Comprehensive Study Guide: Kinematics and One-Dimensional Motion

Fundamentals of Motion and Rest

• State of a Body:

  • Rest: A body is said to be at rest if its position does not change with respect to time and its frame of reference.
  • Motion: A body is said to be in motion if its position changes continuously with respect to time and its frame of reference.

• Branches of Mechanics in Motion:

  • Kinematics: The study of motion of objects without considering the causes or forces causing the motion. Key variables include speed, velocity, displacement, and acceleration.
  • Dynamics: The study of motion taking into account the forces and causes responsible for the motion (e.g., frictional force, gravitational force).

• Frame of Reference (FOR):

  • Defined as a coordinate system or specific point/location from which an observer observes and measures physical quantities related to motion.

• Key Principles of Rest and Motion:

  • Rest and motion are relative terms; they are not absolute.
  • An object can be simultaneously at rest with respect to one frame of reference and in motion with respect to another.
  • There is no object in the universe that is in a state of absolute rest.

• Position:

  • The specific location or point of an object relative to a chosen frame of reference.

Relative Velocity and Reference Frames

• Concept of Relative Velocity:

  • Relative velocity is the velocity of an object as observed from a particular frame of reference.
  • Mathematical definition: The relative velocity of body $A$ with respect to body $B$ ($V_{AB}$) is given by:     $V_{AB} = V_A - V_B$
  • $V_A$ = Velocity of body $A$
  • $V_B$ = Velocity of body $B$
  • $V_{AB}$ = Velocity of body $A$ with respect to body $B$
  • Here, body $A$ acts as the target object and body $B$ acts as the observer (frame of reference).

• Case Study / Reference Frame Activity:

  • Setup:
  • Sagar Sir at Police Station: $V_s = 0\,\text{m/s}$
  • Om Havaldar: $V_o = 12\,\text{m/s}$
  • Motorcycle with Argan ($V_a = 15\,\text{m/s}$) and passenger R ($V_r = 15\,\text{m/s}$)
  1. Frame of Reference of Sagar Sir ($V_s = 0\,\text{m/s}$):
  • Relative velocity of Om Havaldar: $V_{os} = V_o - V_s = 12 - 0 = 12\,\text{m/s}$
  • Relative velocity of Argan: $V_{as} = V_a - V_s = 15 - 0 = 15\,\text{m/s}$
  • Relative velocity of Passenger R: $V_{rs} = V_r - V_s = 15 - 0 = 15\,\text{m/s}$
  1. Frame of Reference of Om Havaldar ($V_o = 12\,\text{m/s}$):
  • Relative velocity of Sagar Sir: $V_{so} = V_s - V_o = 0 - 12 = -12\,\text{m/s}$
  • Relative velocity of Argan: $V_{ao} = V_a - V_o = 15 - 12 = 3\,\text{m/s}$
  • Relative velocity of Passenger R: $V_{ro} = V_r - V_o = 15 - 12 = 3\,\text{m/s}$
  1. Frame of Reference of Argan ($V_a = 15\,\text{m/s}$):
  • Relative velocity of Sagar Sir: $V_{sa} = V_s - V_a = 0 - 15 = -15\,\text{m/s}$
  • Relative velocity of Passenger R: $V_{ra} = V_r - V_a = 15 - 15 = 0\,\text{m/s}$
  • Relative velocity of Om Havaldar: $V_{oa} = V_o - V_a = 12 - 15 = -3\,\text{m/s}$

Classification of Motion

• One-Dimensional (1D) Motion:

  • Motion restricted along a straight line.
  • Also termed linear or rectilinear motion.
  • Example: A vehicle moving along a straight path.

• Two-Dimensional (2D) Motion:

  • Motion confined to a plane.
  • Examples: Circular motion, projectile motion.

• Three-Dimensional (3D) Motion:

  • Motion occurring freely in three-dimensional space.
  • Examples: Random movement of gas molecules, flight of a bird.

Kinematic Parameters: Distance and Displacement

• Distance ($x$):

  • Definition: The total path length traversed by a moving body during a given time interval.
  • S.I. Unit: Meter ($\text{m}$)
  • Character: Scalar quantity (has magnitude only, no direction).
  • Magnitude: Always positive ($\text{distance} > 0$ for a moving body).

• Displacement ($\Delta x$ or $s$):

  • Definition: The shortest straight-line distance directed from the initial position to the final position.
  • Formula:     $\Delta x = x_f - x_i$
  • S.I. Unit: Meter ($\text{m}$)
  • Dimensional Formula: $[L^1]$
  • Character: Vector quantity.
  • Direction: Points straight from initial position to final position.
  • Magnitude: Can be positive, negative, or zero.

• Relationship Between Distance and Displacement:

  • General case:     $\text{distance} \ge |\text{displacement}|$
  • For 1D / Linear motion in a single direction (without reversing direction):     $\text{distance} = |\text{displacement}|$
  • Example 1: Straight path from point $A$ to $B$ of length $4\,\text{m}$.
  • $\text{distance} = 4\,\text{m}$
  • $\text{displacement} = 4\,\text{m}$
  • Example 2: Curved or semi-circular path from $A$ to $B$.
  • Path length along arc ($\text{distance}$) is strictly greater than the direct path length ($|\text{displacement}|$).

Kinematic Parameters: Speed and Velocity

• Speed:

  • Measure of how fast or slow a body travels.
  • Definition: The ratio of total distance travelled to the total time taken.
  • Formula:     $\text{Speed} = \frac{\text{distance}}{\text{time}}$
  • S.I. Unit: $\text{m/s}$
  • Dimensional Formula: $[L^1 T^{-1}]$
  • Character: Scalar quantity.
  • Average Speed:     $\text{Speed}_{\text{avg}} = \frac{\text{total distance}}{\text{total time}}$
  • Instantaneous Speed:     $\text{Speed}_{\text{inst}} = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt}$
  • Physical Meaning: A speed of $x\,\text{m/s}$ implies $x\,\text{m}$ of path distance covered every second.

• Velocity:

  • Measure of speed along with directional orientation.
  • Definition: The ratio of total displacement to total time taken, or rate of change of displacement.
  • Formula:     $\text{Velocity} = \frac{\text{displacement}}{\text{time}}$
  • S.I. Unit: $\text{m/s}$
  • Dimensional Formula: $[L^1 T^{-1}]$
  • Character: Vector quantity.
  • Direction: Identical to the direction of displacement.
  • Average Velocity:     $V_{\text{avg}} = \frac{\text{total displacement}}{\text{total time}}$
  • Instantaneous Velocity:     $V_{\text{inst}} = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt}$

• Calculus & Rate Terminology:

  • Rate: Any physical parameter changing with respect to time ($\frac{d}{dt}$).
  • Gradient: Any physical parameter changing with respect to position ($\frac{d}{dx}$).
  • "In time interval": Corresponds to average value.
  • "At specific instant": Corresponds to instantaneous value.
  • Small differential interval:     $\lim_{\Delta x \to 0} f(x) = \frac{d[f(x)]}{dx}$

• Mathematical Relation Between Speed and Velocity:

  • Since $\text{distance} \ge |\text{displacement}|$:     $\frac{\text{distance}}{\text{time}} \ge \frac{|\text{displacement}|}{\text{time}} \implies \text{Speed} \ge |\text{Velocity}|$
  • For 1D motion without change in direction:     $\text{Speed} = |\text{Velocity}|$$(\text{Speed})_{\text{avg}} = |V_{\text{avg}}|$$(\text{Speed})_{\text{inst}} = |V_{\text{inst}}|$
  • Instantaneous speed is always equal to the magnitude of instantaneous velocity:     $(\text{Speed})_{\text{inst}} = |V_{\text{inst}}|$

Solved Problems on Average Speed and Velocity

• Problem 1: An object moves along a straight road without change in direction, covering distance $x_1$ in time $t_1$ and distance $x_2$ in time $t_2$. Find average speed and average velocity.

  • Solution:     Since direction does not change, average speed equals magnitude of average velocity.     $\text{Average Speed} = \frac{x_1 + x_2}{t_1 + t_2}$

• Problem 2: An object moves in a straight line with speed $v_1$ for time $t_1$ and speed $v_2$ for time $t_2$. Find average speed.

  • Solution:     $\text{Total distance} = v_1 t_1 + v_2 t_2$$\text{Total time} = t_1 + t_2$$\text{Average Speed} = \frac{v_1 t_1 + v_2 t_2}{t_1 + t_2}$

• Problem 3: An object moves in a straight line with speed $v_1$ and $v_2$ for equal time intervals $t$. Find average speed and average velocity.

  • Solution:     $V_{\text{avg}} = \frac{v_1 t + v_2 t}{t + t} = \frac{(v_1 + v_2)t}{2t} = \frac{v_1 + v_2}{2}$

• Problem 4: An object covers distance $x_1$ with speed $v_1$ and distance $x_2$ with speed $v_2$. Find average speed of the journey.

  • Solution:     Time intervals: $t_1 = \frac{x_1}{v_1}$, $t_2 = \frac{x_2}{v_2}$$\text{Average Speed} = \frac{x_1 + x_2}{\frac{x_1}{v_1} + \frac{x_2}{v_2}} = \frac{v_1 v_2 (x_1 + x_2)}{x_1 v_2 + x_2 v_1}$

• Problem 5: An object covers distance $x$ with speed $v_1$ and next equal distance $x$ with speed $v_2$. Find average speed.

  • Solution:     $V_{\text{avg}} = \frac{x + x}{\frac{x}{v_1} + \frac{x}{v_2}} = \frac{2x}{x\left(\frac{1}{v_1} + \frac{1}{v_2}\right)} = \frac{2 v_1 v_2}{v_1 + v_2}$

• Problem 6: A car travels from Mumbai to Sangli with a speed of $30\,\text{km/hr}$ and returns along the same path with a speed of $60\,\text{km/hr}$. Calculate the average speed.

  • Solution:     $\text{Total distance} = 2x$$t_1 = \frac{x}{30}$, $t_2 = \frac{x}{60}$$\text{Average Speed} = \frac{2x}{\frac{x}{30} + \frac{x}{60}} = \frac{2x}{\frac{3x}{60}} = \frac{2 \times 60}{3} = 40\,\text{km/hr}$

• Problem 7: Average speed of a car for a complete journey is $200\,\text{m/s}$. It moves with a speed of $100\,\text{m/s}$ for half of the total time. Find its speed during the remaining half of the time.

  • Options: a) $300\,\text{m/s}$, b) $100\,\text{m/s}$, c) $250\,\text{m/s}$, d) $900\,\text{m/s}$
  • Solution:     $V_{\text{avg}} = \frac{v_1 + v_2}{2}$$200 = \frac{100 + v}{2} \implies 400 = 100 + v \implies v = 300\,\text{m/s}$
  • Correct option: a

• Problem 8: A truck moves a distance of $50\,\text{km}$. It covers half of the distance at a speed of $200\,\text{m/s}$ and the second half at a speed of $v$. The average speed of the truck is $100\,\text{m/s}$. Find v$.\n - Options: a) 200\,\text{m/s}$, b)$\frac{200}{5}\,\text{m/s}$, c)$\frac{200}{3}\,\text{m/s}$, d)$10\,\text{m/s}\n - **Solution**:\n    Total distance = 50000\,\text{m}$, half distance =$25000\,\text{m}.\n    V_{\text{avg}} = \frac{2 v_1 v}{v_1 + v}\n    100 = \frac{2(200)v}{200 + v} \implies 100(200 + v) = 400v \implies 20000 + 100v = 400v\n    300v = 20000 \implies v = \frac{200}{3}\,\text{m/s}\n - Correct option: **c**\n\n- **Problem 9**: A body covers the first 1/3^{\text{rd}}$distance with velocity$10\,\text{m/s}$, second$1/3^{\text{rd}}$distance with velocity$20\,\text{m/s}$, and last$1/3^{\text{rd}}$distance with velocity$30\,\text{m/s}. Find average velocity.\n - Options: a) 17.8\,\text{m/s}$, b)$16.4\,\text{m/s}$, c)$18.3\,\text{m/s}$, d)$20.4\,\text{m/s}\n - **Solution**:\n    Let total distance = 3x.\n    t_1 = \frac{x}{10}$,$t_2 = \frac{x}{20}$,$t_3 = \frac{x}{30}\n    V_{\text{avg}} = \frac{3x}{\frac{x}{10} + \frac{x}{20} + \frac{x}{30}} = \frac{3x}{\frac{6x + 3x + 2x}{60}} = \frac{3x \times 60}{11x} = \frac{180}{11} = 16.36 \approx 16.4\,\text{m/s}\n - Correct option: **b**\n\n- **Problem 10**: An object moves along a circular path of radius r$with constant speed$v. Find average velocity and average speed when it completes half a revolution.\n - Options: a) \pi v, v$, b)$2\pi v, \frac{2v}{\pi}$, c)$\frac{2v}{\pi}, v$, d)$\frac{\pi}{2v}, v\n - **Solution**:\n    For half revolution:\n    Displacement = 2r$, Distance =$\pi r\n    Time taken t = \frac{\text{distance}}{\text{speed}} = \frac{\pi r}{v}\n    \text{Average Velocity} = \frac{\text{displacement}}{t} = \frac{2r}{\frac{\pi r}{v}} = \frac{2v}{\pi}\n    \text{Average Speed} = \frac{\text{distance}}{t} = \frac{\pi r}{\frac{\pi r}{v}} = v\n - Correct option: **c**\n\n- **Problem 11**: Calculate the average speed for 1 full revolution given a circular path divided into 3 sectors with angles 90^\circ$,$120^\circ$, and$150^\circ$traversed at speeds$20\,\text{m/s}$,$10\,\text{m/s}$, and$5\,\text{m/s} respectively.\n - Options: a) 2\,\text{m/s}$, b)$4\,\text{m/s}$, c)$8\,\text{m/s}$, d)$16\,\text{m/s}\n - **Solution**:\n    Converting angles to radians: \n    \theta_1 = 90^\circ = \frac{\pi}{2}\,\text{rad}$,$\theta_2 = 120^\circ = \frac{2\pi}{3}\,\text{rad}$,$\theta_3 = 150^\circ = \frac{5\pi}{6}\,\text{rad}\n    Total distance = r(\theta_1 + \theta_2 + \theta_3) = r\left(\frac{\pi}{2} + \frac{2\pi}{3} + \frac{5\pi}{6}\right) = 2\pi r\n    Time for sector 1: t_1 = \frac{r (\pi / 2)}{20} = \frac{\pi r}{40}\n    Time for sector 2: t_2 = \frac{r (2\pi / 3)}{10} = \frac{2\pi r}{30}\n    Time for sector 3: t_3 = \frac{r (5\pi / 6)}{5} = \frac{5\pi r}{30}\n    Total time t = r\left[\frac{\pi}{40} + \frac{2\pi}{30} + \frac{5\pi}{30}\right] = r\left[\frac{\pi}{40} + \frac{7\pi}{30}\right] = r\left[\frac{3\pi + 28\pi}{120}\right] = \frac{31\pi r}{120}\n    \text{Average Speed} = \frac{2\pi r}{\frac{31\pi r}{120}} = \frac{240}{31} = 7.74\,\text{m/s} \approx 8\,\text{m/s}\n - Correct option: **c**\n\n- **Problem 12**: A bus travels half its total journey distance at 5\,\text{m/s}$. The remaining distance is covered in 2 equal time intervals with speeds of$15\,\text{m/s}$and$25\,\text{m/s}. Calculate the average speed of the journey.\n - Options: a) 20\,\text{m/s}$, b)$7.5\,\text{m/s}$, c)$10\,\text{m/s}$, d)$8\,\text{m/s}\n - **Solution**:\n    Let total distance = 2x.\n    First half time: t_1 = \frac{x}{5}.\n    For the second half distance x$, average speed during this portion (equal time intervals) =$\frac{15 + 25}{2} = 20\,\text{m/s}.\n    Time for second half: t_2 = \frac{x}{20}.\n    \text{Average Speed} = \frac{2x}{\frac{x}{5} + \frac{x}{20}} = \frac{2}{\frac{1}{5} + \frac{1}{20}} = \frac{2}{\frac{5}{20}} = 8\,\text{m/s}\n - Correct option: **d**\n\n\n# Calculus Applications in Kinematics\n\n- **Problem 13**: The position of an object is given by x = t^2 + 6t + 5$. Find the instantaneous velocity at$t = 3\,\text{s}.\n - Options: a) 2\,\text{m/s}$, b)$4\,\text{m/s}$, c)$12\,\text{m/s}, d) None of these\n - **Solution**:\n    v = \frac{dx}{dt} = 2t + 6\n    At t = 3\,\text{s}$:$v = 2(3) + 6 = 12\,\text{m/s}\n - Correct option: **c**\n\n- **Problem 14**: If position of an object is x = 2t^2 - 4$, find velocity in$4\,\text{sec}$and at$4\,\text{sec}.\n - Options: a) 16\,\text{m/s}, 16\,\text{m/s}$, b)$8\,\text{m/s}, 16\,\text{m/s}$, c)$16\,\text{m/s}, 8\,\text{m/s}$, d)$8\,\text{m/s}, 8\,\text{m/s}\n - **Solution**:\n    Velocity in 4\,\text{sec}$(Average velocity from$t=0$to$t=4):\n    x(0) = -4$,$x(4) = 2(4)^2 - 4 = 28\n    V_{\text{avg}} = \frac{x(4) - x(0)}{4} = \frac{28 - (-4)}{4} = \frac{32}{4} = 8\,\text{m/s}\n    Velocity at 4\,\text{sec} (Instantaneous velocity):\n    v = \frac{dx}{dt} = 4t \implies v(4) = 4(4) = 16\,\text{m/s}\n - Correct option: **b**\n\n- **Problem 15**: Position of an object is given by x = 5t^3 + 4t - 5$. Find velocity in$2\,\text{sec}$and at$2\,\text{sec}.\n - **Solution**:\n    Velocity in 2\,\text{sec} (Average velocity):\n    x_i = x(0) = -5\n    x_f = x(2) = 5(2)^3 + 4(2) - 5 = 40 + 8 - 5 = 43\n    V_{\text{avg}} = \frac{43 - (-5)}{2} = \frac{48}{2} = 24\,\text{m/s}\n    Velocity at 2\,\text{sec} (Instantaneous velocity):\n    v = \frac{dx}{dt} = 15t^2 + 4\n    At t = 2\,\text{s}$:$v = 15(2)^2 + 4 = 15(4) + 4 = 64\,\text{m/s}\n\n- **Example 1**: Position of a particle is x = \alpha t^2 + \beta t + \gamma. Find initial velocity, initial acceleration, and initial position.\n - Options: a) \alpha, \beta, \gamma$, b)$\beta, 2\alpha, \gamma$, c)$\gamma, 2\alpha, \beta$, d)$2\alpha, \beta, \gamma\n - **Solution**:\n    At t = 0:\n    Initial position x_i = \gamma\n    v = \frac{dx}{dt} = 2\alpha t + \beta \implies \text{initial velocity } v_i = \beta\n    a = \frac{dv}{dt} = 2\alpha \implies \text{initial acceleration } a_i = 2\alpha\n - Correct option: **b** (listing v_i = \beta$,$a_i = 2\alpha$,$x_i = \gamma)\n\n- **Example 2**: Position of object is x = t^2 - 4t + 5$. Find the instant when velocity becomes$0 and displacement when body comes to rest.\n - Options: a) 2\,\text{sec}, 4\,\text{m}$, b)$2\,\text{sec}, -4\,\text{m}$, c)$4\,\text{sec}, 2\,\text{m}$, d)$4\,\text{sec}, -2\,\text{m}\n - **Solution**:\n    v = \frac{dx}{dt} = 2t - 4 = 0 \implies t = 2\,\text{sec}\n    Displacement \Delta x = x(2) - x(0)\n    x(0) = 5$,$x(2) = (2)^2 - 4(2) + 5 = 1\n    \Delta x = 1 - 5 = -4\,\text{m}\n - Correct option: **b**\n\n- **Example 5**: Object moves with velocity v = 4t^2 + 2t + 4$. Find velocity and acceleration at$t = 1\,\text{s}.\n - Options: a) 10, 8$, b)$10, -10$, c)$10, 10$, d)$8, -8\n - **Solution**:\n    At t = 1\,\text{s}$:$v = 4(1)^2 + 2(1) + 4 = 10\,\text{m/s}\n    a = \frac{dv}{dt} = 8t + 2 \implies a(1) = 8(1) + 2 = 10\,\text{m/s}^2\n - Correct option: **c**\n\n- **Example 6**: If velocity v = kx$where$k is constant, find acceleration.\n - Options: a) k^2$, b)$k^2/x$, c)$k^2 x$, d)$kx\n - **Solution**:\n    a = v \frac{dv}{dx} = (kx) \frac{d}{dx}(kx) = kx \cdot k = k^2 x\n - Correct option: **c**\n\n- **Example 7**: If velocity v = k\sqrt{x}$where$k is constant, find acceleration.\n - Options: a) k^2$, b)$k^2/2$, c)$\frac{k^2 x}{2}$, d)$k^2 x\n - **Solution**:\n    a = v \frac{dv}{dx} = k\sqrt{x} \frac{d}{dx}\left(k\sqrt{x}\right) = k\sqrt{x} \cdot k\left(\frac{1}{2\sqrt{x}}\right) = \frac{k^2}{2}\n - Correct option: **b**\n\n- **Example 8**: Velocity v = \beta x^{2n}$, where$\beta$and$n are constants. Find acceleration.\n - Options: a) 2n\beta^2 x^{4n-1}$, b)$2n\beta^{-2} x^{2n-1}$, c)$2n\beta^2 x^{2n+1}$, d)$2n\beta^{-2} x^{2n+1}\n - **Solution**:\n    a = v \frac{dv}{dx} = (\beta x^{2n}) \frac{d}{dx}(\beta x^{2n}) = (\beta x^{2n}) \cdot \beta (2n x^{2n-1}) = 2n \beta^2 x^{2n + 2n - 1} = 2n \beta^2 x^{4n-1}\n - Correct option: **a**\n\n- **Example 9**: Initial velocity of a particle is u$at$t = 0$and acceleration is$a = \alpha t^{3/2}$. Find final velocity at$t = t$.

  • Options: a) $v = u + \alpha t^{3/2}$, b) $v = u + \frac{3}{2}\alpha t^3$, c) $v = u + \frac{2}{5}\alpha t^{5/2}$, d) None of these
  • Solution:     $\frac{dv}{dt} = \alpha t^{3/2} \implies \int_u^v dv = \int_0^t \alpha t^{3/2} dt$$v - u = \alpha \left[ \frac{t^{5/2}}{5/2} \right]_0^t = \frac{2}{5}\alpha t^{5/2}$$v = u + \frac{2}{5}\alpha t^{5/2}$
  • Correct option: c

• Example 10: Acceleration $a = kx$ where $k$ is constant, initial velocity is $u$ at $t = 0$. Find velocity at position x$.\n - Options: a) v^2 = u^2 + 2kx$, b)$v^2 = u^2 + 2kx^2$, c)$v^2 = u^2 + kx^2$, d)$v^2 = -u^2 + 2kx\n - **Solution**:\n    v \frac{dv}{dx} = kx \implies \int_u^v v dv = \int_0^x kx dx\n    \left[ \frac{v^2}{2} \right]_u^v = k \left[ \frac{x^2}{2} \right]_0^x \implies \frac{v^2 - u^2}{2} = \frac{kx^2}{2} \implies v^2 = u^2 + kx^2\n - Correct option: **c**\n\n- **Example 11**: Acceleration a = f_0 \left(1 - \frac{t}{T}\right)$where$f_0$and$T$are constants. If initial velocity at$t = 0$is$0$, find velocity at$t = T$.

  • Options: a) $\frac{f_0 T}{2}$, b) $f_0 T$, c) $\frac{f_0 T^2}{2}$, d) $f_0 T^2$
  • Solution:     $\int_0^v dv = f_0 \int_0^T \left(1 - \frac{t}{T}\right) dt$$v = f_0 \left[ t - \frac{t^2}{2T} \right]_0^T = f_0 \left( T - \frac{T^2}{2T} \right) = \frac{f_0 T}{2}$
  • Correct option: a

• Example 12: Velocity $v = \sqrt{25 - x}$. Find acceleration at position x$.\n - Options: a) 2\,\text{m/s}^2$, b)$3\,\text{m/s}^2$, c)$4\,\text{m/s}^2$, d)$6\,\text{m/s}^2\n - **Solution**:\n    v^2 = 25 - x\n    Differentiating w.r.t x:\n    2v \frac{dv}{dx} = -1 \implies a = v \frac{dv}{dx} = -\frac{1}{2} = -0.5\,\text{m/s}^2\n\n\n# Newton's Kinematical Equations and Derivations\n\n1. **First Equation of Motion: v = u + at**\n - Derivation:\n     a = \frac{dv}{dt} \implies dv = a dt\n     Integrating both sides within limits (t: 0 \to t$,$v: u \to v):\n     \int_u^v dv = a \int_0^t dt\n     [v]u^v = a [t]_0^t\n     v - u = a(t - 0) \implies v = u + at\n\n2. **Second Equation of Motion: v^2 = u^2 + 2as**\n - Derivation:\n     a = v \frac{dv}{dx} \implies a dx = v dv\n     Integrating both sides within limits (x: x_i \to x_f$,$v: u \to v):\n     a \int{x_i}^{x_f} dx = \int_u^v v dv\n     a [x]{x_i}^{x_f} = \left[ \frac{v^2}{2} \right]_u^v\n     a (x_f - x_i) = \frac{v^2 - u^2}{2}\n     Let s = x_f - x_i:\n     2as = v^2 - u^2 \implies v^2 = u^2 + 2as\n\n3. **Third Equation of Motion: s = ut + \frac{1}{2}at^2**\n - Derivation:\n     v = u + at \implies \frac{dx}{dt} = u + at \implies dx = (u + at) dt\n     Integrating both sides within limits (x: x_i \to x_f$,$t: 0 \to t):\n     \int{x_i}^{x_f} dx = u \int_0^t dt + a \int_0^t t dt\n     x_f - x_i = u[t]0^t + a \left[ \frac{t^2}{2} \right]_0^t\n     s = u(t - 0) + a \left( \frac{t^2}{2} - 0 \right) \implies s = ut + \frac{1}{2}at^2\n\n4. **Displacement in n^{\text{th}}$Second:$S{n^{\text{th}}} = u + \frac{a}{2}(2n - 1)**\n - Derivation:\n     S_{n^{\text{th}}} = S_n - S_{n-1}\n     Where S_n$is displacement in$n$seconds and$S_{n-1}$is displacement in$(n-1) seconds.\n     S_n = un + \frac{1}{2}an^2\n     S_{n-1} = u(n - 1) + \frac{1}{2}a(n - 1)^2 = un - u + \frac{a}{2}(n^2 - 2n + 1)\n     Subtracting S_{n-1}$from$S_n:\n     S_{n^{\text{th}}} = \left(un + \frac{1}{2}an^2\right) - \left(un - u + \frac{1}{2}an^2 - an + \frac{a}{2}\right)\n     S_{n^{\text{th}}} = u + an - \frac{a}{2} = u + \frac{a}{2}(2n - 1)\n\n\n# Advanced Distance Fraction & Ratio Problems\n\n- **Example 1**: An object starts motion from u$with constant acceleration$a$. Find the velocity$v_m$at the midpoint of displacement$s$.

  • Options: a) $v_m = \frac{v^2 + u^2}{2}$, b) $v_m = \sqrt{\frac{v^2 + u^2}{2}}$, c) $v_m = v - u$, d) $v_m = 0$
  • Solution:     Total journey: $v^2 = u^2 + 2as \implies 2as = v^2 - u^2$     Midpoint ($s/2$): $v_m^2 = u^2 + 2a\left(\frac{s}{2}\right) = u^2 + as = u^2 + \frac{v^2 - u^2}{2} = \frac{v^2 + u^2}{2}$$v_m = \sqrt{\frac{v^2 + u^2}{2}}$
  • Correct option: b

• Example 2: An object starts motion from $u$ with constant acceleration. Find velocity $v'$ at $\frac{1}{3}^{\text{rd}}$ of total distance $s$.

  • Options: a) $v' = \sqrt{\frac{v^2 + 2u^2}{3}}$, b) $v' = \sqrt{\frac{v^2 + u^2}{3}}$, c) $v' = \sqrt{v^2 + 4u^2}$, d) $v' = \frac{v^2 + 4u^2}{3}$
  • Solution:     $v'^2 = u^2 + 2a\left(\frac{s}{3}\right) = u^2 + \frac{2as}{3} = u^2 + \frac{v^2 - u^2}{3} = \frac{v^2 + 2u^2}{3}$$v' = \sqrt{\frac{v^2 + 2u^2}{3}}$
  • Correct option: a

• Example 3: An object starts motion from $u$ with constant acceleration. Find velocity $v'$ at $\frac{1}{4}^{\text{th}}$ of total distance s$.\n - Options: a) v' = \sqrt{\frac{v^2 + 3u^2}{3}}$, b)$v' = \sqrt{\frac{v^2 + 3u^2}{4}}$, c)$v' = \sqrt{3u^2 + 4u^2}$, d)$v' = \frac{u^2 + v^2}{4}\n - **Solution**:\n    v'^2 = u^2 + 2a\left(\frac{s}{4}\right) = u^2 + \frac{2as}{4} = u^2 + \frac{v^2 - u^2}{4} = \frac{v^2 + 3u^2}{4}\n    v' = \sqrt{\frac{v^2 + 3u^2}{4}}\n - Correct option: **b**\n\n- **Example 4**: An object starts motion from u$with constant retardation$k$. It loses half its velocity after a displacement of$3\,\text{m}. Find further displacement required to bring it to rest.\n - Options: a) 1\,\text{m}$, b)$2\,\text{m}$, c)$3\,\text{m}$, d)$4\,\text{m}\n - **Solution**:\n    First stage: v = \frac{u}{2}$,$s_1 = 3\,\text{m}$,$a = -k\n    \left(\frac{u}{2}\right)^2 - u^2 = 2(-k)(3) \implies -\frac{3u^2}{4} = -6k \implies u^2 = 8k\n    Second stage: Initial velocity = \frac{u}{2}$, final velocity =$0$, displacement =$s\n    0^2 - \left(\frac{u}{2}\right)^2 = 2(-k)s \implies -\frac{u^2}{4} = -2ks\n    Substituting u^2 = 8k:\n    -\frac{8k}{4} = -2ks \implies -2k = -2ks \implies s = 1\,\text{m}\n - Correct option: **a**\n\n- **Example 5**: An object starts from rest with constant acceleration. Find:\n 1. Ratio of displacement in n$sec to$n^{\text{th}}$sec ($S_n : S_{n^{\text{th}}}).\n 2. Ratio of displacement in n^{\text{th}}$sec to$n$sec ($S_{n^{\text{th}}} : S_n).\n - **Solution**:\n    Since u = 0:\n    S_n = \frac{1}{2}an^2$,$S_{n^{\text{th}}} = \frac{a}{2}(2n - 1)\n 1. \frac{S_n}{S_{n^{\text{th}}}} = \frac{\frac{1}{2}an^2}{\frac{a}{2}(2n - 1)} = \frac{n^2}{2n - 1}\n 2. \frac{S_{n^{\text{th}}}}{S_n} = \frac{2n - 1}{n^2} = \frac{2}{n} - \frac{1}{n^2}\n\n- **Example 6**: An object starts from rest. Find the ratio of displacement in 6\,\text{sec}$to displacement in$6^{\text{th}}\,\text{sec}.\n - Options: a) \frac{11}{10}$, b)$\frac{36}{11}$, c)$\frac{15}{11}$, d)$\frac{17}{11}\n - **Solution**:\n    \frac{S_6}{S_{6^{\text{th}}}} = \frac{6^2}{2(6) - 1} = \frac{36}{12 - 1} = \frac{36}{11}\n - Correct option: **b**\n\n\n# Motion Under Gravity (Free Fall)\n\n- **Problem 1**: An object is dropped from rest (u = 0$). If displacement in the first$3\,\text{s} equals displacement in the last second of motion, find total time of flight and height.\n - Options: a) 10\,\text{sec}, 125\,\text{m}$, b)$5\,\text{sec}, 25\,\text{m}$, c)$10\,\text{sec}, 25\,\text{m}$, d)$5\,\text{sec}, 125\,\text{m}\n - **Solution**:\n    Displacement in first 3\,\text{s}$:$S(t=3) = \frac{1}{2}(-10)(3)^2 = -45\,\text{m}\n    Displacement in n^{\text{th}}$sec:$S_{n^{\text{th}}} = \frac{-10}{2}(2n - 1) = -5(2n - 1)\n    Equating magnitudes: 45 = 5(2n - 1) \implies 9 = 2n - 1 \implies n = 5\,\text{sec}\n    Time of flight T = 5\,\text{sec}.\n    Height H = \frac{1}{2}(10)(5)^2 = 5 \times 25 = 125\,\text{m}.\n - Correct option: **d**\n\n- **Problem 2**: An object is dropped (u = 0$). Displacement in the first$5\,\text{s} equals displacement in the last second. Find time of flight and total height.\n - Options: a) 445\,\text{m}, 13\,\text{sec}$, b)$125\,\text{m}, 26\,\text{sec}$, c)$12\,\text{m}, 2\,\text{sec}$, d)$845\,\text{m}, 13\,\text{sec}\n - **Solution**:\n    S(t=5) = \frac{1}{2}(-10)(5)^2 = -125\,\text{m}\n    S_{n^{\text{th}}} = -5(2n - 1)\n    125 = 5(2n - 1) \implies 25 = 2n - 1 \implies 26 = 2n \implies n = 13\,\text{sec}\n    Total height H = \frac{1}{2}(10)(13)^2 = 5 \times 169 = 845\,\text{m}.\n - Correct option: **d**\n\n- **Problem 3**: An object is released from height H$. Time taken to cover half of its height is$10\,\text{sec}. Find total time of flight.\n - Options: a) 10\sqrt{5}\,\text{sec}$, b)$10\sqrt{2}\,\text{sec}$, c)$10\,\text{sec}$, d)$\sqrt{2}\,\text{sec}\n - **Solution**:\n    First half height (H/2$) covered in$10\,\text{s}:\n    \frac{H}{2} = \frac{1}{2}(10)(10)^2 = 500 \implies H = 1000\,\text{m}\n    Full height equation:\n    H = \frac{1}{2} g T^2 \implies 1000 = 5 T^2 \implies T^2 = 200 \implies T = \sqrt{200} = 10\sqrt{2}\,\text{sec}\n - Correct option: **b**\n\n- **Problem 4**: A ball is dropped from rest. If it travels half of its total flight distance in its last 1\,\text{second}, find the total time of flight.\n - Options: a) 2 + \sqrt{3}\,\text{sec}$, b)$2 + \sqrt{2}\,\text{sec}$, c)$1 + \sqrt{2}\,\text{sec}, d) None of these\n - **Solution**:\n    First half height (H/2$) covered in time$(T - 1)$, full height$H$covered in time$T.\n    \frac{H}{2} = \frac{1}{2} g (T - 1)^2 \implies H = g(T - 1)^2\n    H = \frac{1}{2} g T^2\n    Equating both expressions for H:\n    \frac{1}{2} g T^2 = g (T - 1)^2 \implies T^2 = 2(T - 1)^2\n    Taking square root:\n    T = \sqrt{2}(T - 1) = \sqrt{2}T - \sqrt{2}\n    T(\sqrt{2} - 1) = \sqrt{2} \implies T = \frac{\sqrt{2}}{\sqrt{2} - 1} = \frac{\sqrt{2}(\sqrt{2} + 1)}{1} = 2 + \sqrt{2}\,\text{sec}\n - Correct option: **b**\n\n- **Problem 5**: Water drops fall from a tap at regular time intervals. When the 1st drop reaches the ground, the 4th drop is ready to release from the tap. Find the distance between the 2nd and 3rd drop if the tap is 1\,\text{m} above the ground.\n - **Solution**:\n    Let interval between drops = t.\n    1st drop reaches ground at time T = 3t$.     $H = 1\,\text{m} = \frac{1}{2} g (3t)^2 = \frac{9}{2} g t^2 \implies g t^2 = \frac{2}{9}$     Position of 3rd drop from tap (travelled for time $t$):     $y_3 = \frac{1}{2} g t^2 = \frac{1}{2} \left(\frac{2}{9}\right) = \frac{1}{9}\,\text{m}$     Position of 2nd drop from tap (travelled for time $2t$):     $y_2 = \frac{1}{2} g (2t)^2 = 4 \left(\frac{1}{2} g t^2\right) = 4 \left(\frac{1}{9}\right) = \frac{4}{9}\,\text{m}$     Distance between 2nd and 3rd drop:     $\Delta y = y_2 - y_3 = \frac{4}{9} - \frac{1}{9} = \frac{3}{9} = \frac{1}{3}\,\text{m}$