out tire treads (reduces grip)

  • Higher overall vehicle mass (increases momentum)

  • Delayed driver reaction time due to fatigue, fog, or night driving