Physics Lecture 4 Notes: Kinematics and Graphical Analysis

Kinematics: Graphical Analysis (Lecture 4)

Problem 1: Particle with Sudden Acceleration Reversal

Problem Statement A particle starts from the origin from rest ($u = 0$, $x = 0$) and moves along the $x$-axis with a positive acceleration. It travels with a constant acceleration $a_o$ for $t_o$ seconds. Suddenly, the direction of acceleration changes while the magnitude remains the same ($-a_o$), and the particle keeps traveling forever.

Sub-problems:

1. At what time from the start does the particle cross the origin again?

2. How many instants after the start does the particle acquire a speed of $\frac{a_o t_o}{2}$?

Step-by-Step Derivation (Crossing the Origin):

• Phase 1 ($0 \le t \le t_o$):

  • Acceleration = $a_o$

  • Velocity at $t = t_o$: $v = u + at \implies v = 0 + a_o t_o = a_o t_o$

  • Displacement ($S_1$): $S = ut + \frac{1}{2} at^2 \implies S_1 = \frac{1}{2} a_o t_o^2$

• Phase 2 (t > t_o):

  • Acceleration = $-a_o$

  • Initial velocity for this phase ($u'$) = $a_o t_o$

  • To cross the origin, the net displacement must be zero ($S_{total} = 0$).

  • Formula for total displacement: $S_{total} = S_1 + S_2 = 0$

  • Using the area under the Velocity-Time ($v-t$) graph:

    • The area above the time axis (positive displacement) must equal the area below the time axis (negative displacement).

    • Area $A_1$ (from $0$ to $t_o$) = $\frac{1}{2} \times t_o \times (a_o t_o) = \frac{1}{2} a_o t_o^2$

    • Area $A_2$ (from $t_o$ to $2t_o$) = $\frac{1}{2} \times t_o \times (a_o t_o) = \frac{1}{2} a_o t_o^2$. At $t = 2t_o$, the velocity becomes zero.

    • Area $A_3$ (from $2t_o$ onwards): To return to the origin, the particle must accumulate a negative displacement equal to the total positive displacement gained ($A_1 + A_2 = a_o t_o^2$). Let the time taken to return from the stop point ($t=2t_o$) be $t'$.

    • $A_3 = \frac{1}{2} (t') (a_o t') = \frac{1}{2} a_o (t')^2$

    • Setting $A_3 = A_1 + A_2 \implies \frac{1}{2} a_o (t')^2 = a_o t_o^2 \implies (t')^2 = 2 t_{o}^{2} \implies t' = \sqrt{2} t_o$

  • Total Time ($T$):

    • $T = 2 t_o + t' = 2 t_o + \sqrt{2} t_o = (2 + \sqrt{2})t_o$

Speed Instants Calculation:

• The target speed is $v_{target} = \frac{a_o t_o}{2}$.

• Instant 1: During initial acceleration phase.

  • $v = a_o t = \frac{a_o t_o}{2} \implies t = \frac{1}{2} t_o$

• Instant 2: During the deceleration phase while the particle is still moving in the positive direction.

  • $v = u' + at = a_o t_o - a_o(t - t_o) = \frac{a_o t_o}{2}$

  • $1 - \frac{t}{t_o} + 1 = \frac{1}{2} \implies 2 - \frac{t}{t_o} = 0.5 \implies \frac{t}{t_o} = 1.5 \implies t = \frac{3}{2} t_o$

• Instant 3: During the deceleration phase after the velocity has reversed (moving in the negative direction).

  • $v = a_o t_o - a_o(t - t_o) = -\frac{a_o t_o}{2}$

  • $2 - \frac{t}{t_o} = -0.5 \implies t = 2.5 t_o = \frac{5}{2} t_o$

Problem 2: Acceleration followed by Deceleration (Alpha-Beta Problem)

Problem Statement A car accelerates from rest at a constant rate $\alpha$ for some time and then immediately decelerates at a constant rate $\beta$ to come back to rest. The total time elapsed is $t$ seconds.

Calculations:

1. Max Velocity ($V_m$):

   • Let acceleration time be $t_1$ and deceleration time be $t_2$. So, $t_1 + t_2 = t$.

   • From acceleration phase: $V_m = 0 + \alpha t_1 \implies t_1 = \frac{V_m}{\alpha}$

   • From deceleration phase: $0 = V_m - \beta t_2 \implies t_2 = \frac{V_m}{\beta}$

   • Substitute into total time: $\frac{V_m}{\alpha} + \frac{V_m}{\beta} = t$

   • $V_m \left( \frac{\alpha + \beta}{\alpha \beta} \right) = t \implies V_m = \frac{\alpha \beta t}{\alpha + \beta}$

2. Total Distance Travelled ($S$):

   • $S = \text{Area under v-t graph} = \frac{1}{2} \times t \times V_m$

   • $S = \frac{1}{2} \times t \times \left( \frac{\alpha \beta t}{\alpha + \beta} \right) = \frac{\alpha \beta t^2}{2(\alpha + \beta)}$

Graphical Analysis of 1-D Motion: Position-Time ($x-t$) Graphs

Key Principles:

• The slope of the position-time graph ($x-t$) represents velocity: $v = \frac{dx}{dt} = \tan(\theta)$.

• If the slope is constant, velocity is uniform, and acceleration $a = \frac{dv}{dt}$. If $v$ is constant, then $a = 0$.

Specific Cases Analyzed:

• Case 1: Particle at Rest

  • Graph is a horizontal line ($x = x_o$).

  • Slope = $0$, therefore $v = 0$ and $a = 0$.

• Case 2: Constant Positive Velocity from $x = x_o$

  • At $t = 0$, $x = +x_o$.

  • The graph is a straight line sloping upwards.

  • $v = \tan(\theta) = \text{constant}$, $a = 0$.

• Case 3: Constant Positive Velocity from Origin

  • At $t = 0$, $x = 0$.

  • $v = \tan(\theta) = \text{constant}$, $a = 0$.

• Case 4: Constant Positive Velocity from $x = -x_o$

  • At $t = 0$, $x = -x_o$.

  • $v = \tan(\theta) = \text{constant}$, $a = 0$.

• Case 5: Constant Negative Velocity from $x = +x_o$

  • The particle moves toward the origin.

  • The graph is a straight line sloping downwards.

  • $v = \text{slope} = -\tan(\alpha) = \text{negative constant}$.

• Case 6: Constant Negative Velocity from Origin

  • Starts at origin and moves in the negative $x$ direction.

  • $v = -\tan(\alpha) = \text{constant}$.

• Case 7: Constant Negative Velocity from $x = -x_o$

  • Moves further into the negative region.

  • $v = -\tan(\alpha) = \text{constant}$.

Summary of Uniform Motion:

• In all 7 cases above, acceleration $a = 0$.

• Equation: $x = x_i + vt$ or $\Delta x = vt$.

Graphical Analysis: Non-Uniform Motion (Accelerated Motion)

In non-uniform motion, the acceleration is not zero. For constant acceleration, the $x-t$ graph is a parabola described by the equation: $x = x_i + ut + \frac{1}{2} at^2$

Concavity and Acceleration:

• Positive Acceleration (a > 0): The graph is concave up (described colloquially as "Seedhi katori"). The slope (velocity) increases over time.

• Negative Acceleration/Retardation (a < 0): The graph is concave down (described colloquially as "Ulti katori"). The slope (velocity) decreases over time.

Visual Indicators:

• If the slope is becoming steeper, the particle is speeding up.

• If the slope is becoming flatter, the particle is slowing down (retardation).

Velocity-Time ($v-t$) Graphs

1. Uniform Motion ($v = \text{constant}$):

• The graph is a horizontal line.

• Slope = $0$, which means $a = 0$.

• Displacement = Area under the curve = $v_o \times t_o$.

2. Uniform Acceleration ($a = \text{constant}$):

• The graph is a straight line with a constant non-zero slope.

• Slope = Acceleration ($a$).

• Equations involved:

  • $v = u + at$

  • $s = ut + \frac{1}{2} at^2$

  • $v^2 = u^2 + 2as$

Assignment and Administrative Details

Homework:

• Exercise 1 (P-1 & P-2)

• Module: Complete sections up to Section-C (P-2)

Test Schedule:

• Test Name: JEE Main-04

• Scheduled Date: 05 July 2026

• Syllabus:

  • Physics: Units and Dimensions & Vectors

  • Chemistry: Structure of Atom

  • Mathematics: Set Theory, Logarithms

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