Chapter 4: Motion in Two and Three Dimensions

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Motion in two and three dimensions builds on the concepts of one-dimensional motion (Chapter 2) and vectors (Chapter 3), extending kinematics to describe real-world scenarios like the trajectory of a projectile, the circular motion of a car around a curve, or the relative velocity of a boat crossing a river. This chapter is crucial for JEE Main, JEE Advanced, and NEET, as multi-dimensional motion problems frequently appear in questions on projectile motion (e.g., range, maximum height), circular motion (e.g., centripetal acceleration), and relative velocity (e.g., airplane navigation). A deep understanding of these topics is essential for mastering later chapters like force and motion, as well as applications in electromagnetism and orbital mechanics. This chapter covers position, displacement, velocity, and acceleration in 2D/3D, projectile motion, uniform circular motion, relative motion, and applications in physics, providing detailed explanations, derivations, numerous solved examples, and exam-focused strategies to ensure conceptual clarity and problem-solving proficiency.

4.1 Position, Displacement, Velocity, and Acceleration in 2D/3D

In one-dimensional motion, quantities like position, displacement, velocity, and acceleration are scalars along a straight line. In two and three dimensions, these become vector quantities, requiring the use of vector notation and components (Chapter 3). Understanding their vector nature is key to solving JEE/NEET problems involving multi-dimensional motion.

Position and Displacement

The position vector of a particle in 3D space relative to an origin is $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$, where $x$, $y$, $z$ are coordinates. In 2D, the z-component is zero: $\overrightarrow{r}=x\hat{i}+y\hat{j}$. Displacement is the change in position: $\overrightarrow{\mathrm{\Delta }r}={\overrightarrow{r}}_f-{\overrightarrow{r}}_i$. For example, moving from $(1,2,0)$ to $(4,5,0)$ gives $\overrightarrow{\mathrm{\Delta }r}=(4-1)\hat{i}+(5-2)\hat{j}=3\hat{i}+3\hat{j}$.

Velocity

The average velocity is a vector: ${\overrightarrow{v}}_{\text{avg}}=\frac{\overrightarrow{\mathrm{\Delta }r}}{\mathrm{\Delta }t}$. The instantaneous velocity is the time derivative of the position vector:

$$\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}=\frac{dx}{dt}\hat{i}+\frac{dy}{dt}\hat{j}+\frac{dz}{dt}\hat{k}=v_x\hat{i}+v_y\hat{j}+v_z\hat{k}$$

The magnitude of velocity, $|\overrightarrow{v}|=\sqrt{v_x^2+v_y^2+v_z^2}$, is the speed.

Acceleration

The average acceleration is ${\overrightarrow{a}}_{\text{avg}}=\frac{\overrightarrow{\mathrm{\Delta }v}}{\mathrm{\Delta }t}$. The instantaneous acceleration is:

$$\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}=\frac{d{v}_x}{dt}\hat{i}+\frac{d{v}_y}{dt}\hat{j}+\frac{d{v}_z}{dt}\hat{k}=a_x\hat{i}+a_y\hat{j}+a_z\hat{k}$$

Acceleration can change the magnitude or direction of velocity (e.g., in circular motion). Fundamentals of Physics emphasizes the vector nature of these quantities in multi-dimensional motion.

Derivation: Instantaneous Velocity in 2D
The position vector is $\overrightarrow{r}(t)=x(t)\hat{i}+y(t)\hat{j}$. The velocity is the derivative:

$$\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}=\frac{d}{dt}(x\hat{i}+y\hat{j})=\frac{dx}{dt}\hat{i}+\frac{dy}{dt}\hat{j}=v_x\hat{i}+v_y\hat{j}$$

For example, if $\overrightarrow{r}(t)=(2t^2)\hat{i}+(3t)\hat{j}$, then $\overrightarrow{v}=4t\hat{i}+3\hat{j}$.

Derivation: Instantaneous Acceleration in 2D
The velocity vector is $\overrightarrow{v}(t)=v_x\hat{i}+v_y\hat{j}$. The acceleration is:

$$\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}=\frac{d}{dt}(v_x\hat{i}+v_y\hat{j})=\frac{d{v}_x}{dt}\hat{i}+\frac{d{v}_y}{dt}\hat{j}=a_x\hat{i}+a_y\hat{j}$$

If $\overrightarrow{v}(t)=(4t)\hat{i}+(3)\hat{j}$, then $\overrightarrow{a}=4\hat{i}+0\hat{j}=4\hat{i}$.

Solved Example: A JEE Main problem involves a particle moving from position ${\overrightarrow{r}}_1=2\hat{i}+3\hat{j}m$ to ${\overrightarrow{r}}_2=5\hat{i}+7\hat{j}m$ in $2s$. Find the average velocity.

• Solution:
  Displacement: $\overrightarrow{\mathrm{\Delta }r}={\overrightarrow{r}}_2-{\overrightarrow{r}}_1=(5\hat{i}+7\hat{j})-(2\hat{i}+3\hat{j})=3\hat{i}+4\hat{j}m$.
  Average velocity:$${\overrightarrow{v}}_{\text{avg}}=\frac{\overrightarrow{\mathrm{\Delta }r}}{\mathrm{\Delta }t}=\frac{3\hat{i}+4\hat{j}}{2}=1.5\hat{i}+2\hat{j}m/s$$Magnitude: $|{\overrightarrow{v}}_{\text{avg}}|=\sqrt{{1.5}^2+2^2}\approx 2.5m/s$ (2 significant figures, April 14, 2025).
  • JEE Tip: Use vector subtraction for displacement; divide by time for average velocity. Common error: Using distance instead of displacement.

Solved Example: A NEET problem gives a particle’s position as $\overrightarrow{r}(t)=(3t^2)\hat{i}+(4t)\hat{j}m$. Find the velocity at $t=2s$.

• Solution:
  Velocity: $\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}$. Given $\overrightarrow{r}(t)=(3t^2)\hat{i}+(4t)\hat{j}$:$$\overrightarrow{v}=\frac{d}{dt}(3t^2)\hat{i}+\frac{d}{dt}(4t)\hat{j}=(6t)\hat{i}+4\hat{j}$$At $t=2$: $\overrightarrow{v}=6(2)\hat{i}+4\hat{j}=12\hat{i}+4\hat{j}m/s$.
  • NEET Tip: Differentiate each component of the position vector. Common error: Forgetting to differentiate.

Solved Example: A JEE Advanced problem involves a particle with velocity $\overrightarrow{v}(t)=(2t)\hat{i}+(3t^2)\hat{j}m/s$. Find the acceleration at $t=1s$.

• Solution:
  Acceleration: $\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}$. Given $\overrightarrow{v}(t)=(2t)\hat{i}+(3t^2)\hat{j}$:$$\overrightarrow{a}=\frac{d}{dt}(2t)\hat{i}+\frac{d}{dt}(3t^2)\hat{j}=2\hat{i}+(6t)\hat{j}$$At $t=1$: $\overrightarrow{a}=2\hat{i}+6(1)\hat{j}=2\hat{i}+6\hat{j}m/s^2$.
  • JEE Tip: Acceleration is the derivative of velocity; compute each component separately. Common error: Incorrect differentiation.

Solved Example: A JEE Main problem involves a particle’s position in 3D as $\overrightarrow{r}(t)=(t)\hat{i}+(t^2)\hat{j}+(2t)\hat{k}m$. Find the speed at $t=3s$.

• Solution:
  Velocity: $\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}=1\hat{i}+(2t)\hat{j}+2\hat{k}$. At $t=3$: $\overrightarrow{v}=1\hat{i}+6\hat{j}+2\hat{k}m/s$. Speed:$$|\overrightarrow{v}|=\sqrt{1^2+6^2+2^2}=\sqrt{1+36+4}=\sqrt{41}\approx 6.40m/s$$
  • JEE Tip: Speed is the magnitude of velocity; include all components in 3D. Common error: Omitting the z-component.

Application: Position and velocity vectors are used in GPS navigation (e.g., tracking an airplane’s 3D path), while acceleration vectors describe satellite motion in orbits.

4.2 Projectile Motion

Projectile motion is the motion of an object launched into the air, subject only to gravity (ignoring air resistance). Examples include a thrown ball, a bullet fired at an angle, or a javelin in sports. The motion is two-dimensional: horizontal (x-axis, constant velocity) and vertical (y-axis, constant acceleration due to gravity, $g\approx 9.8m/s^2$ downward). JEE/NEET problems often involve calculating the range, maximum height, time of flight, or velocity at a specific point.

Key Equations

• Initial Velocity: ${\overrightarrow{v}}_0=v_0\cos \theta \hat{i}+v_0\sin \theta \hat{j}$, where $v_0$ is the launch speed, $\theta$ is the angle above the horizontal.
• Horizontal Motion: $a_x=0$, so $v_x=v_0\cos \theta$, $x=(v_0\cos \theta )t$.
• Vertical Motion: $a_y=-g$, so $v_y=v_0\sin \theta -gt$, $y=(v_0\sin \theta )t-\frac{1}{2}g{t}^2$.
• Time of Flight: Time to return to the ground ($y=0$): $T=\frac{2v_0\sin \theta }{g}$.
• Maximum Height: At the peak, $v_y=0$: $H=\frac{(v_0\sin \theta {)}^2}{2g}$.
• Range: Horizontal distance at $y=0$: $R=\frac{v_0^2\sin 2\theta }{g}$.

Derivation: Time of Flight for Projectile Motion
Set $y=0$ at landing: $y=(v_0\sin \theta )t-\frac{1}{2}g{t}^2$.

$$0=(v_0\sin \theta )t-\frac{1}{2}g{t}^2$$

Factor out $t$: $t(v_0\sin \theta -\frac{1}{2}gt)=0$. Solutions: $t=0$ (launch) or $v_0\sin \theta -\frac{1}{2}gt=0$, so $t=\frac{2v_0\sin \theta }{g}$.

Derivation: Maximum Height of a Projectile
At maximum height, $v_y=0$. Use $v_y=v_0\sin \theta -gt$. Set $v_y=0$:

$$0=v_0\sin \theta -gt\Rightarrow t=\frac{v_0\sin \theta }{g}$$

Substitute $t$ into $y=(v_0\sin \theta )t-\frac{1}{2}g{t}^2$:

$$H=(v_0\sin \theta )\left(\frac{v_0\sin \theta }{g}\right)-\frac{1}{2}g{\left(\frac{v_0\sin \theta }{g}\right)}^2=\frac{(v_0\sin \theta {)}^2}{g}-\frac{(v_0\sin \theta {)}^2}{2g}=\frac{(v_0\sin \theta {)}^2}{2g}$$

Solved Example: A JEE Main problem involves a projectile launched at $v_0=20m/s$ at $\theta ={30}^{\circ }$ ($g=9.8m/s^2$). Find the time of flight.

• Solution:
  Time of flight: $T=\frac{2v_0\sin \theta }{g}$. Given $v_0=20m/s$, $\theta ={30}^{\circ }$, $\sin {30}^{\circ }=0.5$, $g=9.8m/s^2$:$$T=\frac{2\times 20\times 0.5}{9.8}=\frac{20}{9.8}\approx 2.04s$$Round to 2 significant figures: $2.0s$.
  • JEE Tip: Use the time of flight formula directly; ensure consistent units. Common error: Using $\cos \theta$ instead of $\sin \theta$.

Solved Example: A NEET problem involves a ball thrown at $v_0=30m/s$ at ${45}^{\circ }$ ($g=9.8m/s^2$). Find the maximum height.

• Solution:
  Maximum height: $H=\frac{(v_0\sin \theta {)}^2}{2g}$. Given $v_0=30m/s$, $\theta ={45}^{\circ }$, $\sin {45}^{\circ }=\frac{\sqrt{2}}{2}\approx 0.707$:$$H=\frac{(30\times 0.707{)}^2}{2\times 9.8}=\frac{(21.21{)}^2}{19.6}=\frac{450}{19.6}\approx 22.96m$$Round to 2 significant figures: $23m$.
  • NEET Tip: Maximum height depends on the vertical component; square the component first. Common error: Using total velocity.

Solved Example: A JEE Advanced problem involves a projectile launched at $v_0=40m/s$ at $\theta ={60}^{\circ }$ ($g=9.8m/s^2$). Find the range.

• Solution:
  Range: $R=\frac{v_0^2\sin 2\theta }{g}$. Given $v_0=40m/s$, $\theta ={60}^{\circ }$, $2\theta ={120}^{\circ }$, $\sin {120}^{\circ }=\sin {60}^{\circ }=\frac{\sqrt{3}}{2}\approx 0.866$:$$R=\frac{(40{)}^2\times 0.866}{9.8}=\frac{1600\times 0.866}{9.8}\approx 141.63m$$Round to 2 significant figures: $140m$.
  • JEE Tip: Use $\sin 2\theta$ for range; maximum range occurs at $\theta ={45}^{\circ }$. Common error: Using $\sin \theta$.

Solved Example: A JEE Main problem involves a projectile launched at $v_0=25m/s$ at $\theta ={30}^{\circ }$ ($g=9.8m/s^2$). Find the velocity at $t=1s$.

• Solution:
  Horizontal: $v_x=v_0\cos \theta =25\cos {30}^{\circ }=25\cdot \frac{\sqrt{3}}{2}\approx 21.65m/s$. Vertical: $v_y=v_0\sin \theta -gt$, $\sin {30}^{\circ }=0.5$, so $v_y=25\cdot 0.5-9.8\times 1=12.5-9.8=2.7m/s$. Velocity: $\overrightarrow{v}=21.65\hat{i}+2.7\hat{j}m/s$.
  • JEE Tip: Horizontal velocity is constant; vertical velocity changes due to gravity. Common error: Forgetting gravity’s effect on $v_y$.

Application: Projectile motion applies to sports (e.g., basketball shots), military ballistics (e.g., artillery range), and space science (e.g., satellite launches).

4.3 Uniform Circular Motion

Uniform circular motion occurs when an object moves in a circle at constant speed, such as a car on a circular track or a satellite in orbit. Although the speed is constant, the velocity changes direction, resulting in a centripetal acceleration toward the center of the circle, given by $a_c=\frac{v^2}{r}$, where $v$ is the speed and $r$ is the radius. The velocity vector is tangent to the circle, and the acceleration vector points radially inward. JEE/NEET problems often involve calculating centripetal acceleration, period, or relating to centripetal force (Chapter 5).

Key Equations

• Centripetal Acceleration: $a_c=\frac{v^2}{r}$, directed toward the center.
• Angular Velocity: $\omega =\frac{v}{r}$, in radians per second.
• Period: $T=\frac{2\pi r}{v}=\frac{2\pi }{\omega }$.
• Velocity Vector: For position $\overrightarrow{r}=r\cos \theta \hat{i}+r\sin \theta \hat{j}$, $\overrightarrow{v}=-v\sin \theta \hat{i}+v\cos \theta \hat{j}$, where $v=\omega r$, $\theta =\omega t$.

Derivation: Centripetal Acceleration
Consider an object in uniform circular motion with radius $r$ and speed $v$. The velocity vector changes direction but not magnitude. Over a small time $\mathrm{\Delta }t$, the velocity changes by $\mathrm{\Delta }\overrightarrow{v}$, with magnitude approximately $v\mathrm{\Delta }\theta$ (where $\mathrm{\Delta }\theta$ is the angle subtended). The acceleration $a=\frac{\mathrm{\Delta }v}{\mathrm{\Delta }t}$, and for small $\mathrm{\Delta }\theta$, $\mathrm{\Delta }v\approx v\mathrm{\Delta }\theta$. Since $\mathrm{\Delta }\theta =\frac{v\mathrm{\Delta }t}{r}$ (arc length $v\mathrm{\Delta }t=r\mathrm{\Delta }\theta$):

$$a=\frac{\mathrm{\Delta }v}{\mathrm{\Delta }t}=\frac{v\mathrm{\Delta }\theta }{\mathrm{\Delta }t}=\frac{v\left(\frac{v\mathrm{\Delta }t}{r}\right)}{\mathrm{\Delta }t}=\frac{v^2}{r}$$

This acceleration points toward the center, hence called centripetal acceleration.

Derivation: Period of Circular Motion
The circumference of the circle is $2\pi r$. The object travels this distance in one period $T$ at speed $v$:

$$v=\frac{\text{distance}}{\text{time}}=\frac{2\pi r}{T}\Rightarrow T=\frac{2\pi r}{v}$$

Alternatively, since $v=\omega r$, $T=\frac{2\pi }{\omega }$.

Solved Example: A JEE Main problem involves a car moving in a circle of radius $50m$ at a speed of $10m/s$. Find the centripetal acceleration.

• Solution:
  Centripetal acceleration: $a_c=\frac{v^2}{r}$. Given $v=10m/s$, $r=50m$:$$a_c=\frac{(10{)}^2}{50}=\frac{100}{50}=2m/s^2$$The acceleration is $2m/s^2$ toward the center.
  • JEE Tip: Centripetal acceleration always points inward; ensure units are consistent. Common error: Using diameter instead of radius.

Solved Example: A NEET problem involves a satellite in circular orbit with radius $r=7\times {10}^{6}m$ and period $T=6000s$. Find the speed.

• Solution:
  Speed: $v=\frac{2\pi r}{T}$. Given $r=7\times {10}^{6}m$, $T=6000s$:$$v=\frac{2\times 3.14\times 7\times {10}^6}{6000}=\frac{43.96\times {10}^6}{6000}\approx 7327m/s$$Round to 2 significant figures: $7300m/s$.
  • NEET Tip: Use the period formula to find speed; round appropriately (April 14, 2025). Common error: Forgetting $2\pi$.

Solved Example: A JEE Advanced problem involves a particle in circular motion with $r=2m$ and $\omega =5rad/s$. Find the magnitude of the centripetal acceleration.

• Solution:
  Speed: $v=\omega r=5\times 2=10m/s$. Centripetal acceleration:$$a_c=\frac{v^2}{r}=\frac{(10{)}^2}{2}=\frac{100}{2}=50m/s^2$$Alternatively, $a_c={\omega }^{2}r=(5{)}^2\times 2=25\times 2=50m/s^2$.
  • JEE Tip: Use either $v^2/r$ or ${\omega }^{2}r$ for centripetal acceleration; both are equivalent. Common error: Using linear acceleration.

Solved Example: A JEE Main problem involves a stone tied to a string of length $1m$, rotating at $v=4m/s$. Find the angular velocity.

• Solution:
  Angular velocity: $\omega =\frac{v}{r}$. Given $v=4m/s$, $r=1m$:$$\omega =\frac{4}{1}=4rad/s$$
  • JEE Tip: Angular velocity relates linear speed to radius; units are rad/s. Common error: Using degrees instead of radians.

Application: Uniform circular motion applies to planetary orbits (e.g., Earth around the Sun), amusement park rides (e.g., Ferris wheels), and vehicle dynamics (e.g., cars on curved roads).

4.4 Relative Motion in 2D and 3D

Relative motion describes the motion of one object as observed from another moving object. In 2D/3D, this involves vector subtraction of velocities. For objects A and B, the velocity of A relative to B is ${\overrightarrow{v}}_{A/B}={\overrightarrow{v}}_A-{\overrightarrow{v}}_B$. JEE/NEET problems often involve relative velocity in contexts like boats crossing rivers, airplanes in wind, or objects in moving frames (e.g., rain seen from a moving car).

Key Concept

• Relative Velocity: ${\overrightarrow{v}}_{A/B}={\overrightarrow{v}}_A-{\overrightarrow{v}}_B$, where ${\overrightarrow{v}}_A$ and ${\overrightarrow{v}}_B$ are the velocities of A and B relative to a common reference frame (e.g., the ground).
• Relative Position: Position of A relative to B: ${\overrightarrow{r}}_{A/B}={\overrightarrow{r}}_A-{\overrightarrow{r}}_B$.
• Applications: Common in boat-river problems, airplane navigation, and collision scenarios.

Derivation: Relative Velocity in 2D
The velocity of A relative to the ground is ${\overrightarrow{v}}_A$, and B relative to the ground is ${\overrightarrow{v}}_B$. The velocity of A as observed by B is the difference:

$${\overrightarrow{v}}_{A/B}={\overrightarrow{v}}_A-{\overrightarrow{v}}_B$$

Components: $(v_{A/B}{)}_x=(v_A{)}_x-(v_B{)}_x$, $(v_{A/B}{)}_y=(v_A{)}_y-(v_B{)}_y$. Magnitude: $|{\overrightarrow{v}}_{A/B}|=\sqrt{(v_{A/B}{)}_x^2+(v_{A/B}{)}_y^2}$.

Derivation: Relative Position Over Time
The position of A relative to B changes with time: ${\overrightarrow{r}}_{A/B}(t)={\overrightarrow{r}}_A(t)-{\overrightarrow{r}}_B(t)$. Differentiate to find relative velocity:

$${\overrightarrow{v}}_{A/B}=\frac{d}{dt}({\overrightarrow{r}}_{A/B})=\frac{d{\overrightarrow{r}}_A}{dt}-\frac{d{\overrightarrow{r}}_B}{dt}={\overrightarrow{v}}_A-{\overrightarrow{v}}_B$$

Solved Example: A JEE Main problem involves a boat with velocity ${\overrightarrow{v}}_b=5\hat{i}m/s$ crossing a river with current ${\overrightarrow{v}}_r=3\hat{j}m/s$. Find the boat’s velocity relative to the ground.

• Solution:
  Velocity relative to the ground: ${\overrightarrow{v}}_{b/g}={\overrightarrow{v}}_b+{\overrightarrow{v}}_r=5\hat{i}+3\hat{j}m/s$. Magnitude:$$|{\overrightarrow{v}}_{b/g}|=\sqrt{5^2+3^2}=\sqrt{25+9}=\sqrt{34}\approx 5.83m/s$$Direction: $\theta ={\tan }^{-1}\left(\frac{3}{5}\right)\approx {30.96}^{\circ }$ north of east.
  • JEE Tip: Add velocity vectors component-wise; the current affects the y-direction. Common error: Adding magnitudes directly.

Solved Example: A NEET problem involves a car moving at ${\overrightarrow{v}}_c=20\hat{i}m/s$ and rain falling at ${\overrightarrow{v}}_r=-10\hat{j}m/s$. Find the velocity of rain relative to the car.

• Solution:
  Relative velocity: ${\overrightarrow{v}}_{r/c}={\overrightarrow{v}}_r-{\overrightarrow{v}}_c=-10\hat{j}-20\hat{i}=-20\hat{i}-10\hat{j}m/s$. Magnitude:$$|{\overrightarrow{v}}_{r/c}|=\sqrt{(-20{)}^2+(-10{)}^2}=\sqrt{400+100}=\sqrt{500}\approx 22.36m/s$$Direction: $\theta ={\tan }^{-1}\left(\frac{-10}{-20}\right)={26.57}^{\circ }$ below the negative x-axis.
  • NEET Tip: Subtract the observer’s velocity; negative components indicate direction. Common error: Reversing the order (${\overrightarrow{v}}_c-{\overrightarrow{v}}_r$).

Solved Example: A JEE Advanced problem involves two particles A and B with velocities ${\overrightarrow{v}}_A=10\hat{i}+5\hat{j}m/s$ and ${\overrightarrow{v}}_B=5\hat{i}-5\hat{j}m/s$. Find the velocity of A relative to B.

• Solution:
  ${\overrightarrow{v}}_{A/B}={\overrightarrow{v}}_A-{\overrightarrow{v}}_B=(10\hat{i}+5\hat{j})-(5\hat{i}-5\hat{j})=(10-5)\hat{i}+(5-(-5))\hat{j}=5\hat{i}+10\hat{j}m/s$. Magnitude:$$|{\overrightarrow{v}}_{A/B}|=\sqrt{5^2+{10}^2}=\sqrt{125}\approx 11.18m/s$$
  • JEE Tip: Use vector subtraction for relative velocity; compute components separately. Common error: Forgetting negative signs.

Solved Example: A JEE Main problem involves an airplane with velocity ${\overrightarrow{v}}_a=300\hat{i}m/s$ in a wind ${\overrightarrow{v}}_w=50\hat{j}m/s$. Find the airplane’s velocity relative to the ground.

• Solution:
  ${\overrightarrow{v}}_{a/g}={\overrightarrow{v}}_a+{\overrightarrow{v}}_w=300\hat{i}+50\hat{j}m/s$. Magnitude:$$|{\overrightarrow{v}}_{a/g}|=\sqrt{{300}^2+{50}^2}=\sqrt{90000+2500}=\sqrt{92500}\approx 304.14m/s$$Direction: $\theta ={\tan }^{-1}\left(\frac{50}{300}\right)\approx {9.46}^{\circ }$ north of east.
  • JEE Tip: Wind adds to the airplane’s velocity; calculate the resultant vector. Common error: Ignoring the wind’s effect.

Application: Relative motion applies to navigation (e.g., airplanes adjusting for crosswinds), sports (e.g., a swimmer crossing a river), and physics problems (e.g., relative velocity in collision scenarios).

4.5 Applications of 2D/3D Motion in Physics

Multi-dimensional motion concepts are foundational in physics, connecting to dynamics, rotational motion, and beyond. This section explores applications in various contexts, reinforcing their importance in JEE/NEET exams.

Projectile Motion in Sports

In sports like basketball or javelin throw, projectile motion determines the optimal angle for maximum range ($\theta ={45}^{\circ }$ on level ground) or height (e.g., high jump).

Circular Motion in Mechanics

Centripetal acceleration underlies problems like banking of roads, where the angle ensures vehicles don’t skid: $\tan \theta =\frac{v^2}{rg}$.

Relative Motion in Navigation

Relative velocity is critical in navigation, such as a boat crossing a river, where the boat’s velocity relative to the water and the river’s current combine to determine the path.

3D Motion in Orbital Mechanics

Satellites in 3D orbits involve position, velocity, and acceleration vectors, often simplified to 2D circular motion for basic problems (e.g., geostationary orbits).

Derivation: Banking Angle for a Curved Road
For a car on a banked curve of radius $r$ at speed $v$, the normal force provides centripetal force. No friction: $N\sin \theta =\frac{m{v}^2}{r}$, $N\cos \theta =mg$. Divide equations:

$$\tan \theta =\frac{v^2}{rg}$$

Solved Example: A JEE Main problem involves a projectile launched at $v_0=50m/s$ at $\theta ={45}^{\circ }$ ($g=9.8m/s^2$). Find the velocity at maximum height.

• Solution:
  At maximum height, $v_y=0$. Horizontal: $v_x=v_0\cos \theta =50\cos {45}^{\circ }=50\cdot \frac{\sqrt{2}}{2}\approx 35.36m/s$. Velocity: $\overrightarrow{v}=35.36\hat{i}m/s$, magnitude $35m/s$ (2 significant figures).
  • JEE Tip: Only the horizontal component remains at the peak. Common error: Assuming vertical velocity is non-zero.

Solved Example: A NEET problem involves a car on a banked road of radius $100m$ at $v=20m/s$ ($g=9.8m/s^2$). Find the banking angle.

• Solution:
  Banking angle: $\tan \theta =\frac{v^2}{rg}$. Given $v=20m/s$, $r=100m$, $g=9.8m/s^2$:$$\tan \theta =\frac{(20{)}^2}{100\times 9.8}=\frac{400}{980}\approx 0.408$$$\theta ={\tan }^{-1}(0.408)\approx {22.2}^{\circ }$, round to ${22}^{\circ }$.
  • NEET Tip: Banking angle ensures centripetal force; use $\tan \theta$ formula. Common error: Using degrees for $g$.

Solved Example: A JEE Advanced problem involves two airplanes with velocities ${\overrightarrow{v}}_1=400\hat{i}m/s$ and ${\overrightarrow{v}}_2=300\hat{i}+100\hat{j}m/s$. Find the relative velocity of the second relative to the first.

• Solution:
  ${\overrightarrow{v}}_{2/1}={\overrightarrow{v}}_2-{\overrightarrow{v}}_1=(300\hat{i}+100\hat{j})-(400\hat{i})=-100\hat{i}+100\hat{j}m/s$. Magnitude:$$|{\overrightarrow{v}}_{2/1}|=\sqrt{(-100{)}^2+{100}^2}=\sqrt{20000}\approx 141.42m/s$$
  • JEE Tip: Relative velocity in 2D requires component subtraction. Common error: Forgetting direction.

Solved Example: A JEE Main problem involves a satellite at $r=6.4\times {10}^{6}m$ with speed $v=8000m/s$. Find the centripetal acceleration.

• Solution:
  $a_c=\frac{v^2}{r}$. Given $v=8000m/s$, $r=6.4\times {10}^{6}m$:$$a_c=\frac{(8000{)}^2}{6.4\times {10}^6}=\frac{64\times {10}^6}{6.4\times {10}^6}=10m/s^2$$
  • JEE Tip: Centripetal acceleration matches gravitational acceleration in orbits. Common error: Incorrect powers of 10.

Application: 2D/3D motion applies to projectile motion (e.g., sports), circular motion (e.g., satellites), relative motion (e.g., navigation), and orbital mechanics (e.g., space missions).

Summary and Quick Revision

• Position and Displacement: Position vector: $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$ (3D). Displacement: $\overrightarrow{\mathrm{\Delta }r}={\overrightarrow{r}}_f-{\overrightarrow{r}}_i$. Magnitude: $|\overrightarrow{r}|=\sqrt{x^2+y^2+z^2}$.
• Velocity and Acceleration: Average velocity: ${\overrightarrow{v}}_{\text{avg}}=\frac{\overrightarrow{\mathrm{\Delta }r}}{\mathrm{\Delta }t}$. Instantaneous velocity: $\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}$. Average acceleration: ${\overrightarrow{a}}_{\text{avg}}=\frac{\overrightarrow{\mathrm{\Delta }v}}{\mathrm{\Delta }t}$. Instantaneous acceleration: $\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}$. Speed: $|\overrightarrow{v}|=\sqrt{v_x^2+v_y^2+v_z^2}$.
• Projectile Motion: Horizontal: $v_x=v_0\cos \theta$, $x=(v_0\cos \theta )t$. Vertical: $v_y=v_0\sin \theta -gt$, $y=(v_0\sin \theta )t-\frac{1}{2}g{t}^2$. Time of flight: $T=\frac{2v_0\sin \theta }{g}$. Maximum height: $H=\frac{(v_0\sin \theta {)}^2}{2g}$. Range: $R=\frac{v_0^2\sin 2\theta }{g}$ (maximum at $\theta ={45}^{\circ }$).
• Uniform Circular Motion: Centripetal acceleration: $a_c=\frac{v^2}{r}={\omega }^{2}r$, directed inward. Angular velocity: $\omega =\frac{v}{r}$. Period: $T=\frac{2\pi r}{v}=\frac{2\pi }{\omega }$. Velocity is tangent to the circle.
• Relative Motion: Relative velocity: ${\overrightarrow{v}}_{A/B}={\overrightarrow{v}}_A-{\overrightarrow{v}}_B$. Relative position: ${\overrightarrow{r}}_{A/B}={\overrightarrow{r}}_A-{\overrightarrow{r}}_B$. Use components for 2D/3D calculations.
• Applications: Projectile motion (e.g., sports: $H$, $R$), circular motion (e.g., orbits: $a_c=\frac{v^2}{r}$), relative motion (e.g., navigation: ${\overrightarrow{v}}_{A/B}$), orbital mechanics (e.g., satellites: $v=\sqrt{\frac{GM}{r}}$ in later chapters).
• SI Units: Displacement ($m$), velocity ($m/s$), acceleration ($m/s^2$), time ($s$), angular velocity ($rad/s$).
• JEE/NEET Tips: Resolve motion into independent x- and y-components, use $g$ downward in projectile motion, ensure centripetal acceleration points inward, subtract velocities for relative motion, verify significant figures (April 14, 2025), check directions in 2D/3D problems.
• Applications: Sports, navigation, orbital mechanics, vehicle dynamics, space science.

Practice Problems

Explore our extensive problem set with 100 problems inspired by JEE Main, JEE Advanced, and NEET patterns to test your understanding of motion in two and three dimensions.

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Note: Content regularly updated to align with current JEE/NEET syllabi.