PhysicsCBSE Class 11 · Chapter 3

Motion in a Plane Class 11: Vectors, Projectiles and Circular Motion

Updated 14 October 2026 · Fact-checked

Motion in a Plane is Chapter 3 of CBSE Class 11 Physics. It covers scalars and vectors, adding, subtracting and resolving vectors, unit vectors, scalar and vector products, motion in a plane with constant acceleration, projectile motion and uniform circular motion with centripetal acceleration.

What Motion in a Plane is about

In one dimension a plus or minus sign was enough to show direction. In two dimensions it is not: a velocity of 10 m s $^{- 1}$ could point anywhere in the plane. So this chapter begins with vectors, quantities with both magnitude and direction, and the rules for handling them: adding by the triangle or parallelogram law, subtracting, multiplying by a number, and splitting a vector into perpendicular components along unit vectors $\hat{i}$ and $\hat{j}$ . The CBSE syllabus also asks for the scalar (dot) and vector (cross) products, which you will use for work, torque and angular momentum.

With vectors in hand, motion in a plane splits into two independent straight-line motions. The equations of the last chapter apply to the x and y components separately. Two important cases follow. In projectile motion the horizontal velocity stays constant while the vertical motion is free fall, giving a parabolic path with time of flight $\frac{2 u s i n θ}{g}$ , maximum height $\frac{u ^{2} s i n ^{2} θ}{2 g}$ and range $\frac{u ^{2} s i n 2 θ}{g}$ . In uniform circular motion the speed is constant but the direction changes, so there is a centripetal acceleration $\frac{v ^{2}}{r}$ towards the centre.

Relative velocity in two dimensions is not in the 2026-27 syllabus. Vectors here are the toolkit for Laws of Motion, Work, Energy and Power, Rotational Motion and the whole of Class 12 electromagnetism.

Weightage: This chapter is part of Unit II, Kinematics. The syllabus groups Unit II with Unit I, Physical World and Measurement and Unit III, Laws of Motion: together they carry 23 of the 70 theory marks in the school annual exam.Source: CBSE Physics (042) syllabus 2026-27, Class XI course structure, Unit II Kinematics

Key concepts

Scalars, vectors, position and displacement vectors

NCERT section 3.2 and 3.3

A scalar has only magnitude (mass, time, speed, work, temperature). Scalars add by ordinary arithmetic. A vector has magnitude and direction and obeys the triangle law of addition (displacement, velocity, acceleration, force, momentum). We write a vector as $A$ and its magnitude as $∣ A ∣$ or A.

The position vector $r$ of a point P is the arrow from the origin O to P.
The displacement vector from P to P′ is $Δ r = r^{'} - r$ . It depends only on the end points, not on the path.
Two vectors are equal if they have the same magnitude and direction, wherever they are drawn (a vector may be shifted parallel to itself).
Multiplying a vector by a positive number λ changes its length by λ and keeps its direction; a negative number also reverses it. $λ A$ can have different units, e.g. velocity × time = displacement.

A quantity is a vector only if it adds by the triangle law. Electric current has a direction but adds like a scalar, so it is a scalar.

Worked example

Classify as scalar or vector: (a) impulse (b) work (c) electric current (d) displacement (e) speed. [1 mark]

Try it first, then show the solutionSolution

Impulse ( $F Δ t$ ) and displacement have direction and add by the triangle law: vectors
Work, current and speed add as numbers: scalars

Answer: Vectors: (a), (d). Scalars: (b), (c), (e).

Adding and subtracting vectors

NCERT section 3.4

Triangle law: place the tail of $B$ at the head of $A$ ; the arrow from the tail of $A$ to the head of $B$ is $R = A + B$ . The parallelogram law gives the same result with both tails together; the diagonal is the resultant.

Vector addition is commutative ( $A + B = B + A$ ) and associative.
$A + (- A) = 0$ , the null (zero) vector.
Subtraction: $A - B = A + (- B)$ .

For two vectors at angle θ:
$R = A^{2} + B^{2} + 2 A B cos θ$ and $tan α = \frac{B s i n θ}{A + B c o s θ}$ ,
where α is the angle R makes with $A$ . R is largest (A + B) at θ = 0 and smallest (|A − B|) at θ = 180°.

Worked example

Two forces of 3 N and 4 N act at a point at 60° to each other. Find the magnitude of the resultant and its angle with the 3 N force. [2 marks]

Try it first, then show the solutionSolution

$R = 9 + 16 + 2 \times 3 \times 4 \times cos 60° = 37 \approx 6.08 N$
$tan α = \frac{4 s i n 60°}{3 + 4 c o s 60°} = \frac{3.464}{5} = 0.693$ , so $α \approx 34.7°$

Answer: About 6.1 N at about 35° to the 3 N force

Unit vectors and resolution into components

NCERT section 3.5 and 3.6

A unit vector has magnitude 1 and only shows a direction: $\hat{A} = \frac{A}{A}$ . Along the x, y and z axes we use $\hat{i}$ , $\hat{j}$ and $\hat{k}$ .

Any vector in the x-y plane can be resolved into rectangular components:
$A = A_{x} \hat{i} + A_{y} \hat{j}$ , with $A_{x} = A cos θ$ and $A_{y} = A sin θ$ ,
where θ is the angle with the x-axis. Going back: $A = A_{x}^{2} + A_{y}^{2}$ and $tan θ = \frac{A _{y}}{A _{x}}$ .

Adding by components (analytical method) is the easiest way: $A + B = (A_{x} + B_{x}) \hat{i} + (A_{y} + B_{y}) \hat{j}$ . In three dimensions, $A = A_{x}^{2} + A_{y}^{2} + A_{z}^{2}$ and the direction cosines are $cos α = \frac{A _{x}}{A}$ , and so on.

Worked example

A force of 10 N acts at 30° above the x-axis. Write it in terms of $\hat{i}$ and $\hat{j}$ . [1 mark]

Try it first, then show the solutionSolution

$F_{x} = 10 cos 30° = 8.66 N$ ; $F_{y} = 10 sin 30° = 5 N$
$F = 8.66 \hat{i} + 5 \hat{j}$ N

Answer: $F = (8.66 \hat{i} + 5 \hat{j})$ N

Scalar (dot) and vector (cross) products

NCERT section Syllabus topic (NCERT uses the products in Chapters 5 and 6)

Scalar product: $A \cdot B = A B cos θ$ , a scalar. It is commutative. In components, $A \cdot B = A_{x} B_{x} + A_{y} B_{y} + A_{z} B_{z}$ , with $\hat{i} \cdot \hat{i} = 1$ and $\hat{i} \cdot \hat{j} = 0$ . It is zero for perpendicular vectors and is used for work: $W = F \cdot d$ . The angle between two vectors: $cos θ = \frac{A \cdot B}{A B}$ .

Vector product: $A \times B$ has magnitude $A B sin θ$ and direction perpendicular to both, by the right-hand rule. It is not commutative: $B \times A = - A \times B$ . It is zero for parallel vectors. $\hat{i} \times \hat{j} = \hat{k}$ , $\hat{j} \times \hat{k} = \hat{i}$ , $\hat{k} \times \hat{i} = \hat{j}$ , and $\hat{i} \times \hat{i} = 0$ . In components:
$A \times B = (A_{y} B_{z} - A_{z} B_{y}) \hat{i} + (A_{z} B_{x} - A_{x} B_{z}) \hat{j} + (A_{x} B_{y} - A_{y} B_{x}) \hat{k}$

Its magnitude is the area of the parallelogram with sides $A$ and $B$ . It is used for torque $τ = r \times F$ and angular momentum.

Worked example

For $A = 2 \hat{i} + 3 \hat{j} - \hat{k}$ and $B = \hat{i} - 2 \hat{j} + 2 \hat{k}$ , find $A \cdot B$ and $A \times B$ . [3 marks]

Try it first, then show the solutionSolution

$A \cdot B = 2 (1) + 3 (- 2) + (- 1) (2) = 2 - 6 - 2 = - 6$
x: $3 (2) - (- 1) (- 2) = 6 - 2 = 4$ ; y: $(- 1) (1) - 2 (2) = - 5$ ; z: $2 (- 2) - 3 (1) = - 7$
$A \times B = 4 \hat{i} - 5 \hat{j} - 7 \hat{k}$

Answer: $A \cdot B = - 6$ ; $A \times B = 4 \hat{i} - 5 \hat{j} - 7 \hat{k}$

Motion in a plane with constant acceleration

NCERT section 3.7 and 3.8

Position, velocity and acceleration become vectors:
$v = \frac{d r}{d t} = v_{x} \hat{i} + v_{y} \hat{j}$ , $a = \frac{d v}{d t}$ .
The velocity at any point is along the tangent to the path.

For constant acceleration the straight-line equations hold in vector form:
$v = v_{0} + a t$ and $r = r_{0} + v_{0} t + \frac{1}{2} a t^{2}$ .

The big idea: the x and y motions are independent. Write
$x = x_{0} + v_{0 x} t + \frac{1}{2} a_{x} t^{2}$ and $y = y_{0} + v_{0 y} t + \frac{1}{2} a_{y} t^{2}$
and solve each as a one-dimensional problem; time t is the only link between them.

Worked example

A particle starts from the origin with velocity $3 \hat{i}$ m s $^{- 1}$ and has a constant acceleration $2 \hat{j}$ m s $^{- 2}$ . Find its position and speed at t = 2 s. [2 marks]

Try it first, then show the solutionSolution

$x = 3 t = 6 m$ ; $y = \frac{1}{2} (2) t^{2} = 4 m$ , so $r = 6 \hat{i} + 4 \hat{j}$ m
$v_{x} = 3$ , $v_{y} = 2 t = 4$ , so speed $= 9 + 16 = 5 m s^{- 1}$

Answer: $r = (6 \hat{i} + 4 \hat{j})$ m; speed 5 m s $^{- 1}$

Projectile motion

NCERT section 3.9

A projectile is an object thrown into the air that then moves under gravity alone (air resistance ignored). Launch it with speed u at angle θ above the horizontal:

Horizontal: $u_{x} = u cos θ$ , $a_{x} = 0$ , so $x = (u cos θ) t$ ; the horizontal velocity never changes.
Vertical: $u_{y} = u sin θ$ , $a_{y} = - g$ , so $y = (u sin θ) t - \frac{1}{2} g t^{2}$ .

Results:

Path: eliminating t, $y = x tan θ - \frac{g x ^{2}}{2 u ^{2} c o s ^{2} θ}$ , a parabola.
Time to the top $= \frac{u s i n θ}{g}$ ; time of flight $T = \frac{2 u s i n θ}{g}$ .
Maximum height $H = \frac{u ^{2} s i n ^{2} θ}{2 g}$ .
Horizontal range $R = \frac{u ^{2} s i n 2 θ}{g}$ , largest at θ = 45°, $R_{ma x} = \frac{u ^{2}}{g}$ .
Angles θ and 90° − θ give the same range (since sin 2θ = sin(180° − 2θ)), but different heights and times.

At the top the velocity is horizontal ( $u cos θ$ ) and the acceleration is still g downwards, so they are at 90°.

Worked example

A ball is projected at 20 m s $^{- 1}$ at 30° above the horizontal. Find the time of flight, maximum height and range. (g = 10 m s $^{- 2}$ ) [3 marks]

Try it first, then show the solutionSolution

$T = \frac{2 \times 20 \times s i n 30°}{10} = 2 s$
$H = \frac{400 \times ( 0.5 ) ^{2}}{20} = 5 m$
$R = \frac{400 \times s i n 60°}{10} = 34.6 m$

Answer: T = 2 s, H = 5 m, R ≈ 34.6 m

Uniform circular motion and centripetal acceleration

NCERT section 3.10

In uniform circular motion an object moves in a circle at constant speed. Its velocity keeps changing direction, so it is accelerating.

Angular speed $ω = \frac{Δ θ}{Δ t}$ (rad s $^{- 1}$ ); $v = ω r$ .
Time period T (one revolution) and frequency $ν = \frac{1}{T}$ : $ω = \frac{2 π}{T} = 2 π ν$ .
Centripetal acceleration $a_{c} = \frac{v ^{2}}{r} = ω^{2} r = 4 π^{2} ν^{2} r$ , always directed towards the centre.

Why towards the centre: in a short time Δt the velocity turns through Δθ. The triangle of velocity vectors is similar to the triangle of position vectors, so $\frac{∣Δ v ∣}{v} = \frac{∣Δ r ∣}{r}$ , giving $a = \frac{v}{r} \cdot \frac{∣Δ r ∣}{Δ t} \to \frac{v ^{2}}{r}$ , and $Δ v$ points to the centre as Δt → 0.

The magnitude of $a_{c}$ is constant but its direction keeps changing, so it is not a constant vector: the equations of motion for constant acceleration do not apply.

Worked example

A stone tied to a 0.5 m string is whirled in a horizontal circle at 2 revolutions per second. Find its angular speed, speed and centripetal acceleration. [2 marks]

Try it first, then show the solutionSolution

$ω = 2 π ν = 4 π \approx 12.6 r a d s^{- 1}$ ; $v = ω r = 6.28 m s^{- 1}$
$a_{c} = ω^{2} r = (4 π)^{2} \times 0.5 \approx 79 m s^{- 2}$ , towards the centre

Answer: 12.6 rad s $^{- 1}$ ; 6.28 m s $^{- 1}$ ; about 79 m s $^{- 2}$

Physics in real life: Motion in a Plane

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Horizontal projection

Suppose a relief helicopter flying horizontally at 50 m s $^{- 1}$ , 125 m above flat ground, releases a food packet towards a village.

How long does the packet take to reach the ground, and how far ahead of the release point (horizontally) does it land? Ignore air resistance; g = 10 m s $^{- 2}$ .

Try it first, then show the solutionSolution

Vertical motion starts from zero: $125 = \frac{1}{2} \times 10 \times t^{2} \Rightarrow t = 5 s$
Horizontal velocity stays 50 m s $^{- 1}$ : distance $= 50 \times 5 = 250 m$

Answer: 5 s; 250 m ahead

Memory hook: Height decides the time; horizontal speed decides how far.

Maximum range at 45°

Imagine you are watering a lawn with a hose whose water leaves the nozzle at 10 m s $^{- 1}$ , and you want to reach the far flower bed.

Suppose the nozzle is near the ground. At what angle should you hold it for the greatest reach, and how far will the water go? (g = 10 m s $^{- 2}$ )

Try it first, then show the solutionSolution

$R = \frac{u ^{2} s i n 2 θ}{g}$ is largest when $sin 2 θ = 1$ , i.e. θ = 45°
$R_{ma x} = \frac{u ^{2}}{g} = \frac{100}{10} = 10 m$

Answer: At 45°; about 10 m

Memory hook: Half of a right angle gives the full range.

Centripetal acceleration

Suppose the drum of a washing machine of radius 0.25 m spins at 20 revolutions per second in the spin cycle.

Find the centripetal acceleration of clothes pressed against the drum, and compare it with g (9.8 m s $^{- 2}$ ).

Try it first, then show the solutionSolution

$ω = 2 π \times 20 = 40 π \approx 125.7 r a d s^{- 1}$
$a_{c} = ω^{2} r = (125.7)^{2} \times 0.25 \approx 3950 m s^{- 2}$
$\frac{3950}{9.8} \approx 400$ , so about 400 times g

Answer: About $3.9 \times 1 0^{3}$ m s $^{- 2}$ , roughly 400 g; the water is flung out through the holes

Memory hook: Acceleration grows with ω squared: double the spin, four times the squeeze.

Topics in this chapter

Motion in a Plane formula sheet

Resultant of two vectors: $R = A^{2} + B^{2} + 2 A B cos θ, tan α = \frac{B s i n θ}{A + B c o s θ}$
Components and magnitude: $A_{x} = A cos θ, A_{y} = A sin θ, A = A_{x}^{2} + A_{y}^{2}$
Scalar product: $A \cdot B = A B cos θ = A_{x} B_{x} + A_{y} B_{y} + A_{z} B_{z}$
Vector product: $∣ A \times B ∣ = A B sin θ, \hat{i} \times \hat{j} = \hat{k}$
Constant acceleration in a plane: $v = v_{0} + a t, r = r_{0} + v_{0} t + \frac{1}{2} a t^{2}$
Projectile: equation of path: $y = x tan θ - \frac{g x ^{2}}{2 u ^{2} c o s ^{2} θ}$
Projectile: time, height, range: $T = \frac{2 u s i n θ}{g}, H = \frac{u ^{2} s i n ^{2} θ}{2 g}, R = \frac{u ^{2} s i n 2 θ}{g}$
Maximum range: $R_{ma x} = \frac{u ^{2}}{g} at θ = 45°$
Circular motion: $v = ω r, ω = \frac{2 π}{T} = 2 π ν, a_{c} = \frac{v ^{2}}{r} = ω^{2} r$

How the school exam asks Motion in a Plane

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Vector basics	Classify scalars and vectors; equal, negative and null vectors; can three vectors of given sizes add to zero?	Use the triangle law; three vectors can add to zero only if each is no larger than the sum of the other two.
Resultant and components	Resultant of two forces at an angle; angle for a given resultant; components of a force; adding vectors in $\hat{i}$ , $\hat{j}$ form.	Law of cosines formula for R, or resolve every vector and add components.
Dot and cross products	Angle between vectors, condition for perpendicular or parallel vectors, area of a parallelogram, cross product in components.	$cos θ = A \cdot B / A B$ ; perpendicular ⇔ dot = 0; parallel ⇔ cross = 0; area = \|cross\|.
Projectile derivations	3-5 marks: show the path is a parabola; derive T, H and R; show range is maximum at 45°.	Split the motion; write x and y equations; eliminate t; set y = 0 for T, $v_{y} = 0$ for H.
Projectile numericals	Oblique launch from ground; horizontal throw from a height; angle for equal H and R; two angles for one range.	Horizontal throw: time from the height alone, $t = 2 h / g$ , then $x = u t$ .
Circular motion	Derive $a_{c} = v^{2} / r$ ; find ω, v, a for a fan, wheel or satellite; change in velocity over part of a circle.	Convert rpm to rad s $^{- 1}$ ( $\times 2 π /60$ ); $a_{c} = ω^{2} r$ ; change in velocity is a vector difference.

Tricks to remember Motion in a Plane

Short memory aids. Say each one aloud once, then use it on a question the same day.

Cos for the side you start from
The component next to the angle uses cos; the one opposite uses sin: $A_{x} = A cos θ$ when θ is measured from x.
Use it when: Resolving vectors.
Cross product cycle i → j → k → i
Going round the cycle gives +: $\hat{i} \times \hat{j} = \hat{k}$ . Going backwards gives −: $\hat{j} \times \hat{i} = - \hat{k}$ .
Use it when: Cross products in components.
THR: 2, 2, 1
T has 2u sin θ over g; H has u² sin² θ over 2g; R has u² sin 2θ over g.
Use it when: Projectile formulae.
45 for most, pairs add to 90
Maximum range at 45°; angles that add up to 90° (30° and 60°) give the same range.
Use it when: Range MCQs.
Circle: speed same, velocity changing
Constant speed, changing direction, so acceleration $v^{2} / r$ always points to the centre.
Use it when: Assertion-reason on circular motion.

Motion in a Plane: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Angle for maximum range · 1 mark · MCQExam level

For a given speed of projection, the range of a projectile on level ground is greatest when the angle of projection is:
(a) 30°
(b) 45°
(c) 60°
(d) 90°

Try it first, then show the solutionSolution

$R \propto sin 2 θ$ , largest when 2θ = 90°.

Answer: (b) 45°

Example 2: Horizontal velocity of a projectile · 1 mark · Assertion-reasonExam level

Assertion (A): The horizontal component of a projectile's velocity stays constant throughout its flight.
Reason (R): Gravity acts vertically, so there is no horizontal acceleration (air resistance neglected).
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

Try it first, then show the solutionSolution

With $a_{x} = 0$ , $v_{x} = u cos θ$ never changes. R is the reason for A.

Answer: (a)

Example 3: Two equal forces · 2 marks · Very short answerHarder exam level

Two forces, each of magnitude F, act at a point. Their resultant also has magnitude F. Find the angle between them. [2 marks]

Try it first, then show the solutionSolution

$F^{2} = F^{2} + F^{2} + 2 F^{2} cos θ \Rightarrow cos θ = - \frac{1}{2}$
$θ = 120°$

Answer: 120°

Example 4: Ball thrown horizontally from a cliff · 3 marks · Short answerHarder exam level

A ball is thrown horizontally at 20 m s $^{- 1}$ from the top of a cliff 45 m high. Find (a) the time to reach the ground, (b) the horizontal distance, (c) the speed and direction of the velocity on hitting the ground. (g = 10 m s $^{- 2}$ ) [3 marks]

Try it first, then show the solutionSolution

(a) $45 = \frac{1}{2} \times 10 \times t^{2} \Rightarrow t = 3 s$
(b) $x = 20 \times 3 = 60 m$
(c) $v_{x} = 20$ , $v_{y} = 10 \times 3 = 30$ m s $^{- 1}$ ; $v = 400 + 900 \approx 36.1 m s^{- 1}$ at $tan^{- 1} (30/20) \approx 56.3°$ below the horizontal

Answer: (a) 3 s (b) 60 m (c) about 36.1 m s $^{- 1}$ at about 56° below the horizontal

Example 5: Angle between two vectors · 3 marks · Short answerHarder exam level

Find the angle between $A = \hat{i} + 2 \hat{j} + 2 \hat{k}$ and $B = 2 \hat{i} + \hat{j} - 2 \hat{k}$ , and the area of the parallelogram they form. [3 marks]

Try it first, then show the solutionSolution

$A \cdot B = 2 + 2 - 4 = 0$ , so the vectors are perpendicular: θ = 90°
$∣ A ∣ = 1 + 4 + 4 = 3$ , $∣ B ∣ = 4 + 1 + 4 = 3$
Area $= ∣ A \times B ∣ = A B sin 90° = 9$ square units

Answer: 90°; area 9 square units

Example 6: Projectile: full derivation and numbers · 5 marks · Long answerHarder exam level

(a) Derive the horizontal range of a projectile launched with speed u at angle θ. (b) Show that angles θ and (90° − θ) give the same range. (c) A ball launched at 20 m s $^{- 1}$ lands 34.6 m away on level ground. Find the two possible angles of projection and the greater of the two maximum heights. (g = 10 m s $^{- 2}$ ) [5 marks]

Try it first, then show the solutionSolution

(a) Time of flight $T = \frac{2 u s i n θ}{g}$ ; $R = (u cos θ) T = \frac{2 u ^{2} s i n θ c o s θ}{g} = \frac{u ^{2} s i n 2 θ}{g}$
(b) $sin 2 (90° - θ) = sin (180° - 2 θ) = sin 2 θ$ , so the ranges are equal
(c) $sin 2 θ = \frac{R g}{u ^{2}} = \frac{34.6 \times 10}{400} \approx 0.866$
$2 θ = 60°$ or 120°, so θ = 30° or 60°
Greater height at 60°: $H = \frac{400 \times ( 0.866 ) ^{2}}{20} = 15 m$

Answer: θ = 30° or 60°; greater height 15 m

Example 7: Height equal to range · 2 marks · Very short answerHOTS / competency

At what angle must a projectile be launched so that its maximum height equals its horizontal range? [2 marks]

Try it first, then show the solutionSolution

$\frac{u ^{2} s i n ^{2} θ}{2 g} = \frac{2 u ^{2} s i n θ c o s θ}{g}$
$sin θ = 4 cos θ \Rightarrow tan θ = 4$ , θ ≈ 76°

Answer: $θ = tan^{- 1} 4 \approx 76°$

Example 8: Car on a curve: change in velocity · 3 marks · Short answerHOTS / competency

A car goes round a circular curve of radius 50 m at a steady 36 km h $^{- 1}$ . (a) Find its centripetal acceleration. (b) Find the magnitude of the change in its velocity as it goes a quarter of the way round. (c) Find the magnitude of its average acceleration over that quarter circle. [3 marks]

Try it first, then show the solutionSolution

(a) v = 10 m s $^{- 1}$ ; $a_{c} = \frac{100}{50} = 2 m s^{- 2}$
(b) The velocity turns through 90° with the same size: $∣Δ v ∣ = 1 0^{2} + 1 0^{2} \approx 14.1 m s^{- 1}$
(c) Time $= \frac{π \times 50/2}{10} \approx 7.85 s$ ; average acceleration $= \frac{14.1}{7.85} \approx 1.8 m s^{- 2}$ (less than the instantaneous 2 m s $^{- 2}$ )

Answer: (a) 2 m s $^{- 2}$ (b) about 14.1 m s $^{- 1}$ (c) about 1.8 m s $^{- 2}$

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 1 + 1 marks.

A football kick

Suppose a footballer kicks a ball from the ground at 20 m s $^{- 1}$ at 45° to the horizontal. Ignore air resistance and take g = 10 m s $^{- 2}$ and $sin 45° = 0.707$ .

(i) · 1 mark
Find the time of flight.
Answer
Answer: About 2.83 s
$\frac{2 \times 20 \times 0.707}{10}$
(ii) · 1 mark
Find the maximum height.
Answer
Answer: 10 m
$\frac{400 \times 0.5}{20}$
(iii) · 1 mark
Find the range.
Answer
Answer: 40 m
$\frac{400 \times s i n 90°}{10}$
(iv) · 1 mark
What is the ball's speed at the highest point?
Answer
Answer: About 14.1 m s $^{- 1}$
Only $u cos 45° = 20 \times 0.707$ remains.

Racing round a track

Suppose a racing car moves round a circular track of radius 100 m at a constant speed of 20 m s $^{- 1}$ .

(i) · 1 mark
Find its centripetal acceleration.
Answer
Answer: 4 m s $^{- 2}$
$\frac{400}{100}$
(ii) · 1 mark
Find its angular speed.
Answer
Answer: 0.2 rad s $^{- 1}$
$ω = v / r$
(iii) · 1 mark
Find the time for one lap.
Answer
Answer: About 31.4 s
$\frac{2 π \times 100}{20}$
(iv) · 1 mark
If the speed is doubled, what is the new centripetal acceleration?
Answer
Answer: 16 m s $^{- 2}$
$a_{c} \propto v^{2}$

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Show that, for a given speed of projection, the horizontal range is maximum when the angle of projection is 45°. [2 marks] (2 marks)

Step	Marks
$R = \frac{u ^{2} s i n 2 θ}{g}$ ; for fixed u, R is greatest when $sin 2 θ$ is greatest, i.e. 1	1
$2 θ = 90° \Rightarrow θ = 45°$ , and $R_{ma x} = \frac{u ^{2}}{g}$	1

Derive an expression for the centripetal acceleration of a body in uniform circular motion. [3 marks] (3 marks)

Step	Marks
Diagram: position vectors $r$ , $r^{'}$ and velocity vectors $v$ , $v^{'}$ at P and P′, with $v ⊥ r$ ; the velocity triangle is similar to the position triangle	1
$\frac{∣Δ v ∣}{v} = \frac{∣Δ r ∣}{r} \Rightarrow a = lim \frac{∣Δ v ∣}{Δ t} = \frac{v}{r} lim \frac{∣Δ r ∣}{Δ t}$	1
$lim \frac{∣Δ r ∣}{Δ t} = v$ , so $a_{c} = \frac{v ^{2}}{r}$ , directed towards the centre	1

A projectile is fired with speed u at angle θ to the horizontal. (a) Show that its path is a parabola. (b) Derive expressions for its time of flight and maximum height. (c) A ball is projected at 20 m s $^{- 1}$ at 30°. Find its range (g = 10 m s $^{- 2}$ ). [5 marks] (5 marks)

Step	Marks
(a) $x = (u cos θ) t$ and $y = (u sin θ) t - \frac{1}{2} g t^{2}$	1
Eliminate t: $y = x tan θ - \frac{g x ^{2}}{2 u ^{2} c o s ^{2} θ}$ , of the form $y = a x - b x^{2}$ : a parabola	1
(b) y = 0 again when $t = \frac{2 u s i n θ}{g}$ = time of flight	1
At the top $v_{y} = 0$ : $0 = u^{2} sin^{2} θ - 2 g H$ , so $H = \frac{u ^{2} s i n ^{2} θ}{2 g}$	1
(c) $R = \frac{400 \times s i n 60°}{10} \approx 34.6 m$	1

Where students lose marks in Motion in a Plane

The slips examiners see most often, and what to write instead.

Mistake: Adding magnitudes of vectors at an angle
Treating vectors like scalars.
Wrong working
3 N + 4 N at 90° = 7 N
Right working
$9 + 16 = 5$ N
Do this: Use the resultant formula or add components.
Usually costs: 1-2 marks.
Mistake: Using the horizontal velocity to find the time of a horizontal throw
Forgetting the independence of x and y motion.
Do this: The time of fall depends only on the height: $t = 2 h / g$ .
Usually costs: Whole numerical.
Mistake: Saying the velocity at the top of a projectile's path is zero
Confusing it with a vertical throw.
Do this: Only the vertical component is zero; the horizontal component $u cos θ$ remains.
Usually costs: 1 mark.
Mistake: Getting the cross-product sign wrong
Mixing up the order.
Do this: Write the determinant with $\hat{i}, \hat{j}, \hat{k}$ in the top row, then $A$ , then $B$ ; $B \times A = - A \times B$ .
Usually costs: 1 mark.
Mistake: Using rpm directly as ω
Forgetting units.
Wrong working
300 rpm, so ω = 300
Right working
$ω = 300 \times \frac{2 π}{60} = 10 π r a d s^{- 1}$
Do this: Convert revolutions per minute to radians per second.
Usually costs: 1 mark.
Mistake: Applying equations of motion to circular motion
Seeing 'constant acceleration' in uniform circular motion.
Do this: Only the magnitude of $a_{c}$ is constant; its direction changes, so $v = v_{0} + a t$ does not apply.
Usually costs: Concept marks.

NCERT exercise guide: Motion in a Plane

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercises 3.1-3.8 and 3.19 (vectors)

What it tests: Classifying scalars and vectors, legitimate algebraic operations, statements about vector addition, and resultants of several displacements.

How to approach it: Check units and the triangle law; for several displacements resolve each one and add components.

Same kind as Exercise 3.4

Can three forces of 3 N, 4 N and 8 N acting at a point be in equilibrium? Explain.

Show the solutionSolution

For three vectors to add to zero they must form a closed triangle.
In a triangle each side is less than the sum of the other two, but 8 > 3 + 4 = 7.

Answer: No: the largest force exceeds the sum of the other two.

Exercises 3.9-3.13, 3.17, 3.18 and 3.22 (motion in a plane and projectiles)

What it tests: Velocity and acceleration from position vectors, motion with constant acceleration, and projectile range, height and time.

How to approach it: Treat x and y separately; for projectiles use T, H, R formulae or the component equations directly.

Same kind as Exercise 3.13

A girl can throw a stone to a maximum horizontal distance of 80 m on level ground. How high can she throw it straight up with the same speed? (g = 10 m s $^{- 2}$ )

Show the solutionSolution

$R_{ma x} = \frac{u ^{2}}{g} = 80 \Rightarrow u^{2} = 800$
Thrown straight up: $H = \frac{u ^{2}}{2 g} = \frac{800}{20} = 40 m$

Answer: 40 m (half the maximum range)

Exercises 3.14-3.16, 3.20 and 3.21 (circular motion and concepts)

What it tests: Centripetal acceleration of a stone or an aircraft in a loop, true/false statements on circular motion, and velocity vectors on curved paths.

How to approach it: $a_{c} = v^{2} / r$ towards the centre; velocity is tangent to the path; in uniform circular motion the average acceleration over a full circle is zero.

Same kind as Exercise 3.14

The blade of a ceiling fan is 0.6 m long and turns at 300 rpm. Find the speed and centripetal acceleration of its tip.

Show the solutionSolution

$ω = 300 \times \frac{2 π}{60} = 10 π \approx 31.4 r a d s^{- 1}$
$v = ω r = 31.4 \times 0.6 \approx 18.8 m s^{- 1}$
$a_{c} = ω^{2} r = (31.4)^{2} \times 0.6 \approx 592 m s^{- 2}$

Answer: About 18.8 m s $^{- 1}$ ; about 592 m s $^{- 2}$

Practice questions on Motion in a Plane

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQExam level
Which of these is a vector quantity?
(a) work
(b) pressure
(c) impulse
(d) time
Hint
Force × time.
Answer
(c) impulse
Q2 · 1 mark · MCQExam level
At the highest point of a projectile's path, the angle between its velocity and acceleration is:
(a) 0°
(b) 45°
(c) 90°
(d) 180°
Hint
Velocity horizontal, g vertical.
Answer
(c) 90°
Q3 · 1 mark · Assertion-reasonExam level
Assertion (A): The scalar product of two perpendicular vectors is zero.
Reason (R): $cos 90° = 0$ .
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
$A \cdot B = A B cos θ$ .
Answer
(a)
Q4 · 2 marks · Very short answerExam level
Find the unit vector along $A = 3 \hat{i} + 4 \hat{j}$ and the angle $A$ makes with the x-axis.
Hint
Divide by the magnitude.
Answer
$A = 5$ ; $\hat{A} = 0.6 \hat{i} + 0.8 \hat{j}$ ; $θ = tan^{- 1} (4/3) \approx 53.1°$ .
Q5 · 2 marks · Very short answerHarder exam level
If $∣ A + B ∣ = ∣ A - B ∣$ , find the angle between $A$ and $B$ .
Hint
Square both sides.
Answer
$A^{2} + B^{2} + 2 A B cos θ = A^{2} + B^{2} - 2 A B cos θ \Rightarrow cos θ = 0$ , so θ = 90°.
Q6 · 2 marks · Very short answerHarder exam level
The position of a particle is $r = 3 t \hat{i} + 2 t^{2} \hat{j}$ (m). Find its velocity and acceleration, and its speed at t = 1 s.
Hint
Differentiate each component.
Answer
$v = 3 \hat{i} + 4 t \hat{j}$ ; $a = 4 \hat{j}$ m s $^{- 2}$ ; at t = 1 s, speed $= 9 + 16 = 5$ m s $^{- 1}$ .
Q7 · 3 marks · Short answerHarder exam level
A projectile has a range of 30 m and a maximum height of 10 m. Find the angle of projection and the speed of projection. (g = 10 m s $^{- 2}$ )
Hint
$\frac{H}{R} = \frac{t a n θ}{4}$ .
Answer
$\frac{H}{R} = \frac{t a n θ}{4} = \frac{1}{3} \Rightarrow tan θ = \frac{4}{3}$ , θ ≈ 53°. Then $sin 2 θ = 2 \times \frac{4}{5} \times \frac{3}{5} = 0.96$ and $u^{2} = \frac{R g}{s i n 2 θ} = \frac{300}{0.96} = 312.5$ , so $u \approx 17.7$ m s $^{- 1}$ (check: $H = \frac{312.5 \times 0.64}{20} = 10$ m).
Q8 · 2 marks · Very short answerHarder exam level
Find $A \times B$ for $A = 3 \hat{i} - \hat{j}$ and $B = \hat{i} + 2 \hat{j}$ , and the area of the triangle with these two vectors as sides.
Hint
Only a $\hat{k}$ part survives.
Answer
$A \times B = (3 \times 2 - (- 1) \times 1) \hat{k} = 7 \hat{k}$ ; triangle area $= \frac{1}{2} \times 7 = 3.5$ square units.
Q9 · 2 marks · Very short answerHarder exam level
A wheel of radius 0.4 m turns at 120 rpm. Find the speed of a point on the rim and its centripetal acceleration.
Hint
120 rpm = 2 rev s $^{- 1}$ .
Answer
$ω = 4 π \approx 12.6$ rad s $^{- 1}$ ; $v = 12.6 \times 0.4 \approx 5.0$ m s $^{- 1}$ ; $a_{c} = ω^{2} r = 158 \times 0.4 \approx 63$ m s $^{- 2}$ .
Q10 · 3 marks · Short answerHOTS / competency
Two balls are thrown with the same speed from the same point, one at 30° and one at 60°. Compare their ranges, maximum heights and times of flight.
Hint
Complementary angles.
Answer
Ranges are equal ( $sin 60° = sin 120°$ ). Heights: $\frac{s i n ^{2} 30°}{s i n ^{2} 60°} = \frac{1/4}{3/4} = 1 : 3$ . Times: $\frac{s i n 30°}{s i n 60°} = 1 : 3$ .
Q11 · 3 marks · Short answerHOTS / competency
A stone is thrown horizontally from a tower and lands 40 m from its foot after 2 s. Find the height of the tower, the speed of projection and the speed on landing. (g = 10 m s $^{- 2}$ )
Hint
Time comes from height.
Answer
$h = \frac{1}{2} \times 10 \times 4 = 20$ m; $u = \frac{40}{2} = 20$ m s $^{- 1}$ ; landing: $v_{x} = 20$ , $v_{y} = 20$ , so $v = 202 \approx 28.3$ m s $^{- 1}$ at 45° below the horizontal.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State the parallelogram law of vector addition.Show answer
If two vectors are represented by adjacent sides of a parallelogram from a common point, their resultant is the diagonal from that point.
2.What is a unit vector?Show answer
A vector of magnitude 1 showing direction only: $\hat{A} = A / A$ .
3.When is the dot product of two non-zero vectors zero?Show answer
When they are perpendicular.
4.Write the range of a projectile.Show answer
$R = u^{2} sin 2 θ / g$ .
5.At what angle is the range maximum?Show answer
45°, giving $R_{ma x} = u^{2} / g$ .
6.What is the angle between velocity and acceleration at the top of a projectile's path?Show answer
90°.
7.Write the centripetal acceleration in terms of ω and r.Show answer
$a_{c} = ω^{2} r$ , directed towards the centre.
8.Is $A \times B$ equal to $B \times A$ ?Show answer
No: $B \times A = - A \times B$ .

Best order to study it

Learn scalars and vectors, position and displacement vectors, equality and multiplication by a number.
Add and subtract vectors graphically, then with the resultant formula.
Resolve vectors into components and add them analytically using unit vectors.
Learn the dot and cross products and their component forms.
Learn motion in a plane with constant acceleration as two independent 1-D motions.
Derive the projectile results and practise oblique and horizontal throws.
Finish with uniform circular motion and the derivation of $a_{c} = v^{2} / r$ .

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you explain why current is a scalar although it has direction?
Can you find the resultant of two vectors at any angle and its direction?
Can you resolve a vector into $\hat{i}$ and $\hat{j}$ components?
Can you find the angle between two vectors using the dot product?
Can you compute $A \times B$ in components?
Can you write x(t) and y(t) for a particle with constant acceleration in a plane?
Can you show a projectile's path is a parabola?
Can you derive T, H and R and show R is maximum at 45°?
Can you solve a horizontal-throw problem from a cliff?
Can you derive $a_{c} = v^{2} / r$ and convert rpm to rad s $^{- 1}$ ?

Motion in a Plane: frequently asked questions

Are dot and cross products part of this chapter for 2026-27?

Yes. The CBSE content for the unit lists the scalar and vector product of vectors. The NCERT book introduces them where they are first used (work in Chapter 5, torque in Chapter 6), so learn them here with the vectors and practise them there.

Is relative velocity in two dimensions in the syllabus?

No. It was removed in the NCERT rationalisation and is not in the 2026-27 CBSE content. River-crossing and rain-umbrella problems are therefore not needed for the school exam, though JEE still uses them.

Why is the path of a projectile a parabola?

Its x-coordinate grows in proportion to t while its y-coordinate has a term in t^2. Eliminating t gives y as a quadratic in x, y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}, which is a parabola.

Is uniform circular motion accelerated motion?

Yes. The speed is constant but the direction of velocity changes continuously, so there is an acceleration v^2/r towards the centre.

How does this chapter link to Class 12?

Vector addition and resolution are used for electric fields of several charges, the dot product for flux, and the cross product for magnetic force q\vec v\times\vec B and torque on a dipole. Circular motion of a charge in a magnetic field uses mv^2/r.

Based on: CBSE Physics (042) syllabus 2026-27 (Classes XI-XII); NCERT Physics Part I, Textbook for Class XI, Chapter 3 (Reprint 2026-27)