SCALARS AND VECTORS
SCALAR QUANTITIES
A scalar quantity is a physical quantity that is completely specified by magnitude only. It does not require any direction for its description. Scalars are expressed by a single numerical value along with the appropriate unit.
Scalars follow the rules of ordinary algebra. They can be added, subtracted, multiplied or divided using normal arithmetic operations.
Examples of Scalar Quantities
- Distance between two points
- Mass of an object
- Temperature of a body
- Time of occurrence of an event
- Speed
- Energy
- Work
- Power
Example
If the temperature of a room is 30°C, only the magnitude is required. There is no direction associated with temperature, therefore it is a scalar quantity.
Illustration of Scalar Quantity
VECTOR QUANTITIES
A vector quantity is a physical quantity that possesses both magnitude and direction and obeys the triangle law of vector addition or equivalently the parallelogram law of vector addition.
Because direction is involved, vectors cannot be added using ordinary algebra. Instead, special rules of vector algebra are used.
Examples of Vector Quantities
- Displacement
- Velocity
- Acceleration
- Force
- Momentum
- Electric field
Representation of Vectors
A vector is usually represented by a directed line segment. The length of the arrow represents the magnitude, while the arrowhead indicates the direction.
In printed text, vectors are written in bold letters such as \( \mathbf{v} \). When writing by hand, a vector is commonly represented with an arrow over the symbol, for example \( \vec{v} \).
The magnitude (absolute value) of a vector is written as \( |\vec{v}| \) or simply \( v \).
Vector Representation Diagram
Difference Between Scalars and Vectors
- Scalar: Has magnitude only.
- Vector: Has both magnitude and direction.
- Scalar Addition: Ordinary algebra.
- Vector Addition: Triangle law / parallelogram law.
- Example Scalar: Mass, Time, Energy.
- Example Vector: Velocity, Force, Acceleration.
Significance for Exams
- This concept forms the foundation of the entire chapter “Motion in a Plane.”
- Important for understanding vector addition, resolution of vectors, and projectile motion.
- Frequently tested in JEE Main, NEET, BITSAT, and board examinations.
- Many numerical problems in mechanics require distinguishing between scalar and vector quantities.
Position and Displacement Vectors
Position Vector
To describe the location of a particle in a plane, we first choose a reference point, usually the origin \(O\) of a coordinate system. The position of the particle is then specified relative to this origin.
- The position vector of a particle is the vector drawn from the origin to the point where the particle is located.
- It completely specifies the location of the particle in the coordinate system.
- Position vectors change if the origin of the coordinate system changes.
Definition
The position vector \( \vec{r} \) is the vector joining the origin to the position of the particle.
Mathematical Form
If a particle is located at point \(P(x,y)\), its position vector is
\[ \boxed{\vec{r}=x\hat{i}+y\hat{j}} \]where,
\(x\) → x-coordinate of the particle\(y\) → y-coordinate of the particle
\(\hat{i}\) → unit vector along x-axis
\(\hat{j}\) → unit vector along y-axis
Magnitude of Position Vector
\[ |\vec{r}|=\sqrt{x^2+y^2} \]This magnitude represents the distance of the particle from the origin.
Displacement Vector
When a particle moves from one point to another, the change in its position is described by the displacement vector.
Definition
The displacement vector is the vector drawn from the initial position of the particle to its final position.
Mathematical Form
If
Initial position vector = \( \vec{r_1} \)Final position vector = \( \vec{r_2} \)
Then the displacement vector is
\[ \boxed{\vec{s}=\vec{r_2}-\vec{r_1}} \]Component Form
If the particle moves from \( (x_1,y_1) \) to \( (x_2,y_2) \), then
\[ \vec{s}=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j} \]Example (Important for JEE/NEET)
A particle moves from point \(A(2,3)\) to point \(B(7,6)\).
Displacement vector: \[ \vec{s}=(7-2)\hat{i}+(6-3)\hat{j} \] \[ \vec{s}=5\hat{i}+3\hat{j} \] Magnitude: \[ |\vec{s}|=\sqrt{5^2+3^2}=\sqrt{34} \]Position Vector vs Displacement Vector
| Position Vector | Displacement Vector |
|---|---|
| Describes the location of a particle | Describes the change in position |
| Drawn from origin to particle | Drawn from initial point to final point |
| Depends on choice of origin | Independent of origin |
| Changes if origin changes | Remains same for the same motion |
| Represents absolute location | Represents relative motion |
Significance for Exams
- Position vectors are essential for describing motion in two-dimensional coordinate systems.
- Displacement vectors form the basis for understanding velocity and acceleration vectors.
- Frequently tested in JEE Main, NEET, BITSAT and board exams.
- Many vector problems in mechanics require converting coordinates into vector form.
- Understanding displacement helps distinguish between distance and displacement, a common conceptual question.
Equality of Vectors
Blue vectors → equal vectors (same magnitude & direction)
Magenta vectors → unequal vectors
Definition
Two vectors are said to be equal if they have the same magnitude and the same direction, even if their initial points are different.
In vector algebra, vectors are considered free vectors, meaning their position in space can change without affecting their identity as long as their magnitude and direction remain unchanged.
Conditions for Equality of Vectors
Two vectors \( \vec{A} \) and \( \vec{B} \) are equal if:
- Their magnitudes are equal \[ |\vec{A}| = |\vec{B}| \]
- They have the same direction (parallel and pointing in the same sense)
- If both conditions are satisfied \[ \vec{A} = \vec{B} \]
Equality Using Components
If vectors are expressed in component form, equality requires that the corresponding components be equal.
Suppose
\[ \vec{A} = A_x\hat{i} + A_y\hat{j} \] \[ \vec{B} = B_x\hat{i} + B_y\hat{j} \]Then vectors are equal if
\[ A_x = B_x \quad \text{and} \quad A_y = B_y \]Example
Consider two vectors:
\[ \vec{A} = 3\hat{i} + 4\hat{j} \] \[ \vec{B} = 3\hat{i} + 4\hat{j} \]Since their components are identical, therefore
\[ \vec{A} = \vec{B} \]Common Student Mistakes
- Assuming vectors are equal simply because their magnitudes are equal.
- Ignoring the direction of the vectors.
- Confusing vectors that are parallel but opposite in direction. Such vectors are not equal.
Significance for Exams
- This concept forms the foundation of vector algebra used throughout mechanics.
- Frequently tested in JEE Main, NEET, Olympiads, and board examinations.
- Essential for solving problems involving vector addition, displacement, velocity, and forces.
- Component equality is widely used in coordinate geometry and projectile motion problems.
MULTIPLICATION OF VECTORS BY REAL NUMBERS
Scalar multiplication changes the magnitude of a vector. Positive scalar → same direction. Negative scalar → opposite direction.
Concept
Multiplying a vector \( \vec{A} \) by a real number \( \lambda \) produces a new vector whose magnitude is multiplied by \( |\lambda| \).
The direction of the new vector depends on the sign of \( \lambda \).
Mathematical Relation
\[ |\lambda \vec{A}| = |\lambda|\,|\vec{A}| \]Case 1: Positive Scalar
If \( \lambda > 0 \), the direction of the vector remains the same as that of \( \vec{A} \).
\[ |\lambda \vec{A}| = \lambda |\vec{A}| \]Example: \(2\vec{A}\) has twice the magnitude of \( \vec{A} \) but the same direction.
Case 2: Negative Scalar
If \( \lambda < 0 \), the direction of the vector reverses.
Example: \(-\vec{A}\) has the same magnitude as \( \vec{A} \) but points in the opposite direction.
Component Form
If a vector is expressed in component form:
\[ \vec{A} = A_x\hat{i} + A_y\hat{j} \]Then multiplying by scalar \( \lambda \) gives:
\[ \lambda \vec{A} = (\lambda A_x)\hat{i} + (\lambda A_y)\hat{j} \]Thus each component of the vector is multiplied by the scalar.
Worked Example
If
\[ \vec{A} = 3\hat{i} + 4\hat{j} \]Then
\[ 2\vec{A} = 6\hat{i} + 8\hat{j} \]Magnitude of \( \vec{A} \):
\[ |\vec{A}|=\sqrt{3^2+4^2}=5 \]Magnitude of \( 2\vec{A} \):
\[ |2\vec{A}|=2|\vec{A}|=10 \]Physical Meaning
In physics, scalar multiplication often represents scaling of physical quantities.
- Doubling a force → \(2\vec{F}\)
- Reversing velocity → \(-\vec{v}\)
- Tripling displacement → \(3\vec{s}\)
Significance for Exams
- Essential for understanding vector algebra.
- Used frequently in problems involving velocity, acceleration, force and displacement.
- Important for solving vector addition and resolution problems.
- Frequently tested in JEE Main, NEET, Olympiads and board exams.
ADDITION AND SUBTRACTION OF VECTORS — GRAPHICAL METHOD
Vector Addition (Graphical Method)
Triangle law (head-to-tail method)
Vector addition means combining two or more vectors to obtain a single equivalent vector called the resultant vector.
The graphical method of vector addition represents vectors as directed line segments and uses geometric constructions to determine the resultant.
Triangle Law of Vector Addition
If two vectors are represented in magnitude and direction by two sides of a triangle taken in order, then the third side taken in the opposite order represents their resultant.
Method
- Draw the first vector \( \vec{A} \).
- From the head of \( \vec{A} \), draw the second vector \( \vec{B} \).
- Join the tail of \( \vec{A} \) to the head of \( \vec{B} \).
- This joining vector gives the resultant vector \( \vec{R} \).
Parallelogram law of vector addition
Parallelogram Law of Vector Addition
If two vectors acting at a point are represented by the two adjacent sides of a parallelogram, then the diagonal of the parallelogram passing through that point represents their resultant.
Method
- Draw vectors \( \vec{A} \) and \( \vec{B} \) from the same origin.
- Complete the parallelogram using these vectors.
- The diagonal from the origin gives the resultant vector.
Vector Subtraction (Graphical Method)
Vector subtraction is defined as the addition of the negative of a vector.
Basic Idea
\[ \vec{A} - \vec{B} = \vec{A} + (-\vec{B}) \]The vector \( -\vec{B} \) has the same magnitude as \( \vec{B} \) but points in the opposite direction.
Graphical Procedure
- Reverse the direction of vector \( \vec{B} \) to obtain \( -\vec{B} \).
- Add \( \vec{A} \) and \( -\vec{B} \) using the triangle law.
- The resultant vector represents \( \vec{A} - \vec{B} \).
Example
If vector \( \vec{A} \) represents displacement of 4 m east and \( \vec{B} \) represents displacement of 3 m north, the resultant displacement can be obtained graphically using the triangle or parallelogram law.
Significance for Exams
- Vector addition is fundamental for solving problems involving displacement, velocity, acceleration and force.
- The triangle and parallelogram laws form the basis for vector resolution and projectile motion.
- Frequently tested in JEE Main, NEET, BITSAT, Olympiads and board exams.
- Understanding graphical addition helps build intuition before using the analytical (component) method.
RESOLUTION OF VECTORS
A vector resolved into horizontal and vertical components
Concept of Resolution
Resolution of a vector is the process of splitting a single vector into two or more component vectors such that their combined effect is exactly the same as the original vector.
These components are usually chosen along convenient directions — most commonly along the x-axis (horizontal) and y-axis (vertical).
Instead of working with one slanted vector, we analyze two perpendicular components. Together they reproduce the original vector exactly.
Why do we resolve vectors?
- Motion in two dimensions can be treated as two independent one-dimensional motions.
- Forces acting at angles can be analyzed using simple algebra.
- Equations of motion can be applied separately along each axis.
Thus resolution simplifies complicated vector problems into smaller and manageable parts.
Resolution of a Vector in a Plane
Consider a vector \( \vec{A} \) making an angle \( \theta \) with the positive x-axis.
- Component along x-axis → horizontal component
- Component along y-axis → vertical component
Mathematical Expression
- Horizontal component \[ A_x = A\cos\theta \]
- Vertical component \[ A_y = A\sin\theta \]
- \( \theta \) = angle with x-axis
Unit Vectors
A unit vector is a vector of magnitude 1 that specifies direction.
- \( \hat{i} \) → direction of x-axis
- \( \hat{j} \) → direction of y-axis
- \( \hat{k} \) → direction of z-axis
These vectors are mutually perpendicular and form the basic directional framework used in vector algebra.
Vector Representation Using Unit Vectors
\[ \vec{A}=A_x\hat{i}+A_y\hat{j}+A_z\hat{k} \]Each coefficient represents how much of the vector acts along that direction.
Resolution in Two Dimensions
\[ A_x=A\cos\theta,\qquad A_y=A\sin\theta \] \[ \vec{A}=A\cos\theta\,\hat{i}+A\sin\theta\,\hat{j} \]Magnitude of the vector:
\[ |\vec{A}|=\sqrt{A_x^2+A_y^2} \]Worked Example
A vector of magnitude 10 makes an angle \(30^\circ\) with the x-axis.
\[ A_x = 10\cos30^\circ = 8.66 \] \[ A_y = 10\sin30^\circ = 5 \] So, \[ \vec{A} = 8.66\hat{i}+5\hat{j} \]Resolution in Three Dimensions
\[ \vec{A}=A_x\hat{i}+A_y\hat{j}+A_z\hat{k} \] \[ |\vec{A}|=\sqrt{A_x^2+A_y^2+A_z^2} \]Here each component represents the projection of the vector along the corresponding axis.
Common Student Mistakes
- Using \( \sin\theta \) instead of \( \cos\theta \) for the x-component.
- Forgetting that components must be perpendicular.
- Ignoring negative signs when vectors lie in different quadrants.
Significance for Exams
- Resolution of vectors is fundamental for solving projectile motion problems.
- Used extensively in force analysis and equilibrium.
- Essential for velocity and acceleration components.
- Very frequently tested in JEE, NEET, BITSAT and board exams.
VECTOR ADDITION – ANALYTICAL METHOD
Analytical addition uses vector components
Concept
The analytical method of vector addition uses algebra and vector components instead of graphical construction. Each vector is resolved into its components along coordinate axes, and corresponding components are added.
Vectors in Component Form
Consider two vectors \( \vec{A} \) and \( \vec{B} \) in the xy-plane.
\[ \vec{A}=A_x\hat{i}+A_y\hat{j} \] \[ \vec{B}=B_x\hat{i}+B_y\hat{j} \]Addition of Vectors
Let the resultant vector be \( \vec{R} \).
\[ \vec{R}=\vec{A}+\vec{B} \] Substituting the component form: \[ \vec{R}=(A_x\hat{i}+A_y\hat{j})+(B_x\hat{i}+B_y\hat{j}) \] Rearranging terms, \[ \vec{R}=(A_x+B_x)\hat{i}+(A_y+B_y)\hat{j} \] Since \[ \vec{R}=R_x\hat{i}+R_y\hat{j} \] Therefore, \[ \begin{aligned} R_x &= A_x + B_x \\ R_y &= A_y + B_y \end{aligned} \]Magnitude of Resultant
\[ R=\sqrt{R_x^2+R_y^2} \]Direction of Resultant
If \( \theta \) is the angle made by the resultant with the x-axis, \[ \tan\theta=\frac{R_y}{R_x} \]Extension to Three Dimensions
\[ \vec{A}=A_x\hat{i}+A_y\hat{j}+A_z\hat{k} \] \[ \vec{B}=B_x\hat{i}+B_y\hat{j}+B_z\hat{k} \] Adding the vectors, \[ \vec{R}=(A_x+B_x)\hat{i}+(A_y+B_y)\hat{j}+(A_z+B_z)\hat{k} \] Thus, \[ \begin{aligned} R_x &= A_x + B_x \\ R_y &= A_y + B_y \\ R_z &= A_z + B_z \end{aligned} \] Magnitude in three dimensions: \[ |\vec{R}|=\sqrt{R_x^2+R_y^2+R_z^2} \]Worked Example
If
\[ \vec{A}=3\hat{i}+4\hat{j},\quad \vec{B}=5\hat{i}+2\hat{j} \] Then \[ R_x=3+5=8 \] \[ R_y=4+2=6 \] Therefore, \[ \vec{R}=8\hat{i}+6\hat{j} \] Magnitude: \[ R=\sqrt{8^2+6^2}=10 \]Significance for Exams
- This method is more accurate than graphical addition.
- Used extensively in problems involving forces, velocities, and accelerations.
- Essential for solving projectile motion and equilibrium problems.
- Very frequently used in JEE, NEET, BITSAT and board exam numericals.
Resultant of Two Vectors \(\vec{A}\) and \(\vec{B}\) with Angle \(\theta\) Between Them
Resultant of vectors A and B separated by angle θ
Concept
Let two vectors \( \vec{A} \) and \( \vec{B} \) act at a point and make an angle \( \theta \) between them.
Using the parallelogram law of vector addition, the diagonal of the parallelogram represents the resultant vector.
\[ \vec{R} = \vec{A} + \vec{B} \]Magnitude of the Resultant
From the geometry of the parallelogram and applying the Pythagoras theorem,
\[ R^2 = (A + B\cos\theta)^2 + (B\sin\theta)^2 \] Expanding, \[ R^2 = A^2 + B^2\cos^2\theta + 2AB\cos\theta + B^2\sin^2\theta \] Using the identity \[ \sin^2\theta + \cos^2\theta = 1 \] we obtain \[ R^2 = A^2 + B^2 + 2AB\cos\theta \] Therefore, \[ R = \sqrt{A^2 + B^2 + 2AB\cos\theta} \]Direction of the Resultant
Let \( \alpha \) be the angle between the resultant \( \vec{R} \) and vector \( \vec{A} \). Then \[ \tan\alpha = \frac{B\sin\theta}{A + B\cos\theta} \]Relation Using Sine Rule
From triangle geometry, \[ \frac{R}{\sin\theta} = \frac{A}{\sin\beta} = \frac{B}{\sin\alpha} \] where \( \alpha \) is angle between \( \vec{R} \) and \( \vec{A} \) \( \beta \) is angle between \( \vec{R} \) and \( \vec{B} \)Important Special Cases
- Vectors in same direction (\(\theta=0^\circ\)) \[ R = A + B \]
- Vectors in opposite direction (\(\theta=180^\circ\)) \[ R = |A - B| \]
- Vectors perpendicular (\(\theta=90^\circ\)) \[ R = \sqrt{A^2 + B^2} \] (Pythagoras theorem)
Worked Example
Two vectors of magnitudes 6 and 8 make an angle of \(60^\circ\).
\[ R=\sqrt{6^2+8^2+2(6)(8)\cos60^\circ} \] \[ R=\sqrt{36+64+48} \] \[ R=\sqrt{148}\approx12.17 \]Significance for Exams
- One of the most frequently used formulas in mechanics.
- Used in problems involving forces, velocities and displacements.
- Appears regularly in JEE Main, NEET, BITSAT and board exams.
- Important for solving equilibrium and projectile motion problems.
MOTION IN A PLANE
Motion of a particle in a plane and displacement vector
Position Vector
The position vector of a particle in an \(x\)-\(y\) plane is written as
\[ \vec{r}=x\hat{i}+y\hat{j} \]where \(x\) and \(y\) are the coordinates of the particle along the x-axis and y-axis respectively.
Displacement
If the particle moves from point \(P\) to point \(P′\), the displacement vector is
\[ \Delta\vec{r}=\vec{r'}-\vec{r} \] Writing in component form: \[ \Delta\vec{r}=(x'-x)\hat{i}+(y'-y)\hat{j} \] or \[ \Delta\vec{r}=\Delta x\hat{i}+\Delta y\hat{j} \]Average Velocity
The average velocity is defined as the ratio of displacement to the time interval.
\[ \vec{v}_{avg}=\frac{\Delta\vec{r}}{\Delta t} \] \[ \vec{v}_{avg}= \frac{\Delta x}{\Delta t}\hat{i}+ \frac{\Delta y}{\Delta t}\hat{j} \] Hence \[ \vec{v}=v_x\hat{i}+v_y\hat{j} \] where \[ v_x=\frac{\Delta x}{\Delta t},\quad v_y=\frac{\Delta y}{\Delta t} \]Instantaneous Velocity
The instantaneous velocity is obtained by taking the limit as the time interval approaches zero.
\[ \vec{v}=\lim_{\Delta t\to0}\frac{\Delta\vec{r}}{\Delta t} \] \[ \boxed{\vec{v}=\frac{d\vec{r}}{dt}} \] Component form: \[ \vec{v}=\frac{dx}{dt}\hat{i}+\frac{dy}{dt}\hat{j} \]The magnitude of velocity is
\[ v=\sqrt{v_x^2+v_y^2} \] Direction: \[ \tan\theta=\frac{v_y}{v_x} \]Acceleration
Average acceleration is defined as the rate of change of velocity.
\[ \vec{a}_{avg}=\frac{\Delta\vec{v}}{\Delta t} \] Instantaneous acceleration: \[ \vec{a}=\lim_{\Delta t\to0}\frac{\Delta\vec{v}}{\Delta t} \] \[ \boxed{\vec{a}=\frac{d\vec{v}}{dt}} \] Component form: \[ \vec{a}=\frac{dv_x}{dt}\hat{i}+\frac{dv_y}{dt}\hat{j} \] or \[ \vec{a}=a_x\hat{i}+a_y\hat{j} \]Common Student Mistakes
- Confusing displacement with distance.
- Assuming velocity direction is always same as motion path.
- Ignoring component form in 2-D motion problems.
Significance for Exams
- Forms the basis for projectile motion.
- Used in analysing relative motion and circular motion.
- Important for solving problems involving velocity and acceleration components.
- Very frequently tested in JEE Main, NEET, BITSAT and board exams.
MOTION IN A PLANE
Motion of a particle in a plane and displacement vector
Position Vector
The position vector of a particle in an \(x\)-\(y\) plane is written as
\[ \vec{r}=x\hat{i}+y\hat{j} \]where \(x\) and \(y\) are the coordinates of the particle along the x-axis and y-axis respectively.
Displacement
If the particle moves from point \(P\) to point \(P′\), the displacement vector is
\[ \Delta\vec{r}=\vec{r'}-\vec{r} \] Writing in component form: \[ \Delta\vec{r}=(x'-x)\hat{i}+(y'-y)\hat{j} \] or \[ \Delta\vec{r}=\Delta x\hat{i}+\Delta y\hat{j} \]Average Velocity
The average velocity is defined as the ratio of displacement to the time interval.
\[ \vec{v}_{avg}=\frac{\Delta\vec{r}}{\Delta t} \] \[ \vec{v}_{avg}= \frac{\Delta x}{\Delta t}\hat{i}+ \frac{\Delta y}{\Delta t}\hat{j} \] Hence \[ \vec{v}=v_x\hat{i}+v_y\hat{j} \] where \[ v_x=\frac{\Delta x}{\Delta t},\quad v_y=\frac{\Delta y}{\Delta t} \]Instantaneous Velocity
The instantaneous velocity is obtained by taking the limit as the time interval approaches zero.
\[ \vec{v}=\lim_{\Delta t\to0}\frac{\Delta\vec{r}}{\Delta t} \] \[ \boxed{\vec{v}=\frac{d\vec{r}}{dt}} \] Component form: \[ \vec{v}=\frac{dx}{dt}\hat{i}+\frac{dy}{dt}\hat{j} \]The magnitude of velocity is
\[ v=\sqrt{v_x^2+v_y^2} \] Direction: \[ \tan\theta=\frac{v_y}{v_x} \]Acceleration
Average acceleration is defined as the rate of change of velocity.
\[ \vec{a}_{avg}=\frac{\Delta\vec{v}}{\Delta t} \] Instantaneous acceleration: \[ \vec{a}=\lim_{\Delta t\to0}\frac{\Delta\vec{v}}{\Delta t} \] \[ \boxed{\vec{a}=\frac{d\vec{v}}{dt}} \] Component form: \[ \vec{a}=\frac{dv_x}{dt}\hat{i}+\frac{dv_y}{dt}\hat{j} \] or \[ \vec{a}=a_x\hat{i}+a_y\hat{j} \]Common Student Mistakes
- Confusing displacement with distance.
- Assuming velocity direction is always same as motion path.
- Ignoring component form in 2-D motion problems.
Significance for Exams
- Forms the basis for projectile motion.
- Used in analysing relative motion and circular motion.
- Important for solving problems involving velocity and acceleration components.
- Very frequently tested in JEE Main, NEET, BITSAT and board exams.
PROJECTILE MOTION
A projectile is any object that, after being projected with an initial velocity, moves only under the influence of gravity. In ideal projectile motion, air resistance is neglected.
Common examples include a stone thrown into the air, a football kicked at an angle, a bullet fired from a gun, or a ball released from a moving vehicle.
Basic Idea Behind Projectile Motion
Projectile motion can be understood by resolving the motion into two independent components:
- Horizontal motion: Uniform motion with constant velocity
- Vertical motion: Uniformly accelerated motion under gravity
These two motions occur simultaneously but are independent of each other.
Trajectory of a projectile launched at angle θ₀
Acceleration Acting on Projectile
After projection, the only acceleration acting on the projectile is due to gravity.
\[ \vec{a}=-g\hat{j} \] or \[ a_x=0,\quad a_y=-g \]Initial Velocity Components
If the projectile is launched with velocity \(v_0\) making an angle \( \theta_0 \) with the horizontal axis:
\[ v_{0x}=v_0\cos\theta_0 \] \[ v_{0y}=v_0\sin\theta_0 \]Position Equations
Assuming initial position at origin: \[ x_0=0,\quad y_0=0 \] Horizontal motion: \[ x=v_0\cos\theta_0\;t \] Vertical motion: \[ y=v_0\sin\theta_0\;t-\frac{1}{2}gt^2 \]Velocity Components
\[ v_x=v_0\cos\theta_0 \] \[ v_y=v_0\sin\theta_0-gt \]Equation of Trajectory
Eliminating time from the equations: \[ y=x\tan\theta_0-\frac{g x^2}{2v_0^2\cos^2\theta_0} \] This represents a parabolic path.Important Results
Time of Flight
\[ T=\frac{2v_0\sin\theta_0}{g} \]Maximum Height
\[ H=\frac{v_0^2\sin^2\theta_0}{2g} \]Horizontal Range
\[ R=\frac{v_0^2\sin2\theta_0}{g} \] Maximum range occurs when \[ \theta_0=45^\circ \]Worked Example
A projectile is launched with speed \(20\;m/s\) at an angle \(30^\circ\). Find the time of flight.
\[ T=\frac{2(20)\sin30^\circ}{9.8} \] \[ T\approx2.04\;s \]Significance for Exams
- Projectile motion is one of the most important topics in mechanics.
- Used extensively in JEE Main, NEET, Olympiads and board exams.
- Helps understand parabolic motion and motion in two dimensions.
- Forms the basis for many advanced problems in mechanics and engineering.
Equation of Path of a Projectile
Eliminating time between the equations
\[ x=v_0\cos\theta_0\,t \] \[ t=\frac{x}{v_0\cos\theta_0} \]Substituting this value of \(t\) in the equation of \(y\):
\[ y=v_0\sin\theta_0\left(\frac{x}{v_0\cos\theta_0}\right)-\frac{1}{2}g\left(\frac{x}{v_0\cos\theta_0}\right)^2 \] \[ y=x\tan\theta_0-\frac{g x^2}{2v_0^2\cos^2\theta_0} \] \[ \boxed{y=x\tan\theta_0-\frac{g x^2}{2v_0^2\cos^2\theta_0}} \]This equation shows that the path of a projectile is a parabola.
Time of Maximum Height
At the highest point, the vertical velocity becomes zero.
\[ v_y=v_0\sin\theta_0-gt \] Setting \(v_y=0\): \[ t_m=\frac{v_0\sin\theta_0}{g} \]The total time of flight is twice this value:
\[ T_f=2t_m \] \[ \boxed{T_f=\frac{2v_0\sin\theta_0}{g}} \]Maximum Height of Projectile
Substituting \(t=t_m\) into the vertical displacement equation:
\[ y=v_0\sin\theta_0\,t-\frac{1}{2}gt^2 \] \[ h_m=\frac{(v_0\sin\theta_0)^2}{2g} \] \[ \boxed{h_m=\frac{v_0^2\sin^2\theta_0}{2g}} \]Horizontal Range of Projectile
The horizontal range is the distance travelled during the total time of flight.
\[ R=(v_0\cos\theta_0)T_f \] \[ R=v_0\cos\theta_0\left(\frac{2v_0\sin\theta_0}{g}\right) \] \[ R=\frac{v_0^2\sin2\theta_0}{g} \] \[ \boxed{R=\frac{v_0^2\sin2\theta_0}{g}} \]The range is maximum when \( \sin2\theta_0=1 \), i.e., when \( \theta_0=45^\circ \).
\[ \boxed{R_{max}=\frac{v_0^2}{g}} \]UNIFORM CIRCULAR MOTION
When an object moves along a circular path with constant speed, its motion is called uniform circular motion. Although the speed remains constant, the velocity keeps changing because its direction changes continuously.
Velocity in Circular Motion
- Velocity is always directed along the tangent to the circular path.
- The magnitude of velocity remains constant.
- The direction of velocity continuously changes.
This change in direction causes acceleration even when speed is constant.
Centripetal Acceleration
To keep a particle moving in a circular path, an acceleration directed towards the centre of the circle is required. This is called centripetal acceleration.
\[ a_c=\frac{v^2}{r} \]Direction: always toward the centre of the circle.
Centripetal Force
According to Newton’s second law, this acceleration must be produced by a force called centripetal force.
\[ F_c=m\frac{v^2}{r} \]Examples:
- Tension in a rotating string
- Gravitational force in planetary motion
- Friction for a vehicle on a curved road
Angular Variables in Circular Motion
- Angular displacement: \[ \theta \]
- Angular velocity: \[ \omega=\frac{v}{r} \]
- Angular velocity definition: \[ \omega=\frac{\Delta\theta}{\Delta t} \]
- Time period: \[ T=\frac{2\pi r}{v} \]
- Centripetal acceleration in angular form: \[ a_c=\omega^2 r \]
Significance for Exams
- Important for understanding planetary motion.
- Frequently asked in JEE, NEET and board exams.
- Forms the basis for rotational motion concepts.
UNIFORM CIRCULAR MOTION
When an object moves along a circular path with constant speed, its motion is called uniform circular motion. Although the speed remains constant, the velocity keeps changing because its direction changes continuously.
Velocity in Circular Motion
- Velocity is always directed along the tangent to the circular path.
- The magnitude of velocity remains constant.
- The direction of velocity continuously changes.
This change in direction causes acceleration even when speed is constant.
Centripetal Acceleration
To keep a particle moving in a circular path, an acceleration directed towards the centre of the circle is required. This is called centripetal acceleration.
\[ a_c=\frac{v^2}{r} \]Direction: always toward the centre of the circle.
Centripetal Force
According to Newton’s second law, this acceleration must be produced by a force called centripetal force.
\[ F_c=m\frac{v^2}{r} \]Examples:
- Tension in a rotating string
- Gravitational force in planetary motion
- Friction for a vehicle on a curved road
Angular Variables in Circular Motion
- Angular displacement: \[ \theta \]
- Angular velocity: \[ \omega=\frac{v}{r} \]
- Angular velocity definition: \[ \omega=\frac{\Delta\theta}{\Delta t} \]
- Time period: \[ T=\frac{2\pi r}{v} \]
- Centripetal acceleration in angular form: \[ a_c=\omega^2 r \]
Significance for Exams
- Important for understanding planetary motion.
- Frequently asked in JEE, NEET and board exams.
- Forms the basis for rotational motion concepts.
🚀 Interactive Practice Test – Motion in a Plane
Test your understanding of Vectors, Projectile Motion and Circular Motion. Select answers and click Check Score.
⭐ Interactive Vector Diagram Builder (Drag-and-Learn)
Drag the blue vector heads to change vectors. The magenta vector shows the resultant instantly.
Vector Components
🚀 Advanced Vector Builder – JEE Practice Mode
Drag the blue vectors. Try to place them so the resultant equals the target vector.
⭐ Physics Sandbox Mode – Mini Simulator
Explore core concepts from Motion in a Plane. Choose a mode and adjust parameters to see how the motion changes in real time.
⭐ Vector Physics Lab – Graph Mode
Adjust the vector magnitude and angle. Observe the motion and real-time graphs of x(t), y(t), and v(t).
⭐ JEE Challenge Mode – Projectile Motion
Solve randomly generated projectile problems. Enter your answer and click Check Answer. Use \(g = 9.8\,\text{m/s}^2\).
Score: 0
🔥 Daily JEE Challenge – Motion in a Plane
Solve today's challenge problem. The question changes automatically every day.
🏆 Leaderboard
🏆 Weekly Physics Challenge Tournament
Solve 5 timed questions from Motion in a Plane. You have 60 seconds. Earn badges and rank!
⏱ Time: 60
🏅 Your Rank
🎖 Achievement Badges
🏆 Leaderboard
⚡ Motion in a Plane – 60-Second Revision
Revise the entire chapter in one minute. Perfect for JEE / NEET quick revision.
Vector Basics
- Position vector \[ \vec r = x\hat i + y\hat j \]
- Velocity \[ \vec v = \frac{d\vec r}{dt} \]
- Acceleration \[ \vec a = \frac{d\vec v}{dt} \]
Projectile Motion
- Time of flight \[ T=\frac{2v_0\sin\theta}{g} \]
- Maximum height \[ H=\frac{v_0^2\sin^2\theta}{2g} \]
- Range \[ R=\frac{v_0^2\sin2\theta}{g} \]
- Trajectory \[ y=x\tan\theta-\frac{g x^2}{2v_0^2\cos^2\theta} \]
Uniform Circular Motion
- Centripetal acceleration \[ a_c=\frac{v^2}{r} \]
- Centripetal force \[ F_c=\frac{mv^2}{r} \]
- Angular velocity \[ \omega=\frac{v}{r} \]
- Time period \[ T=\frac{2\pi r}{v} \]
JEE / NEET Quick Tips
- Maximum range occurs at 45°
- Complementary angles produce equal ranges
- Velocity at highest point = horizontal component only
- Projectile trajectory is a parabola
Next Topics:
❓ Motion in a Plane – Most Asked Questions (JEE / NEET)
What is motion in a plane?
Motion in a plane is motion in two dimensions where the position of an object is described using two coordinates (x, y).
What is projectile motion?
Projectile motion is the motion of an object projected into the air under the influence of gravity alone.
Why is projectile motion parabolic?
Because horizontal motion is uniform and vertical motion is uniformly accelerated due to gravity.
What is the time of flight formula?
\[ T=\frac{2v_0\sin\theta}{g} \]What is the range formula?
\[ R=\frac{v_0^2\sin2\theta}{g} \]At what angle is range maximum?
Maximum range occurs when the projectile is launched at 45°.
What is centripetal acceleration?
\[ a_c=\frac{v^2}{r} \]What direction does centripetal acceleration act?
It always acts towards the centre of the circular path.
Recent posts
Share this Chapter
Found this helpful? Share this chapter with your friends and classmates.
💡 Exam Tip: Share helpful notes with your study group. Teaching others is one of the fastest ways to reinforce your own understanding.