Motion in 1D

Motion in 1D (Kinematics)

Fundamentals of Motion

Dimensionality of Motion

• One-Dimensional (1D) Motion: Motion along a straight line (e.g., purely along the x, y, or z-axis).
• Two-Dimensional (2D) Motion: Motion within a plane (e.g., the x-y plane). Examples include projectile motion and circular motion.
• Three-Dimensional (3D) Motion: Motion in space requiring all three coordinates (x, y, z) to specify position (e.g., a flying bird or a gas molecule).

Rest, Motion, and Reference Frames

• Rest and Motion are Relative: An object can be at rest relative to one observer but in motion relative to another. For example, passengers in a moving train are at rest relative to each other but in motion relative to a person standing on the platform.
• Reference Frame: The coordinate system (and observer) from which the position and time of events are measured.

The Absolute Truth

> There is no such thing as **absolute rest** or **absolute motion** in the universe. Everything is in motion relative to something else.

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Distance and Displacement

Definitions

• Distance ($d$): The actual total path length covered by a moving particle.
  • It is a Scalar quantity (only has magnitude).
  • Unit: $m$ (SI), $cm$ (CGS).
• Displacement ($\vec{s}$ or $\Delta x$): The shortest straight-line path directed from the initial position to the final position.
  • It is a Vector quantity (has both magnitude and direction).
  • Unit: $m$ (SI), $cm$ (CGS).

Key Differences

Feature	Distance	Displacement
Sign	Always positive ($+ve$) or zero. Can never be negative.	Can be positive ($+ve$), negative ($-ve$), or zero.
Time Dependence	For a moving body, distance never decreases with time. It only increases or stays constant.	Displacement can increase, decrease, or become zero.
Magnitude Relation	$\text{Distance}\ge \Vert \text{Displacement}\Vert$	$\Vert \text{Displacement}\Vert \le \text{Distance}$

Displacement in Circular Paths

If a particle moves along a circular path of radius $R$ and subtends an angle $\theta$ at the center, the displacement between the initial and final points is:
$\boxed{|\vec{s}|=2R\sin \left(\frac{\theta }{2}\right)}$

Example

Q1. A particle moves in a semicircle of radius $R$ from point A to point B. What are the distance and displacement?

• $\text{Distance}=\frac{2\pi R}{2}=\pi R$
• $\text{Displacement}=2R\sin \left(\frac{180^{\circ }}{2}\right)=2R\sin (90^{\circ })=2R$ (Straight across the diameter)

Q2. A particle moves 8m East, 4m North, 4m West, and 1m North. Find distance and displacement.

• $\text{Distance}=8+4+4+1=17\text{ m}$.
• Final position relative to start: $x=8-4=4\text{ m}$ East, $y=4+1=5\text{ m}$ North.
• $\text{Displacement}=\sqrt{4^2+5^2}=\sqrt{16+25}=\sqrt{41}\text{ m}$.

Q3. A pendulum of length $28\text{ cm}$ swings through an angle of $60^{\circ }$. Find distance and displacement.

• $\text{Displacement}=2R\sin \left(\frac{\theta }{2}\right)=2(28)\sin (30^{\circ })=2(28)\left(\frac{1}{2}\right)=28\text{ cm}$.
• $\text{Distance (Arc Length)}=R\theta =28\times \left(\frac{60\times \pi }{180}\right)=28\times \frac{\pi }{3}\approx 29.3\text{ cm}$.

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Speed and Velocity

Definitions

• Speed: The rate of distance covered with respect to time. (Scalar)
  • $\text{Speed}=\frac{\text{Distance}}{\text{Time}}$
• Velocity: The rate of change of position (displacement) with respect to time. (Vector)
  • $\vec{v}=\frac{\text{Displacement}}{\text{Time}}$

Uniform Speed vs Uniform Velocity

• Uniform Velocity means both magnitude (speed) and direction are constant. The particle must move in a straight line.
• Uniform Speed means only magnitude is constant. A particle can have uniform speed but variable velocity if its direction changes (e.g., uniform circular motion).

Average Values

Average values are calculated over a finite time interval $\Delta t$.

• Average Speed: $v_{avg,speed}=\frac{\text{Total Distance}}{\text{Total Time}}$
• Average Velocity: ${\vec{v}}_{avg}=\frac{\text{Total Displacement}}{\text{Total Time}}=\frac{\Delta x}{\Delta t}$
• Note: The formula $v_{avg}=\frac{v_1+v_2}{2}$ is only valid for motion with constant acceleration.

Example

Q1. A train travels from station A to B at $40\text{ km/h}$ and returns from B to A at $60\text{ km/h}$. Find its average speed and average velocity.

• Since initial and final positions are the same (A), $\text{Displacement}=0⟹\text{Average Velocity}=0$.
• Let distance A to B be $x$. Total distance = $2x$.
• Total time = $t_1+t_2=\frac{x}{40}+\frac{x}{60}=x\left(\frac{3+2}{120}\right)=\frac{5x}{120}=\frac{x}{24}$.
• $\text{Average Speed}=\frac{2x}{x/24}=2\times 24=48\text{ km/h}$.

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Relations between Speed and Velocity

• Signs: Speed is always positive or zero ($v\ge 0$). Velocity can be negative depending on direction.
• Uniformity: Uniform Velocity guarantees Uniform Speed, but the reverse is only true for straight-line motion.
• Zero Averages: If $v_{avg,speed}=0$, then ${\vec{v}}_{avg}$ must be $0$. But if ${\vec{v}}_{avg}=0$, $v_{avg,speed}$ can still be non-zero (e.g., returning to start).
• Instantaneous vs Average:
  • Instantaneous: Magnitude of instantaneous velocity always equals instantaneous speed ($|\vec{v}|=v$).
  • Average: Magnitude of average velocity $\le$ average speed ($|{\vec{v}}_{avg}|\le v_{avg,speed}$). They are equal only in straight-line, one-way motion.

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Calculus in Kinematics

When position, velocity, or acceleration are given as continuous functions of time ($t$), we use differentiation and integration to analyze the motion.

Instantaneous Values (Differentiation)

Instantaneous values are calculated at a specific, exact instant in time (where $\Delta t\to 0$).

• Instantaneous Velocity (FYI just ts works for speed too): $\boxed{v=\frac{dx}{dt}}$
• Instantaneous Acceleration: $\boxed{a=\frac{dv}{dt}=\frac{d^{2}x}{d{t}^2}}$
• Chain Rule for Acceleration: $\boxed{a=v\frac{dv}{dx}}$

Finding Averages via Integration

When velocity is a function of time $v(t)$, displacement is the integral of velocity. When velocity is a function of position $v(t)$, it’ll be same, just replace the $t$ with $x$ in the below equations.

• Displacement: $\Delta x={\int }_{t_1}^{t_2}vdt$
• Average Velocity: $\boxed{v_{avg}=\frac{{\int }_{t_1}^{t_2}vdt}{t_2-t_1}}$

Example

Q1. A particle’s position is given by $x=8t-3t^2$. Find the average velocity between $t=0$ and $t=1\text{ s}$, and the instantaneous velocity at $t=1\text{ s}$.

• Average Velocity:
  • At $t=0$, $x_1=0$.
  • At $t=1$, $x_2=8(1)-3(1{)}^2=5\text{ m}$.
  • $v_{avg}=\frac{x_2-x_1}{t_2-t_1}=\frac{5-0}{1-0}=5\text{ m/s}$.
• Instantaneous Velocity:
  • $v=\frac{dx}{dt}=8-6t$.
  • At $t=1\text{ s}$, $v=8-6(1)=2\text{ m/s}$.

Q2. If velocity is given by $v=2t^2+3t+1$, find the average velocity for the interval $0\le t\le 2\text{ s}$.

• $v_{avg}=\frac{{\int }_0^2(2t^2+3t+1)dt}{2-0}$
• Integral = ${\left[\frac{2t^3}{3}+\frac{3t^2}{2}+t\right]}_0^2=\left(\frac{16}{3}+6+2\right)-0=\frac{16}{3}+8=\frac{40}{3}$.
• $v_{avg}=\frac{40/3}{2}=\frac{20}{3}\text{ m/s}$.

Q3. The instantaneous velocity is given by $v=\alpha t+\beta t^2$. Find the distance travelled between $t=1\text{ s}$ and $t=2\text{ s}$. (Assume 1D motion without turning).

• Distance = ${\int }_1^{2}vdt={\int }_1^2(\alpha t+\beta t^2)dt$
• $={\left[\frac{\alpha t^2}{2}+\frac{\beta t^3}{3}\right]}_1^2=\left(\frac{4\alpha }{2}+\frac{8\beta }{3}\right)-\left(\frac{\alpha }{2}+\frac{\beta }{3}\right)=\frac{3\alpha }{2}+\frac{7\beta }{3}$.

Q4. The position of a particle as a function of time $t$ is given by $x(t)=at+b{t}^2-c{t}^3$, where $a,b,$ and $c$ are constants. When the particle attains zero acceleration, what will its velocity be?

• Velocity: $v=\frac{dx}{dt}=a+2bt-3c{t}^2$
• Acceleration: $a_{\text{acc}}=\frac{dv}{dt}=2b-6ct$
• For zero acceleration: $2b-6ct=0⟹t=\frac{b}{3c}$
• Substitute $t$ into the velocity equation:
  $v=a+2b\left(\frac{b}{3c}\right)-3c{\left(\frac{b}{3c}\right)}^2=a+\frac{2b^2}{3c}-\frac{3c{b}^2}{9c^2}=a+\frac{2b^2}{3c}-\frac{b^2}{3c}=\mathbf{a}+\frac{b^2}{3c}$

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Graphical Analysis of Motion

Slope of Graphs

• Position-Time ($x-t$) Graph: The slope (or gradient) gives the Velocity.
  • $\text{Slope}=\frac{dx}{dt}=v$
  • Constant slope = Constant velocity.
  • Zero slope = Body is at rest.

• Velocity-Time ($v-t$) Graph: The slope gives the Acceleration.
  • $\text{Slope}=\frac{dv}{dt}=a$
  • Positive slope = Increasing velocity (positive acceleration).
  • Negative slope = Decreasing velocity (retardation/deceleration).

• Acceleration-Time ($a-t$) Graph: The slope gives the Jerk (rate of change of acceleration).

Area Under Graphs

• Velocity-Time ($v-t$) Graph: The area under the curve gives the Displacement (if proper signs are considered) and Distance (if only magnitudes/absolute areas are summed).
  • $\text{Displacement}=\int vdt=\text{Area under }v-t\text{ graph}$
  • Note: Area above the time axis is taken as positive, area below is negative for displacement. For distance, all areas are added positively.
• Acceleration-Time ($a-t$) Graph: The area under the curve gives the Change in Velocity ($\Delta v$).
  • $\Delta v=\int adt=\text{Area under }a-t\text{ graph}$

Example

Q1. The $v-t$ plot of a moving object during the first $10\text{ s}$ is a triangle above the axis (height $5\text{ m/s}$, base $5\text{ s}$) and a symmetric triangle below the axis (depth $-5\text{ m/s}$, base from $5\text{ s}$ to $10\text{ s}$). Find the average velocity.

• $\text{Area 1 (Positive)}=\frac{1}{2}\times 5\times 5=12.5\text{ m}$
• $\text{Area 2 (Negative)}=\frac{1}{2}\times 5\times (-5)=-12.5\text{ m}$
• $\text{Total Displacement}=12.5-12.5=0\text{ m}$
• $\text{Average Velocity}=\frac{0}{10}=0\text{ m/s}$.

Q2. The position, velocity, and acceleration of a particle moving with a constant acceleration can be represented by:

• Since $a=\text{constant}$, the $a-t$ graph is a horizontal straight line.
• $v=u+at$, so the $v-t$ graph is a straight line with a constant slope.
• $x=ut+\frac{1}{2}a{t}^2$, so the $x-t$ graph is a parabola.

Q3. If a velocity-time graph has a V-shape (shape AMB, where AM has a negative slope and MB has a positive slope), what is the shape of the acceleration-time graph?

• Path AM: Constant negative slope $⟹$ constant negative acceleration.
• Path MB: Constant positive slope $⟹$ constant positive acceleration.
• Result: A step function showing a constant negative value followed instantly by a constant positive value.

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Equations of Motion (Constant Acceleration)

When the acceleration of a particle is constant (uniform), the following kinematic equations directly relate initial velocity ($u$), final velocity ($v$), acceleration ($a$), displacement ($s$), and time ($t$).

The Standard Equations

1. Velocity-Time Relation:
   $\boxed{v=u\pm at}$
2. Position-Time Relation:
   $\boxed{s=ut\pm \frac{1}{2}a{t}^2}$
3. Velocity-Position Relation:
   $\boxed{v^2=u^2\pm 2as}$

Note: Use $+a$ for acceleration (velocity increasing) and $-a$ for retardation (velocity decreasing).

Displacement in the $n^{\text{th}}$ second

The displacement covered specifically during the $n^{\text{th}}$ second of motion (from $t=n-1$ to $t=n$) is given by:
$\boxed{s_n=u+\frac{a}{2}(2n-1)}$

Example

Q1. A particle starts with initial velocity $u$ and constant acceleration $a$. Show the derivation of displacement in $n^{\text{th}}$ second.

• Displacement in $n$ seconds: $s_n=un+\frac{1}{2}a{n}^2$
• Displacement in $(n-1)$ seconds: $s_{n-1}=u(n-1)+\frac{1}{2}a(n-1{)}^2$
• $s_{n^{\text{th}}}=s_n-s_{n-1}$
• $s_{n^{\text{th}}}=\left[un+\frac{1}{2}a{n}^2\right]-\left[u(n-1)+\frac{1}{2}a(n^2-2n+1)\right]$
• $s_{n^{\text{th}}}=u+\frac{a}{2}(2n-1)$.

Q2. A body starts from rest and moves with constant acceleration. What is the ratio of distance covered in the $n^{\text{th}}$ second to the distance covered in $n$ seconds?

• $s_{n^{\text{th}}}=u+\frac{a}{2}(2n-1)=\frac{a}{2}(2n-1)$ (since $u=0$)
• $s_n=ut+\frac{1}{2}a{t}^2=\frac{1}{2}a{n}^2$
• Ratio $=\frac{\frac{a}{2}(2n-1)}{\frac{1}{2}a{n}^2}=\frac{2n-1}{n^2}=\frac{2}{n}-\frac{1}{n^2}$

Q3. A train’s initial velocity is $48\text{ km/h}$. After travelling $108\text{ m}$ with uniform retardation, its velocity becomes $24\text{ km/h}$. How much further will it travel before coming to rest?

• Let’s find acceleration ($a$) first. $u=48\text{ km/h}$, $v=24\text{ km/h}$, $s=0.108\text{ km}$.
• $v^2=u^2+2as⟹24^2=48^2+2a(0.108)$
• $a=\frac{576-2304}{0.216}=-8000{\text{ km/h}}^2$
• Now, to stop: $u^{\prime }=24\text{ km/h}$, $v^{\prime }=0$, $s^{\prime }=?$
• $0=24^2+2(-8000)s^{\prime }⟹s^{\prime }=\frac{576}{16000}=0.036\text{ km}=36\text{ m}$.

Q4. A car moving at $50\text{ km/h}$ stops after $6\text{ m}$ by applying brakes. If it moves at $100\text{ km/h}$, what is the minimum stopping distance?

• $v^2=u^2-2as⟹0=u^2-2as⟹s\propto u^2$ (for same constant deceleration)
• If speed doubles ($50\to 100$), stopping distance becomes $2^2=4$ times.
• $s_{\text{new}}=4\times 6=24\text{ m}$.

Q5. A particle starts from rest, accelerates at $2{\text{ m/s}}^2$ for $10\text{ s}$, goes for a constant speed for $30\text{ s}$, and then decelerates at $4{\text{ m/s}}^2$ till it stops. What is the total distance travelled?

• Stage 1 (Acceleration): $u=0$, $a=2{\text{ m/s}}^2$, $t=10\text{ s}$.
  • Distance $s_1=0+\frac{1}{2}(2)(10{)}^2=100\text{ m}$.
  • Final velocity $v_1=0+2(10)=20\text{ m/s}$.
• Stage 2 (Constant Speed): $v=20\text{ m/s}$, $t=30\text{ s}$.
  • Distance $s_2=20\times 30=600\text{ m}$.
• Stage 3 (Deceleration): $u_3=20\text{ m/s}$, $a_3=-4{\text{ m/s}}^2$, $v_3=0$.
  • $v_3^2=u_3^2+2a_3{s}_3⟹0=400-8s_3⟹s_3=50\text{ m}$.
• Total Distance: $S=s_1+s_2+s_3=100+600+50=750\text{ m}$.

Q6. A car starting from rest accelerates at rate $f$ through a distance $S$, then continues at constant speed for time $t$, and then decelerates at rate $f/2$ to come to rest. If the total distance is $15S$, find $S$ in terms of $f$ and $t$.

• Stage 1: $v^2=0+2fS⟹v=\sqrt{2fS}$.
• Stage 2: $S_2=v\times t=\sqrt{2fS}\cdot t$.
• Stage 3: $0=v^2-2(f/2)S_3⟹S_3=\frac{v^2}{f}=\frac{2fS}{f}=2S$.
• Total Distance: $S+S_2+S_3=15S$
  • $S+\sqrt{2fS}\cdot t+2S=15S$
  • $\sqrt{2fS}\cdot t=12S$
  • $2fS{t}^2=144S^2⟹2f{t}^2=144S⟹\mathbf{S}=\frac{f{t}^2}{72}$.

Q7. Body A starts from rest with acceleration $a_1$. After $2\text{ s}$, Body B starts from rest with acceleration $a_2$. If they travel equal distances in the $5^{\text{th}}$ second (after the start of A), what is $a_1:a_2$?

• Distance by A in its $5^{\text{th}}$ second: $s_{A,5}=0+\frac{a_1}{2}(2(5)-1)=\frac{9a_1}{2}$.
• Since B starts $2\text{ s}$ later, A’s $5^{\text{th}}$ second is B’s $3^{\text{rd}}$ second.
• Distance by B in its $3^{\text{rd}}$ second: $s_{B,3}=0+\frac{a_2}{2}(2(3)-1)=\frac{5a_2}{2}$.
• Equating them: $\frac{9a_1}{2}=\frac{5a_2}{2}⟹9a_1=5a_2⟹\frac{a_1}{a_2}=\frac{5}{9}$.

Q8. A $200\text{ m/s}$ bullet just penetrates a $4\text{ cm}$ plank. Assuming uniform resistance, what velocity is required to penetrate a $10\text{ cm}$ plank?

• First, find the retardation ($a$): $v^2=u^2+2as⟹0=(200{)}^2+2a(0.04)$
  • $a=\frac{-40000}{0.08}=-500000{\text{ m/s}}^2$
• Now for the $10\text{ cm}$ plank: $0=u_{\text{new}}^2+2(-500000)(0.1)$
  • $u_{\text{new}}^2=100000⟹u_{\text{new}}=\sqrt{10^5}=100\sqrt{10}\text{ m/s}$.

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Vertical Motion under Gravity

When an object is thrown upwards or dropped, it experiences a constant downward acceleration due to gravity ($g$). The kinematic equations can be directly applied.

Sign Convention

• Upward Direction: Taken as Positive ($+ve$)
• Downward Direction: Taken as Negative ($-ve$)
• Acceleration: Always acts downwards, so $a=-g\approx -9.8{\text{ m/s}}^2$ (or $-10{\text{ m/s}}^2$ for simpler calculations).

Key Properties (Equations of Vertical Motion)

When an object is thrown vertically upwards with initial velocity $u$ from the ground ($u\ne 0$):

• Speed Symmetry: At the same height level, the speed (magnitude of velocity) of the object is exactly the same while going up and coming down, but the direction is opposite.
• Time Symmetry: The time taken to reach a particular height during ascent is exactly equal to the time taken to fall from that maximum height back to the same level during descent.
  • $\text{Time of Ascent }(t_a)=\text{Time of Descent }(t_d)$
• Total Time of Flight ($T$): The total time the object stays in the air.
  • $\boxed{T=t_a+t_d=2t_a=2t_d=\frac{2u}{g}}$
• Maximum Height ($H$): At the highest point, final velocity $v=0$.
  • $\boxed{H=\frac{u^2}{2g}}$

Practice Numericals

Q1. A body is thrown vertically upwards with an initial velocity of $40\text{ m/s}$. ($g=10{\text{ m/s}}^2$)
(i) How long will it take to reach the highest point?

• $v=u+at⟹0=40-10t⟹t=4\text{ s}$

(ii) What is the maximum height reached?

• $v^2=u^2+2as⟹0=(40{)}^2-2(10)H⟹20H=1600⟹H=80\text{ m}$

(iii) What is the total time to return to the ground?

• $T=2t_a=2(4)=8\text{ s}$

(iv) At what times will the body be at a height of $25\text{ m}$?

• $s=ut+\frac{1}{2}a{t}^2⟹25=40t-5t^2⟹t^2-8t+5=0$
• $t=\frac{8\pm \sqrt{64-20}}{2}=\frac{8\pm \sqrt{44}}{2}⟹t\approx 0.68\text{ s}\text{ and }7.32\text{ s}$

(v) What will be its velocity after $2\text{ s}$?

• $v=u+at⟹v=40-10(2)=20\text{ m/s}$

Q2. A stone is dropped from a height of $19.6\text{ m}$. How much time will it take to cover the last meter?

• Total time $t$ to fall $19.6\text{ m}$: $s=ut+\frac{1}{2}a{t}^2⟹-19.6=0-\frac{1}{2}(9.8)t^2⟹t^2=4⟹t=2\text{ s}$
• Time $t^{\prime }$ to fall the first $18.6\text{ m}$: $-18.6=0-\frac{1}{2}(9.8)t^{\prime 2}⟹t^{\prime 2}=\frac{18.6}{4.9}⟹t^{\prime }\approx 1.949\text{ s}$
• Time for the last meter $=t-t^{\prime }=2-1.949=0.051\text{ s}$

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