Motion in a Plane Class 11 Notes | Projectile Motion | IIT JEE | NEET

Motion in a plane extends straight-line kinematics to two dimensions, the kind of motion seen when a cricket ball is hit for a six or a satellite orbits the Earth. These motion in a plane class 11 notes explain each concept with its formulas, so you can revise the chapter quickly before tests. For solved exercises, use the NCERT Solutions for Class 11 Physics Motion in a Plane

Motion in a Plane Class 11 Notes: What You Will Learn

• Scalars and vectors, vector addition and resolution
• Equations of motion in a plane with constant acceleration
• Relative velocity in two dimensions
• Projectile motion and its formulas
• Uniform circular motion and centripetal acceleration

India's Best Exam Preparation for Class 11th - Download Now

Motion in a Plane Class 11 Notes

India's Best Exam Preparation for Class 11th - Download Now

India's Best Exam Preparation for Class 11th - Download Now

India's Best Exam Preparation for Class 11th - Download Now

India's Best Exam Preparation for Class 11th - Download Now

India's Best Exam Preparation for Class 11th - Download Now

 For a concept-first explanation of how scalar and vector quantities become separate x- and y-component equations, read our guide to vector-based two dimensional motion.

India's Best Exam Preparation for Class 11th - Download Now

Scalars and Vectors in Motion in a Plane

A scalar has only magnitude, such as speed, mass or time. A vector has both magnitude and direction, such as displacement, velocity, acceleration and force. For a concept-first explanation of how vectors become separate x and y component equations, read our guide to vector-based two dimensional motion.

Position Vector and Displacement

The position vector of a point is $\vec{r} = x\hat{i} + y\hat{j}$. The displacement between two points is $\Delta\vec{r} = \vec{r}_2 - \vec{r}_1$. Displacement can be zero even if the path is long, as long as the object returns to its starting point.

Vector Addition: Triangle and Parallelogram Law

By the triangle law, if two vectors are drawn head to tail, the resultant is the third side of the triangle. By the parallelogram law, the resultant is the diagonal from the common point. For vectors $\vec{A}$ and $\vec{B}$ with angle $\theta$ between them:

$$R = \sqrt{A^2 + B^2 + 2AB\cos\theta}$$ $$\tan\alpha = \dfrac{B\sin\theta}{A + B\cos\theta}$$

Here $\alpha$ is the angle the resultant makes with $\vec{A}$.

How to Resolve a Vector into Components

For a vector $\vec{A}$ making angle $\theta$ with the x-axis:

$$A_x = A\cos\theta, \qquad A_y = A\sin\theta$$

Resolving vectors turns a two-dimensional problem into two independent one-dimensional problems, one along each axis.

Equations of Motion in a Plane with Constant Acceleration

With constant acceleration $\vec{a}$, the one-dimensional equations extend to vector form:

$$\vec{v} = \vec{v}_0 + \vec{a}t$$ $$\vec{r} = \vec{r}_0 + \vec{v}_0 t + \dfrac{1}{2}\vec{a}t^2$$

Each equation splits into an x equation and a y equation, which is exactly how projectile motion is solved.

Relative Velocity in Two Dimensions

The velocity of object A as seen from object B is found by vector subtraction:

$$\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$$

Resolve both velocities into components before subtracting.

River-Boat Problems

Let the boat speed in still water be $v_b$, the river speed be $v_r$ and the river width be $d$. To cross in the shortest time, head perpendicular to the bank, so $t = \dfrac{d}{v_b}$ and the boat drifts downstream by $v_r t$. To reach the point directly opposite (possible only if $v_b > v_r$), head upstream at angle $\theta$ from the perpendicular, where:

$$\sin\theta = \dfrac{v_r}{v_b}, \qquad t = \dfrac{d}{\sqrt{v_b^2 - v_r^2}}$$

Rain and Man Problems

If rain falls vertically at speed $v_r$ and a man walks at speed $v_m$, he must tilt his umbrella forward at angle $\theta$ from the vertical, where $\tan\theta = \dfrac{v_m}{v_r}$.

Projectile Motion: Formulas and Trajectory

A projectile is launched with speed $u$ at angle $\theta$ to the horizontal. The horizontal velocity $u\cos\theta$ stays constant, and the vertical velocity $u\sin\theta - gt$ changes because of gravity.

Projectile Motion Formulas for Time of Flight, Maximum Height and Range

$$T = \dfrac{2u\sin\theta}{g}$$ $$H = \dfrac{u^2\sin^2\theta}{2g}$$ $$R = \dfrac{u^2\sin 2\theta}{g}$$

The range is maximum at $\theta = 45^\circ$, where $R_{max} = \dfrac{u^2}{g}$. Angles $\theta$ and $90^\circ - \theta$ give the same range. Also, $H = \dfrac{R\tan\theta}{4}$.

Example: for $u = 20\ \text{m/s}$, $\theta = 30^\circ$ and $g = 10\ \text{m/s}^2$, we get $T = 2\ \text{s}$, $H = 5\ \text{m}$ and $R = 20\sqrt{3} \approx 34.6\ \text{m}$.

Why Is the Path of a Projectile a Parabola?

Horizontal motion is uniform, so $x = u\cos\theta\, t$. Vertical motion is uniformly accelerated, so $y = u\sin\theta\, t - \dfrac{1}{2}gt^2$. Eliminating $t$ gives:

$$y = x\tan\theta - \dfrac{gx^2}{2u^2\cos^2\theta}$$

This is the equation of a parabola.

Horizontal Projection from a Height

For a body thrown horizontally with speed $u$ from height $h$:

$$T = \sqrt{\dfrac{2h}{g}}, \qquad \text{Range} = u\sqrt{\dfrac{2h}{g}}$$

Uniform Circular Motion and Centripetal Acceleration

In uniform circular motion, an object moves on a circle of radius $r$ at constant speed. If the period is $T$ and frequency is $f$:

$$\omega = \dfrac{2\pi}{T} = 2\pi f, \qquad v = \omega r$$ $$a_c = \dfrac{v^2}{r} = \omega^2 r, \qquad F_c = \dfrac{mv^2}{r}$$

Example: a car moving at $10\ \text{m/s}$ on a circular track of radius $50\ \text{m}$ has $a_c = \dfrac{10^2}{50} = 2\ \text{m/s}^2$.

Why Is Uniform Circular Motion Accelerated?

The speed is constant, but the direction of velocity keeps changing. A change in a vector means acceleration, and this centripetal acceleration always points towards the centre.

Common Mistakes in Motion in a Plane Class 11 Notes

• Adding magnitudes directly instead of using the triangle or parallelogram law.
• Mixing horizontal and vertical components in projectile motion.
• Treating displacement as distance on curved paths.
• Forgetting that uniform circular motion is accelerated motion.

How to Revise Motion in a Plane Class 11 Notes

• Learn vector addition and resolution first, since every later topic depends on them.
• Split every problem into x and y components before using any formula.
• Practise projectile problems with different angles to see how $T$, $H$ and $R$ change.