Comprehensive Kinematics Study Guide: Rest, Motion, Kinematic Parameters, Graphs, and Equations of Motion

Fundamental Concepts of Rest and Motion

• Definition of Rest: A body is said to be at rest when it does not change its position with respect to time or its surroundings. For example, a chair lying inside a room is in a state of rest because its position remains unchanged relative to the room's surroundings and time.

• Definition of Motion: A body is said to be in motion when it changes its position with respect to time or its surroundings. For example, a car changing its position relative to fixed surroundings such as trees, houses, and road posts is in a state of motion.

• Relativity of Rest and Motion: Rest and motion are relative terms rather than absolute conditions. Although seemingly opposed, they are closely linked. Whether an object is in rest or motion depends entirely on the frame of reference chosen. For example, a person sitting inside a compartment of a moving train is in a state of rest with respect to the immediate surroundings of the train compartment. However, relative to surroundings outside the compartment (such as trees or platforms), the exact same person is in a state of motion.

Scalar and Vector Quantities

• Scalar Quantities: Physical quantities that are completely expressed by their magnitude alone are called scalar quantities. Scalars possess no direction.

  • Examples: Mass, length, time, area, volume, density, energy, power, temperature, electric current. For example, $2\,\text{kg}$ of sugar specifies only the magnitude of mass and contains no directional component.

• Vector Quantities: Physical quantities that require both magnitude and direction for their complete specification are called vector quantities.

  • Examples: Displacement, velocity, acceleration, retardation, momentum, impulse, force. For example, $60\,\text{m}$ towards east is a vector quantity because it specifies both magnitude ($60\,\text{m}$) and direction (east).

• Representation of Vectors: A vector quantity is represented graphically by a straight line with an arrow head. The length of the arrow represents the magnitude of the vector quantity, while the arrow head indicates its direction. For example, vector $\mathbf{AB}$ represents a displacement where the distance $AB$ defines length/magnitude, and the point towards $B$ defines direction.

• Comparison Between Scalar and Vector Quantities:

  • Scalar Quantities:
  • Expressed in magnitude only.
  • Added by simple arithmetic rules.
  • Cannot be easily plotted on graph paper directly without directional axes.
  • Vector Quantities:
  • Expressed in magnitude as well as direction.
  • Cannot be added by simple arithmetic means; require vector algebra.
  • Can be easily plotted on standard coordinate graph paper.

Kinematic Parameters: Distance, Displacement, Speed, and Velocity

• Distance:

  • Definition: The total length of the actual path travelled by a moving body in a given interval of time.
  • Type: Scalar quantity.
  • Units: In the C.G.S. system, distance is measured in centimetres ($\text{cm}$). In the S.I. system, distance is measured in metres ($\text{m}$). It is denoted by the letter $S$

• Displacement:

  • Definition: The shortest straight-line distance measured from the initial position to the final position of a moving body.
  • Type: Vector quantity; directional specifications are mandatory.
  • Examples & Distinction: Moving $10\,\text{m}$ represents a distance of $10\,\text{m}$ in any path or direction. Moving $10\,\text{m}$ towards west specifies displacement as $10\,\text{m}\text{ (west)}$.
  • Zero Displacement Scenario: A moving body can cover a large distance while having zero total displacement. For example, if a body completes one full revolution along a circular path of radius $r$, the distance covered is $2\pi r$ (or $2r$ in simple axial representations), whereas its displacement is zero. Similarly, the Earth completes one rotation about its own axis every 24 hours, covering a huge distance while returning to a net displacement of zero.

• Speed:

  • Definition: The rate of change of motion, defined as distance covered per unit time.
  • Formula: $\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{S}{t}$
  • Type & Units: Scalar quantity. C.G.S. unit is $\text{cm/s}$ (or $\text{cm}\,\text{s}^{-1}$); S.I. unit is $\text{m/s}$ (or $\text{m}\,\text{s}^{-1}$).
  • Uniform Speed: When a body covers equal distances in equal intervals of time (no matter how small these time intervals are). Examples: A rotating ceiling fan operating at a fixed setting, a rocket moving in deep space.
  • Variable Speed: When a body covers unequal distances in equal intervals of time. Examples: A train starting from a station, a dog chasing a cat.
  • Average Speed: The ratio of the total distance travelled to the total time taken.     $\text{Average Speed} = \frac{\text{Total distance travelled}}{\text{Total time taken}}$
  • Instantaneous Speed: The speed of a body measured at a specific instant during continuously changing motion. The speedometer of an automobile measures instantaneous speed.

• Velocity:

  • Definition: The rate of change of position in a specified direction, or the rate of change of displacement.
  • Type & Units: Vector quantity. Units are identical to speed, accompanied by directional specification (e.g., $36\,\text{km/h}$ towards east is a velocity, whereas $36\,\text{km/h}$ is a speed).
  • Uniform Velocity: When a body covers equal distances in equal intervals of time in a specified straight-line direction.
  • Variable Velocity: When a body covers unequal distances in equal intervals of time in a specified direction, or when either the magnitude or the direction of motion changes. For example, a rotating fan turning at constant angular speed has variable linear velocity because its directional vector continuously changes.
  • Average Velocity: The ratio of total displacement (distance in a specified direction) to the total time taken.     $\text{Average Velocity} = \frac{\text{Total distance travelled in specified direction}}{\text{Total time taken}} = \frac{\text{Displacement}}{\text{Time}}$
  • Zero Average Velocity: A body executing complete circular motion experiences zero total displacement after one round; hence, its average velocity is zero, even though its average speed is non-zero.

• Comparison Between Speed and Velocity:

  • Speed:
  • Distance travelled per second.
  • Scalar quantity.
  • Always positive.
  • Average speed after one complete circular round is non-zero.
  • Velocity:
  • Distance travelled per second in a specific direction.
  • Vector quantity.
  • Can be positive or negative depending on directional orientation.
  • Average velocity after one complete circular round is zero.

Acceleration and Acceleration Due to Gravity

• Acceleration:

  • Definition: The rate of change of velocity of a body with respect to time. Non-uniform velocity (e.g., a cyclist navigating a busy street) constitutes accelerated motion.
  • Formula: $\text{Acceleration} = \frac{\text{Change in velocity}}{\text{Time}}$
  • Positive Acceleration: Occurs when velocity increases with time.
  • Negative Acceleration (Deceleration or Retardation): Occurs when velocity decreases with time.
  • Dimensional Derivation & Units:     $\text{Acceleration} = \frac{\text{Velocity}}{\text{Time}} = \frac{\text{Displacement}}{(\text{Time})^2}$     In terms of fundamental dimensions of Length ($L$) and Time ($T$), acceleration is expressed as $LT^{-2}$. In the C.G.S. system, unit is $\text{cm/s}^2$ or $\text{cm}\,\text{s}^{-2}$. In the S.I. system, unit is $\text{m/s}^2$ or $\text{m}\,\text{s}^{-2}$.
  • Vector Character: Represented by vector symbol $a$, requiring magnitude and directional orientation.
  • Uniform Acceleration: Motion in which a body undergoes equal changes in velocity in equal intervals of time.

• Acceleration Due to Gravity:

  • Definition: The uniform acceleration produced in a freely falling body under the sole gravitational attraction of the Earth. It is denoted by $g$.
  • Variations in Value: Constant at a given physical location, but varies across the Earth's surface:
  • Maximum at Earth's poles.
  • Decreases toward the equator.
  • Decreases at high altitudes above Earth's surface.
  • Decreases deep inside mines.
  • Numerical Values:
  • S.I. System: Average value $g = 9.8\,\text{m/s}^2$
  • C.G.S. System: Average value $g = 980\,\text{cm/s}^2$
  • Sign Convention: $g = +9.8\,\text{m/s}^2$ when a body falls downward toward Earth; $g = -9.8\,\text{m/s}^2$ when a body moves vertically upward.

Graphical Representation of Motion

• One-Dimensional (Rectilinear) Motion:

  • Motion along a single straight line path where directional orientation does not flip back and forth sideways (no lateral movement). Example: a freely falling stone.
  • Contrast: Curved path motion is two-dimensional; movement through open space is three-dimensional.

• Displacement-Time Graphs:

  • Displacement ($S$) is plotted on the Y-axis; Time ($t$) on the X-axis.
  • Slope: The slope of a displacement-time graph at any point equals the velocity of the body.
  • Graph Configurations:
  • Parallel to Time Axis: Slope is zero; body is stationary.
  • Straight Line inclined to Time Axis: Uniform velocity. Displacement is directly proportional to time ($S \propto t$).
    • Example calculation: If $\Delta S = 10\,\text{m}$ over $\Delta t = 2\,\text{s}$, velocity $v = \frac{10\,\text{m}}{2\,\text{s}} = 5\,\text{m/s}$. Across intervals like $\Delta S = (8 - 2)\,\text{m} = 6\,\text{m}$ and $\Delta t = (3 - 1)\,\text{s} = 2\,\text{s}$, slope $v = \frac{6\,\text{m}}{2\,\text{s}} = 3\,\text{m/s}$.
  • Curved Line: Variable velocity (accelerated motion). The tangent line's slope at a specific point yields instantaneous velocity.
  • Parallel to Displacement Axis: Physically impossible, as it implies displacement changes instantaneously without time passing.

• Velocity-Time Graphs:

  • Velocity ($v$) is plotted on the Y-axis; Time ($t$) on the X-axis.
  • Slope: Yields acceleration (positive slope = positive acceleration; negative slope = retardation).
  • Area Under Curve: Yields net displacement ($\text{Displacement} = \text{Velocity} \times \text{Time}$).
  • Graph Configurations:
  • Parallel to Time Axis: Zero slope ($a = 0$); constant uniform velocity. Displacement equals area of rectangle $PQRS = PS \times SR = 20\,\text{m/s} \times 4\,\text{s} = 80\,\text{m}$.
  • Straight Line from Origin: Uniform acceleration starting from rest ($u = 0$). Slope $a = \frac{16\,\text{m/s}}{5\,\text{s}} = 3.2\,\text{m/s}^2$. Area of triangle $ABC = \frac{1}{2} \times 16\,\text{m/s} \times 5\,\text{s} = 40\,\text{m}$.
  • Straight Line with Non-Zero Intercept: Uniform acceleration starting with initial velocity ($u \neq 0$). Slope $a = \frac{(25 - 5)\,\text{m/s}}{4\,\text{s}} = 5\,\text{m/s}^2$. Area of trapezium $ECAD = \frac{1}{2}(CE + AD) \times ED = \frac{1}{2}(5 + 25) \times 4 = 60\,\text{m}$.
  • Curved Line: Variable acceleration and variable velocity. Displacement estimated by summing square grid units under the curve (e.g., $12\text{ grid squares} \times 6\,\text{m/square} = 72\,\text{m}$).

• Acceleration-Time Graphs:

  • Acceleration ($a$) is plotted on the Y-axis; Time ($t$) on the X-axis.
  • Line Coinciding with Time Axis: Acceleration is zero; body moves with uniform velocity.
  • Line Parallel to Time Axis: Acceleration is constant (uniform acceleration); body moves with variable velocity. Area under graph ($\text{Acceleration} \times \text{Time}$) gives change in velocity.
  • Inclined Straight/Curved Line: Variable acceleration and variable velocity. Area under curve equals velocity change.

Mathematical Derivations of Equations of Motion

• First Equation of Motion: $v = u + at$

  • Graphical Derivation: Consider a velocity-time graph where initial velocity $u = AB$, final velocity $v = CE$, over time interval $t = AE$.     $\text{Acceleration } a = \text{Slope of line } BC = \frac{CE - DE}{BD} = \frac{v - u}{t}$$a t = v - u \implies v = u + at$
  • Special Cases:
  • Body starting from rest ($u = 0$): $v = at$
  • Freely falling body ($a = +g$): $v = u + gt$
  • Vertically projected upward body ($a = -g$): $v = u - gt$

• Second Equation of Motion: $S = ut + \frac{1}{2}at^2$

  • Graphical Derivation: Distance $S$ equals the total area under trapezium $ABCE$:     $S = \text{Area of rectangle } ABDE + \text{Area of triangle } BCD$$S = (AB \times AE) + \frac{1}{2}(BD \times CD) = (u \times t) + \frac{1}{2}(t \times [v - u])$     Substituting $v - u = at$:     $S = ut + \frac{1}{2}at^2$
  • Algebraic Method:     $\text{Distance } S = \text{Average Velocity} \times t = \left(\frac{u + v}{2}\right) \times t$     Substituting $v = u + at$:     $S = \left(\frac{u + u + at}{2}\right) \times t = \left(\frac{2u + at}{2}\right) \times t = ut + \frac{1}{2}at^2$
  • Special Cases:
  • Body starting from rest ($u = 0$): $S = \frac{1}{2}at^2$
  • Freely falling body ($a = +g$): $S = ut + \frac{1}{2}gt^2$
  • Vertically projected upward body ($a = -g$): $S = ut - \frac{1}{2}gt^2$

• Third Equation of Motion: $v^2 - u^2 = 2aS$

  • Derivation: Squaring the first equation of motion $v = u + at$:     $v^2 = (u + at)^2 = u^2 + 2uat + a^2 t^2 = u^2 + 2a\left(ut + \frac{1}{2}at^2\right)$     Substituting $S = ut + \frac{1}{2}at^2$:     $v^2 = u^2 + 2aS \implies v^2 - u^2 = 2aS$
  • Special Cases:
  • Body starting from rest ($u = 0$): $v^2 = 2aS$
  • Freely falling body ($a = +g$): $v^2 - u^2 = 2gS$
  • Vertically projected upward body ($a = -g$): $v^2 - u^2 = -2gS$

Worked Numerical Problems and Exercises

• Worked Problem 1 (Kinematic Quantities):

  • Problem: An aeroplane flies south and covers $324\,\text{km}$ in $20\,\text{minutes}$. Calculate (i) Displacement of aeroplane, (ii) Velocity in (a) $\text{km/h}$, (b) $\text{m/s}$.
  • Solution:
  • (i) Displacement = $324\,\text{km}\text{ south}$
  • (ii) Time $t = \frac{20}{60}\,\text{h} = \frac{1}{3}\,\text{h}$
    • (a) Velocity = \frac{324\,\text{km}}{1/3\,\text{h}} = 972\,\text{km/h}\text{ south}\n - (b) Velocity in ext{m/s} = 972 \times \frac{5}{18} = 270\,\text{m/s}\text{ south}\n\n- **Worked Problem 2 (Velocity Conversion & Acceleration)**:\n - *Problem*: The velocity of a car changes from 18\,\text{km/h}$to$72\,\text{km/h}$in$30\,\text{s}$. Calculate (i) change in velocity in$ ext{m/s}$, (ii) acceleration in (a)$ ext{km/h}^2$, (b)$ ext{m/s}^2\n - *Solution*:\n - (i) Change in velocity = (72 - 18)\,\text{km/h} = 54\,\text{km/h} = 54 \times \frac{5}{18} = 15\,\text{m/s}\n - (ii) (a) Acceleration in ext{km/h}^2 = \frac{54\,\text{km/h}}{30/3600\,\text{h}} = \frac{54}{1/120} = 6480\,\text{km/h}^2\n - (ii) (b) Acceleration in ext{m/s}^2 = \frac{15\,\text{m/s}}{30\,\text{s}} = 0.5\,\text{m/s}^2\n\n- **Worked Problem 3 (Displacement-Time Graph Analysis)**:\n - *Problem*: Given a displacement-time graph with coordinates (0,0), (4,8), (6,8), (9,0):\n - (i) Velocity between 0 - 4\,\text{s}$:$v = \frac{8\,\text{m} - 0}{4\,\text{s}} = 2\,\text{m/s}\n - (ii) Velocity between 4 - 6\,\text{s}$: Position is constant ($8\,\text{m}$);$v = 0\,\text{m/s}\n - (iii) Velocity between 6 - 9\,\text{s}$:$v = \frac{0 - 8\,\text{m}}{9\,\text{s} - 6\,\text{s}} = \frac{-8}{3} = -2.67\,\text{m/s}$(or$-2\,\text{m/s} over direct interval endpoints)\n - (iv) Average velocities:\n - (a) Between 0 - 4\,\text{s}$:$2\,\text{m/s}\n - (b) Between 0 - 6\,\text{s}$:$\frac{8\,\text{m}}{6\,\text{s}} = 1.33\,\text{m/s}\n - (c) Between 0 - 9\,\text{s}$: Net displacement is zero; Average Velocity =$0\,\text{m/s}\n\n- **Worked Problem 4 (Velocity-Time Graph Analysis - Cyclist)**:\n - *Problem*: A cyclist cycles at uniform rate of 8\,\text{m/s}$for$8\,\text{seconds}$, stops pedalling and comes to rest in the next$10\,\text{seconds}.\n - *Solution*:\n - (i) Retardation = Slope of line BC = \frac{8\,\text{m/s}}{10\,\text{s}} = 0.8\,\text{m/s}^2\n - (ii) Distance covered with uniform velocity = Area of rectangle = 8\,\text{m/s} \times 8\,\text{s} = 64\,\text{m}\n - (iii) Distance covered with variable velocity = Area of triangle = \frac{1}{2} \times 8\,\text{m/s} \times 10\,\text{s} = 40\,\text{m}
  • (iv) Total distance covered = $64\,\text{m} + 40\,\text{m} = 104\,\text{m}$
  • (v) Average velocity = \frac{\text{Total distance}}{\text{Total time}} = \frac{104\,\text{m}}{18\,\text{s}} = 5.77\,\text{m/s}\n\n- **Worked Problem 5 (Velocity-Time Graph Analysis - Trapezoidal Motion)**:\n - *Problem*: Velocity-time graph showing initial velocity 9\,\text{cm/s}$(at$t=0$), peak velocity$21\,\text{cm/s}$at$t=5\,\text{s}$, dropping to rest at$t=20\,\text{s}.\n - *Solution*:\n - (i) Deceleration in AB = Slope = \frac{21 - 12}{5} = 1.8\,\text{cm/s}^2
  • (ii) Acceleration in BC = Slope = \frac{21 - 12}{15} = 0.6\,\text{cm/s}^2\n - (iii) Total distance covered in ABCE = Area trapezium AFEB + Area trapezium CDBE = \frac{1}{2}(21+12) \times 5 + \frac{1}{2}(21+12) \times 15 = 82.5 + 247.5 = 330\,\text{cm}
  • (iv) Average velocity = \frac{330\,\text{cm}}{20\,\text{s}} = 16.5\,\text{cm/s}\n\n- **Worked Problem 6 (Comparative Two-Car Motion Graph)**:\n - *Problem*: Graph of Car A (linear increase from 0$to$80\,\text{m/s}$in$8\,\text{s}$) and Car B (starts at$2\,\text{s}$, rises to$60\,\text{m/s}$at$4\,\text{s}$, constant to$8\,\text{s}).\n - *Solution*:\n - (i) Acceleration of Car A = \frac{80 - 0}{8 - 0} = 10\,\text{m/s}^2
  • (ii) Acceleration of Car B ($2\,\text{s}$ to $4\,\text{s}$) = \frac{60 - 20}{4 - 2} = 20\,\text{m/s}^2\n - (iii) Cars have identical velocity at intersection points: t = 2\,\text{s}$and$t = 6\,\text{s}.\n - (iv) Position after 8\,\text{s}:\n - Distance Car A = \frac{1}{2} \times 80 \times 8 = 320\,\text{m}
    • Distance Car B = \frac{1}{2}(7 + 4) \times 60 = 330\,\text{m}\n - Car B is ahead by 330 - 320 = 10\,\text{m}.\n\n- **Worked Problem 7 (First Equation Application)**:\n - *Problem*: Car initially at rest (u = 0$), picks up velocity$72\,\text{km/h}$($20\,\text{m/s}$) over$25\,\text{m}. Calculate (i) acceleration, (ii) time.\n - *Solution*:\n - (i) v^2 - u^2 = 2aS \implies 20^2 - 0^2 = 2 \times a \times 25 \implies 400 = 50a \implies a = 8\,\text{m/s}^2\n - (ii) v = u + at \implies 20 = 0 + 8t \implies t = 2.5\,\text{s}\n\n- **Worked Problem 8 (Aeroplane Landing Retardation)**:\n - *Problem*: Aeroplane lands at 270\,\text{km/h}$($75\,\text{m/s}$) and stops after$1000\,\text{m}. Calculate (i) retardation, (ii) time taken.\n - *Solution*:\n - (i) v^2 - u^2 = 2aS \implies 0^2 - 75^2 = 2a(1000) \implies -5625 = 2000a \implies a = -2.8125\,\text{m/s}^2$. Retardation =$2.8125\,\text{m/s}^2\n - (ii) v = u + at \implies 0 = 75 - 2.8125t \implies t = 26.67\,\text{s}\n\n- **Worked Problem 9 (Time and Distance Calculations)**:\n - *Problem*: Car at rest picks up 72\,\text{km/h}$($20\,\text{m/s}$) in \frac{1}{4}\,\text{min}$ ($15\,\text{s}$). Calculate (i) acceleration, (ii) distance.
  • Solution:
  • (i) $v = u + at \implies 20 = 0 + a(15) \implies a = 1.33\,\text{m/s}^2$
  • (ii) $S = ut + \frac{1}{2}at^2 = 0(15) + \frac{1}{2}(1.33)(15)^2 = 150\,\text{m}$

• Worked Problem 10 (Aeroplane Runway Touchdown):

  • Problem: Touchdown at $225\,\text{km/h}$ ($62.5\,\text{m/s}$), stops after $2\,\text{minutes}$ ($120\,\text{s}$). Calculate (i) acceleration, (ii) runway length.
  • Solution:
  • (i) $v = u + at \implies 0 = 62.5 + a(120) \implies a = -0.52\,\text{m/s}^2$
  • (ii) $S = ut + \frac{1}{2}at^2 = 62.5(120) + \frac{1}{2}(-0.52)(120)^2 = 7500 - 3744 = 3756\,\text{m}$

• Worked Problem 11 (Two-Stage Deceleration with Same Force):

  • Problem: Car at $72\,\text{km/h}$ ($20\,\text{m/s}$) slows to $18\,\text{km/h}$ ($5\,\text{m/s}$) over $20\,\text{m}$. Calculate deceleration. If same force continues, find (i) total time to stop, (ii) total distance covered.
  • Solution:
  • $v^2 - u^2 = 2aS \implies 5^2 - 20^2 = 2a(20) \implies -375 = 40a \implies a = -9.375\,\text{m/s}^2$. Deceleration = $9.375\,\text{m/s}^2$
  • (i) Total time to stop ($v=0$, $u=20\,\text{m/s}$): $0 = 20 - 9.375t \implies t = 2.13\,\text{s}$
  • (ii) Total distance ($u=20\,\text{m/s}$, $t=2.13\,\text{s}$): $S = 20(2.13) + \frac{1}{2}(-9.375)(2.13)^2 = 42.6 - 21.26 = 21.34\,\text{m}$

• Worked Problem 12 (Free Fall Under Gravity):

  • Problem: Stone dropped from top of tower ($u = 0$) takes $4\,\text{s}$ to reach ground ($g = 10\,\text{m/s}^2$). Calculate (i) final velocity, (ii) height of tower.
  • Solution:
  • (i) $v = u + gt = 0 + 10(4) = 40\,\text{m/s}$
  • (ii) $h = ut + \frac{1}{2}gt^2 = 0(4) + \frac{1}{2}(10)(4)^2 = 80\,\text{m}$

• Worked Problem 13 (Vertical Upward Projection):

  • Problem: Stone projected vertically upward takes $1.5\,\text{s}$ to reach highest point ($v = 0$, $g = -10\,\text{m/s}^2$). Calculate (i) initial velocity, (ii) maximum height.
  • Solution:
  • (i) $v = u + gt \implies 0 = u - 10(1.5) \implies u = 15\,\text{m/s}$
  • (ii) $S = ut + \frac{1}{2}gt^2 = 15(1.5) + \frac{1}{2}(-10)(1.5)^2 = 22.5 - 11.25 = 11.25\,\text{m}$

• Worked Problem 14 (Multi-Phase Spaceship Travel):

  • Problem: Spaceship traveling at $300\,\text{km/s}$ fires engines for $15\,\text{seconds}$ to reach $600\,\text{km/s}$. Calculate total distance covered in $1\,\text{minute}$ ($60\,\text{seconds}$).
  • Solution:
  • Acceleration Phase ($t = 15\,\text{s}$):
    • $v = u + at \implies 600 = 300 + a(15) \implies a = 20\,\text{km/s}^2$
    • $S_1 = ut + \frac{1}{2}at^2 = 300(15) + \frac{1}{2}(20)(15)^2 = 4500 + 2250 = 6750\,\text{km}$
  • Uniform Velocity Phase ($v = 600\,\text{km/s}$, $t = 45\,\text{s}$):
    • $S_2 = \text{Velocity} \times \text{Time} = 600 \times 45 = 27000\,\text{km}$
  • Total Distance = $S_1 + S_2 = 6750\,\text{km} + 27000\,\text{km} = 33750\,\text{km}$

Questions & Discussion

• Practice Problem Set 1 Answers:

  • 1: Car covers $90\,\text{km}$ in $1.5\,\text{hours}$ east: (i) Displacement = $90\,\text{km}\text{ east}$, (ii) Velocity = (a) $60\,\text{km/h}$, (b) $16.67\,\text{m/s}$.
  • 2: Race horse runs north $540\,\text{m}$ in $1\,\text{minute}$: (i) Displacement = $540\,\text{m}\text{ north}$, (ii) Velocity = (a) $9\,\text{m/s}$, (b) $32.4\,\text{km/h}$.

• Practice Problem Set 2 Answers:

  • 1: Motorbike velocity change $54\,\text{km/h}$ in $1\,\text{minute}$: Acceleration = (a) $0.25\,\text{m/s}^2$, (b) $3240\,\text{km/h}^2$
  • 2: Speeding car changes velocity from $108\,\text{km/h}$ to $36\,\text{km/h}$ in $4\,\text{s}$: Deceleration = (i) $5\,\text{m/s}^2$, (ii) $64800\,\text{km/h}^2$
  • 3: Train from rest reaches $20\,\text{m/s}$ in $200\,\text{s}$, constant for $500\,\text{s}$, stops in $100\,\text{s}$: (a) Acceleration = $0.1\,\text{m/s}^2$, (b) Retardation = $0.2\,\text{m/s}^2$, (c) Total distance = $13\,\text{km}$ ($13000\,\text{m}$), (d) Average speed = $16.125\,\text{m/s}$.
  • 4: Ball thrown vertically up returns in $6\,\text{s}$: (i) Deceleration = $-10\,\text{m/s}^2$, Acceleration = $10\,\text{m/s}^2$, (ii) Total distance = $90\,\text{m}$, (iii) Average velocity = $15\,\text{m/s}$.
  • 5: Racing car at $50\,\text{m/s}$ comes to rest in $20\,\text{s}$: Acceleration = $-2.5\,\text{m/s}^2$
  • 6: Free-fall distance vs. square of time plot ($S$ vs $t^2$): The graph is a straight line passing through the origin. The slope of the $S - t^2$ graph equals \frac{1}{2}g, from which $g = 2 \times \text{slope}$.
  • 7: Tower top free-fall graph table values ($g = 10\,\text{m/s}^2$):
  • At $t = 0\,\text{s}$, $S = 0\,\text{m}$
  • At $t = 1\,\text{s}$, $S = \frac{1}{2}(10)(1)^2 = 5\,\text{m}$
  • At $t = 2\,\text{s}$, $S = \frac{1}{2}(10)(2)^2 = 20\,\text{m}$
  • At $t = 3\,\text{s}$, $S = \frac{1}{2}(10)(3)^2 = 45\,\text{m}$

• Practice Problem Set 3 Answers:

  • 1: From graph: Acceleration = $1.5\,\text{m/s}^2$, Deceleration = $2.5\,\text{m/s}^2$, Total distance = $120\,\text{m}$.
  • 2: From graph: Acceleration AB = $1.16\,\text{m/s}^2$, Deceleration BC = $2\,\text{m/s}^2$, Distance ABCE = $124\,\text{m}$, Average velocity CED = $3\,\text{m/s}$.
  • 3: Car P and Car Q graphs: (i) Acceleration Car P = $3.5\,\text{m/s}^2$, (ii) Acceleration Car Q ($2\,\text{s} - 5\,\text{s}$) = $8.33\,\text{m/s}^2$, (iii) Same velocity at $t = 3\,\text{s}$ and $t = 7\,\text{s}$, (iv) Car P is ahead after $10\,\text{s}$ by $12.5\,\text{m}$.

• Practice Problem Set 4 Answers (Equations of Motion):

  • 1: Motorbike at rest reaches $72\,\text{km/h}$ in $40\,\text{m}$: Acceleration = $5\,\text{m/s}^2$, Time = $4\,\text{s}$.
  • 2: Cyclist at $5\,\text{m/s}$ reaches $10\,\text{m/s}$ over $50\,\text{m}$: Acceleration = $0.75\,\text{m/s}^2$, Time = $6.67\,\text{s}$.
  • 3: Aeroplane landing at $216\,\text{km/h}$ stops in $2\,\text{km}$: Acceleration = $-0.9\,\text{m/s}^2$, Time = $66.67\,\text{s}$.
  • 4: Truck at $90\,\text{km/h}$ stops over $25\,\text{m}$: Retardation = $12.5\,\text{m/s}^2$, Time = $2\,\text{s}$.
  • 5: Racing car at rest reaches $180\,\text{km/h}$ in $4.5\,\text{s}$: Acceleration = $11.11\,\text{m/s}^2$, Distance = $112.5\,\text{m}$.
  • 6: Motorbike at $5\,\text{m/s}$ reaches $30\,\text{m/s}$ in $5\,\text{s}$: Acceleration = $5\,\text{m/s}^2$, Distance = $87.5\,\text{m}$.
  • 7: Motorbike at $90\,\text{km/h}$ slows to $18\,\text{km/h}$ in $2.5\,\text{s}$: Acceleration = $-8\,\text{m/s}^2$, Distance = $37.5\,\text{m}$.
  • 8: Cyclist at $36\,\text{km/h}$ stops in $2\,\text{s}$: Retardation = $5\,\text{m/s}^2$, Distance = $10\,\text{m}$.
  • 9: Motorbike at $90\,\text{km/h}$ slows to $54\,\text{km/h}$ over $40\,\text{m}$: Total time to rest = $5\,\text{s}$, Total distance = $62.5\,\text{m}$.
  • 10: Car slows $72\,\text{km/h}$ to $36\,\text{km/h}$ over $25\,\text{m}$: Total time to rest = $3.33\,\text{s}$, Total distance = $33.33\,\text{m}$.
  • 11: Packet dropped from hovering helicopter at height $h$ reaches ground in $12\,\text{s}$ ($g = 9.8\,\text{m/s}^2$): (i) $h = 705.6\,\text{m}$, (ii) Final velocity = $117.6\,\text{m/s}$.
  • 12: Stone dropped from cliff reaches ground in $8\,\text{s}$ ($g = 9.8\,\text{m/s}^2$): (i) Final velocity = $78.4\,\text{m/s}$, (ii) Height = $313.6\,\text{m}$.
  • 13: Stone thrown vertically upwards takes $3\,\text{s}$ to attain max height ($g = 9.8\,\text{m/s}^2$): (i) Initial velocity = $29.4\,\text{m/s}$, (ii) Maximum height = $44.1\,\text{m}$.
  • 14: Stone thrown vertically upwards returns to thrower in $4\,\text{s}$ ($g = 10\,\text{m/s}^2$): Upward time = $2\,\text{s}$, (i) Initial velocity = $20\,\text{m/s}$, (ii) Maximum height = $20\,\text{m}$.
  • 15: Spaceship moving at $50\,\text{km/s}$ fires engine for $10\,\text{s}$ to reach $60\,\text{km/s}$: Total distance in \frac{1}{2}\,\text{minute}$($30\,\text{s}$) =$1750\,\text{km}.\n - 16: Spaceship moving at 60\,\text{km/s}$fires retro-rockets for$20\,\text{s}$reducing velocity to$55\,\text{km/s}$: Total distance in$40\,\text{s}$=$2250\,\text{km}$$.