Motion in a Plane: Class 11 Physics · Chapter 3

Physics Part 1 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 18 Oct 2026 · Updated 18 Oct 2026

95 min
1 simulation
28 practice questions
29 MCQs

What you will learn in Motion in a Plane

Distinguish scalars from vectors and represent position and displacement as vectors
Add, subtract and scale vectors graphically and by components, and resolve a vector along given directions
Use the laws of cosines and sines to find the magnitude and direction of a resultant
Define average and instantaneous velocity and acceleration in a plane and find them by differentiating r(t)
Apply the constant-acceleration equations in vector and component form, treating perpendicular motions independently
Derive the path, time of flight, maximum height and range of a projectile and solve projectile problems
Derive the centripetal acceleration of uniform circular motion and relate v, ω, T, ν and a_c

Original companion text aligned to NCERT Physics Class 11, Chapter 3, pp. 27–48. Not affiliated with or endorsed by NCERT.

3.1 Introduction

An auto-rickshaw weaving across a crossing, a kite drifting over the rooftops on Makar Sankranti, a carrom striker sliding diagonally across the board: none of these keeps to one line. In Chapter 2 a plus or minus sign was enough to show direction, because a line has only two. A plane has infinitely many, and a sign can no longer say which.

So we first learn the language of vectors, quantities that carry a direction along with their size: how to add, subtract and scale them, and how to split them along axes. Then we describe velocity and acceleration in a plane, solve motion with constant acceleration, and apply it to a ball in flight and to steady motion round a circle. All of it extends to three dimensions with one more coordinate.

3.2 Scalars and vectors

Some quantities are completely described by one number and its unit. The mass of your school bag, the time the last bell rings, the temperature of a cup of tea and the volume of milk in a packet point nowhere. Such quantities are scalars (अदिश). They combine by ordinary arithmetic: a 4.5 kg bag plus a 1.2 kg lunchbox makes 5.7 kg, and a 40-minute class followed by a 45-minute class takes 85 minutes. Multiplying or dividing two scalars always makes sense, whatever their units: 120 km divided by 2.0 h is a speed of 60 km/h, a new scalar. Adding or subtracting them does not, unless the units match.

Other quantities make no sense without a direction. “Walk 300 m” could leave you anywhere on a circle of radius 300 m; “walk 300 m towards the temple gate” fixes one spot. A vector (सदिश) has a magnitude and a direction, and it combines with others of its kind by the triangle (or parallelogram) rule of Section 3.4. The last condition matters. An electric current in a wire has a sense along the wire, but two currents meeting at a junction simply add as numbers, so current is a scalar. Displacement, velocity, acceleration and force do obey the triangle rule; they are vectors.

In print we write a vector in bold, $𝐯$ , and its magnitude in ordinary italic type, $v$ , or as $∣ 𝐯 ∣$ . By hand, draw a small arrow over the letter. A magnitude is never negative: it is the length of the arrow on some chosen scale.

Compare

Scalars and vectors

Scalars

Magnitude only: a number with a unit
Combine by ordinary algebra
Mass, time, temperature, speed, energy, current

Vectors

Magnitude and direction
Combine by the triangle law
Displacement, velocity, acceleration, force

3.2.1 Position and displacement vectors

To say where something is in a plane, first pick a reference point, the origin O. Any point P is then reached by an arrow drawn from O to P. That arrow is the position vector $𝐫$ of P. It points from O towards P, and its length is the straight-line distance OP. Move the origin and every position vector changes, so always say where O is.

Now let the object move from P at time $t$ to P′ at a later time $t^{'}$ . The arrow from P to P′ is its displacement (विस्थापन). It records only the change of position, where the object started and where it finished, and says nothing about the route in between. Fig. 3.1 follows three routes through a housing colony from a home H to a school S. The routes have different lengths, yet all three share one displacement: the straight arrow from H to S.

4/4Every route has the same displacement, 500 m from H to S. Its magnitude is never more than the path length.

Fig. 3.1Three routes, three path lengths, one displacement.NCERT Fig. 3.1 · p. 28

A street grid of square blocks seen from above, with home H at the bottom left and school S four blocks east and three blocks north of it. A scale bar marks 100 m per block and an arrow marks north. Step 1: a solid blue route runs east along the bottom street and then north to S, labelled 700 m. Step 2: a dashed orange route overshoots one block east, goes north and then comes back west to S, labelled 900 m. Step 3: a dotted green route curves through the blocks, labelled about 590 m. Step 4: all three routes are faint and a thick vermillion arrow runs straight from H to S, labelled all: 500 m.

This gives a useful inequality. The magnitude of the displacement can never exceed the path length, and the two are equal only for motion along a straight line without turning back. In the same way, over any trip the average speed (path length ÷ time) is at least as large as the magnitude of the average velocity (displacement ÷ time). A delivery rider who finishes the day back at the shop may have ridden 60 km, but her displacement for the day is zero.

Try this outdoors

Pace it out

You’ll need

two easy-to-spot landmarks, such as a gate and a tap
a measuring tape (optional)
a notebook

Predict first

You walk from the gate to the tap by three different routes. Which statement is true?

The longest route gives the largest displacementAll three routes give the same displacementThe displacement equals the step count of the shortest route

Pick two fixed spots on your school ground or in a park, say the gate (G) and a water tap (T).
Walk from G to T along three different routes of your choice and count your steps each time.
Now walk the straight line from G to T and count again.
Measure the length of 10 of your steps with a tape, and convert every count into metres.

Table 3.1

Route	Steps	Metres
Route 1
Route 2
Route 3
Straight line G to T

Your entries are saved automatically.

Can’t do it now? Try the simulation · Fig. 3.10

3.2.2 Equality of vectors

Two vectors are equal when their magnitudes match and their directions match; where they are drawn on the page does not matter. A quick test: slide one arrow, without turning it, until its tail sits on the tail of the other. If the heads also coincide, the vectors are equal (Fig. 3.2). Two arrows of the same length that point different ways fail the test, so they are different vectors even though their magnitudes match.

3/3 $𝐂$ has the same magnitude but a different direction, so its head misses: $𝐂 \neq 𝐀$ .

Fig. 3.2Slide without turning: equal vectors match head to head.NCERT Fig. 3.2 · p. 29

Three arrows of equal length. A blue arrow A rises to the upper right on the left of the picture, a dashed blue arrow B points the same way further right, and an orange arrow C points horizontally to the right at the bottom. Step 1: the three arrows sit in their own places. Step 2: B slides, without turning, until its tail is on the tail of A; its head lands exactly on the head of A and a note says heads coincide: equal. Step 3: B is back in place and C slides to the tail of A instead; the heads end up in different places and the note says heads miss: not equal.

Push a heavy door close to its hinges, then push it just as hard near the handle. The force vector is identical, yet the door barely moves the first time. For a force like this the point of application matters, and a vector tied to a point or a line of action is called a localised vector; you will need the idea for torque in Chapter 6. The vectors in this chapter are free vectors: only magnitude and direction count, which is why sliding them around is allowed.

3.3 Multiplication of vectors by real numbers

Multiplying a vector by an ordinary number λ stretches or shrinks it. If λ is positive, $λ 𝐀$ points the same way as $𝐀$ and is λ times as long. If λ is negative, the arrow also turns round to point the opposite way. In both cases the new magnitude is

∣ λ 𝐀 ∣ = ∣ λ ∣ A

Multiplying by −1 simply reverses a vector: $- 𝐀$ has the same magnitude as $𝐀$ and the opposite direction. Fig. 3.3 lines up $𝐀$ , $2 𝐀$ , $- 𝐀$ and $- 1.5 𝐀$ from a common tail so that you can compare their lengths on the grid.

4/4 $- 1.5 𝐀$ : reversed and 1.5 times as long.

Fig. 3.3Positive multipliers stretch; negative ones also reverse.NCERT Fig. 3.3 · p. 29

Four arrows in rows, all starting from one vertical dashed line, over faint vertical grid lines spaced one length of A apart. Step 1: only the blue arrow A, two grid squares long, pointing right. Step 2: a green arrow 2A appears below it, pointing right and twice as long. Step 3: an orange arrow minus A appears, pointing left with the same length as A. Step 4: a purple arrow minus 1.5A appears, pointing left and one and a half times as long as A.

Multipliers can carry units of their own. Multiply a steady velocity of 15 m/s due north by a time of 4.0 s and you get 60 m due north: a displacement, because $(m / s) \times s = m$ . In Chapter 4, multiplying an acceleration by a mass will give a force in the same way.

Quick check

1 of 2

𝐀

is 4 units long and points east. What is

- 2.5 𝐀

10 units, pointing east10 units, pointing west1.6 units, pointing west6.5 units, pointing west

Show answer

Answer (B) 10 units, pointing west

The magnitude is

∣ - 2.5 ∣ \times 4 = 10

units, and the minus sign reverses the direction, so the result points west. A magnitude is never negative; the sign shows up as a change of direction.

3.4 Addition and subtraction of vectors — graphical method

Walk 4 m east and then 3 m north, and you finish 5 m from your starting point, in a direction between east and north. One straight walk from start to finish would have done the same job as the two walks together. That is what adding vectors means. To add $𝐁$ to $𝐀$ on paper, draw $𝐀$ to scale, then draw $𝐁$ starting where $𝐀$ ends. Join the starting point of $𝐀$ to the end point of $𝐁$ : this arrow is the sum, or resultant (परिणामी), $𝐑 = 𝐀 + 𝐁$ (Fig. 3.4). Drawing the arrows nose to tail closes a triangle, which gives the method both its names: the head-to-tail or triangle method.

4/4Adding $𝐁$ first and then $𝐀$ reaches the same point: addition is commutative, and the two routes make a parallelogram.

Fig. 3.4The triangle method: head to tail, then tail to head.NCERT Fig. 3.4 · p. 30

Step 1: a blue arrow A starts at point O and points to the right and slightly up; an orange arrow B, pointing up and to the right, sits apart from it lower down. Step 2: B slides, without turning, until its tail sits on the head of A. Step 3: a thick vermillion arrow R joins O to the far end of B, closing the triangle, with the label R = A + B. Step 4: dashed copies show B drawn first from O and A drawn after it; they reach the same head, so the two routes form a parallelogram with R as its diagonal, labelled A + B = B + A.

Draw $𝐁$ first and then $𝐀$ , and you arrive at the same final point: the two orders trace the two halves of a parallelogram whose diagonal is $𝐑$ . So vector addition is commutative. With three or more vectors, the grouping does not matter either; this is the associative law.

𝐀 + 𝐁 = 𝐁 + 𝐀

(3.1)

\begin{aligned} (𝐀 + 𝐁) + 𝐂 \\ = 𝐀 + (𝐁 + 𝐂) \end{aligned}

(3.2)

Add a vector to its own reverse and you return to where you began. The result is the null vector (zero vector) $𝟎$ , an arrow of zero length. With no length, it has no definite direction.

\begin{array}{c} 𝐀 + (- 𝐀) = 𝟎 \\ ∣ 𝟎 ∣ = 0 \end{array}

(3.3)

The null vector behaves among vectors the way zero behaves among numbers:

\begin{array}{c} 𝐀 + 𝟎 = 𝐀 \\ λ 𝟎 = 𝟎, 0 𝐀 = 𝟎 \end{array}

(3.4)

Physically, a null vector is the displacement of any round trip. A school bus that leaves its depot, runs its route and parks in the same bay may have covered 40 km, yet its displacement for the day is $𝟎$ .

Subtraction is defined through addition: to subtract $𝐁$ , add its reverse.

𝐀 - 𝐁 = 𝐀 + (- 𝐁)

(3.5)

Fig. 3.5 builds $𝐀 - 𝐁$ in this way and sets it beside $𝐀 + 𝐁$ . If both vectors are drawn from one point and the parallelogram is completed, one diagonal is the sum, and the other, running from the head of $𝐁$ to the head of $𝐀$ , is the difference.

4/4Completing the parallelogram: one diagonal is $𝐀 + 𝐁$ , and the other, from the head of $𝐁$ to the head of $𝐀$ , equals $𝐀 - 𝐁$ .

Fig. 3.5The two diagonals of a parallelogram: sum and difference.NCERT Fig. 3.5 · p. 30

Two arrows start from a common point O: a blue arrow A pointing right and a little up, and an orange arrow B pointing steeply up and to the right. Step 1: only A and B are shown. Step 2: a dashed orange arrow minus B, the same length as B but reversed, is drawn from the head of A, pointing down and to the left. Step 3: a purple arrow A minus B runs from O to the head of minus B. Step 4: the parallelogram on A and B is completed with dashed sides; a vermillion arrow A plus B runs from O to the far corner, and a dashed purple diagonal joins the head of B to the head of A, parallel and equal to A minus B.

The same picture gives a second graphical recipe, the parallelogram method. Put the tails of $𝐀$ and $𝐁$ together at a point O, draw a line through the head of each arrow parallel to the other arrow, and the diagonal from O to the far corner of the parallelogram is $𝐀 + 𝐁$ . Half of the parallelogram is exactly the head-to-tail triangle, so the two methods always agree. You can try both in the vector lab (Fig. 3.8).

Example 3.1NCERT Example 3.1 · p. 31

Swimming across a river

Meera swims across a river 60 m wide. She swims at 1.20 m/s relative to the water, always facing straight across (due north), while the current carries the water due east at 0.50 m/s. (a) What is her velocity relative to the bank? (b) How long does she take to cross, and how far downstream does she land?

Try it yourself first. Then reveal the solution one step at a time.

3.5 Resolution of vectors

Addition turns two vectors into one. The reverse job, splitting one vector into pieces along chosen directions, is called resolution, and it lets us do vector sums with arithmetic instead of a ruler. Pick two directions in the plane, given by vectors $𝐚$ and $𝐛$ that are not parallel, and draw the vector $𝐀$ from O to P. Now rule two guide lines: one through the tail O running parallel to $𝐚$ , and one through the head P running parallel to $𝐛$ . Where they meet, mark Q (Fig. 3.6). Now $𝐀$ is the sum of the arrows OQ and QP. The first lies along $𝐚$ , so it is some multiple λ of $𝐚$ ; the second is a multiple μ of $𝐛$ :

𝐀 = λ 𝐚 + μ 𝐛

(3.8)

3/3OQ $= λ 𝐚$ and QP $= μ 𝐛$ , so $𝐀 = λ 𝐚 + μ 𝐛$ (here λ ≈ 2.0, μ ≈ 2.4).

Fig. 3.6Parallels through the tail and the head split a vector along any two directions.NCERT Fig. 3.8 · p. 31

At the top left are two short reference arrows: a in blue, pointing right and slightly up, and b in orange, pointing steeply up. A long vermillion arrow A runs from point O at the bottom up to point P at the upper right. Step 1: just a, b and A. Step 2: a dashed line through O parallel to a and a dashed line through P parallel to b are drawn; they cross at a marked point Q. Step 3: A fades a little and two thick arrows replace it, a blue one from O to Q labelled lambda a and an orange one from Q to P labelled mu b, with the equation A = lambda a + mu b.

The vectors $λ 𝐚$ and $μ 𝐛$ are called the components of $𝐀$ along $𝐚$ and $𝐛$ . For any pair of non-parallel directions in the plane, the split exists and is unique.

Unit vectors

Directions are easiest to handle with unit vectors (एकांक सदिश): vectors of magnitude exactly 1, with no unit and no dimension, whose only job is to point. In a right-handed set of axes, the three that point along $+ x$ , $+ y$ and $+ z$ are written $\hat{𝐢}$ , $\hat{𝐣}$ and $\hat{𝐤}$ . They are mutually perpendicular, and

∣ \hat{𝐢} ∣ = ∣ \hat{𝐣} ∣ = ∣ \hat{𝐤} ∣ = 1

(3.9)

Every vector is its magnitude times the unit vector $\hat{𝐧}$ along it, so $\hat{𝐧} = 𝐀 / A$ . A velocity of 20 m/s towards the north-east, for example, is $20 m / s$ times the unit vector pointing north-east.

𝐀 = A \hat{𝐧}

(3.10)

Rectangular components

Now resolve along the axes themselves. Let $𝐀$ lie in the $x$ – $y$ plane at an angle θ to the positive $x$ -axis, and drop perpendiculars from its head onto the two axes (Fig. 3.7). The piece along the $x$ -axis is a multiple of $\hat{𝐢}$ , and the piece along the $y$ -axis is a multiple of $\hat{𝐣}$ :

𝐀 = A_{x} \hat{𝐢} + A_{y} \hat{𝐣}

(3.12)

The numbers $A_{x}$ and $A_{y}$ are the $x$ - and $y$ -components (घटक) of $𝐀$ . Keep the distinction clear: $A_{x}$ is a number with a unit, while $A_{x} \hat{𝐢}$ is a vector. Right-angled trigonometry gives the components from the magnitude and the angle:

\begin{array}{c} A_{x} = A \cos θ \\ A_{y} = A \sin θ \end{array}

(3.13)

3/4 $A_{x} = A \cos θ$ and $A_{y} = A \sin θ$ ; together, $𝐀 = A_{x} \hat{𝐢} + A_{y} \hat{𝐣}$ .

Fig. 3.7Components are the shadows of a vector on the axes, and they carry signs.NCERT Fig. 3.9 · p. 32

Coordinate axes with origin O. Step 1: a vermillion arrow A of fixed length leaves O at an angle theta of about 35 degrees above the positive x-axis, with the angle marked by an arc. Step 2: dashed perpendiculars drop from the head of A to both axes. Step 3: a blue arrow along the x-axis, labelled A_x = A cos theta, and a green arrow up the y-axis, labelled A_y = A sin theta, show the two components. Step 4: A is redrawn at 140 degrees, in the second quadrant; its x-component now points left along the negative x-axis, labelled A_x less than 0, while its y-component still points up.

A component may be positive, negative or zero; which one depends on θ. A vector at 150° to the $x$ -axis has $\cos 150^{\circ} < 0$ , so its $x$ -component is negative: the vector leans towards $- x$ . A vector in a plane can therefore be given in two equivalent ways: by its magnitude and angle $(A, θ)$ , or by its components $(A_{x}, A_{y})$ . Going from components back to magnitude and angle uses Pythagoras and the tangent:

A = \sqrt{A_{x}^{2} + A_{y}^{2}}

(3.14)

\tan θ = \frac{A_{y}}{A_{x}}

(3.15)

Vectors in three dimensions

In space a vector has three components. If $𝐀$ makes angles α, β and γ with the $x$ -, $y$ - and $z$ -axes, then

\begin{array}{c} A_{x} = A \cos α \\ A_{y} = A \cos β \\ A_{z} = A \cos γ \end{array}

(3.16a)

\begin{aligned} 𝐀 & = A_{x} \hat{𝐢} + A_{y} \hat{𝐣} + A_{z} \hat{𝐤} \\ A & = \sqrt{A_{x}^{2} + A_{y}^{2} + A_{z}^{2}} \end{aligned}

(3.16b)

Each of these angles is measured in the plane that contains $𝐀$ and that axis, so the three angles do not lie in one flat drawing. The position vector of a point with coordinates $(x, y, z)$ is

𝐫 = x \hat{𝐢} + y \hat{𝐣} + z \hat{𝐤}

(3.17)

A delivery drone hovering 30 m above a spot that lies 40 m east and 20 m north of its launch pad has $𝐫 = (40 \hat{𝐢} + 20 \hat{𝐣} + 30 \hat{𝐤}) m$ , taking $x$ east, $y$ north and $z$ up. Its distance from the pad is $\sqrt{40^{2} + 20^{2} + 30^{2}} \approx 53.9 m$ .

Quick check

1 of 3

A force of 60 N acts at 30° above the

+ x

-axis. What is its

y

-component, in newtons?

Show answer

Answer 30 N

F_{y} = F \sin 30^{\circ} = 60 \times 0.5 = 30 N

. The

x

-component is

60 \cos 30^{\circ} \approx 52 N

3.6 Vector addition — analytical method

With a ruler and a protractor you can add any two vectors, but every sum takes time and carries the errors of your drawing. Working with components is quicker, and as exact as your arithmetic.

DerivationAdding vectors by components

Write the two vectors in terms of their components:

\begin{array}{c} 𝐀 = A_{x} \hat{𝐢} + A_{y} \hat{𝐣} \\ 𝐁 = B_{x} \hat{𝐢} + B_{y} \hat{𝐣} \end{array}

(3.18)

Add them, and use the commutative and associative laws to collect the $\hat{𝐢}$ terms together and the $\hat{𝐣}$ terms together:

\begin{aligned} 𝐑 & = (A_{x} + B_{x}) \hat{𝐢} \\ + (A_{y} + B_{y}) \hat{𝐣} \end{aligned}

Comparing this with $𝐑 = R_{x} \hat{𝐢} + R_{y} \hat{𝐣}$ gives

\begin{array}{c} R_{x} = A_{x} + B_{x} \\ R_{y} = A_{y} + B_{y} \end{array}

(3.21)

So vectors add component by component: $x$ with $x$ , $y$ with $y$ . The rule extends at once. In three dimensions, add the $z$ -components too ( $R_{z} = A_{z} + B_{z}$ ). For any number of vectors, with subtraction allowed, combine the components with the same signs as the vectors. For $𝐓 = 𝐚 + 𝐛 - 𝐜$ ,

\begin{array}{c} T_{x} = a_{x} + b_{x} - c_{x} \\ T_{y} = a_{y} + b_{y} - c_{y} \\ T_{z} = a_{z} + b_{z} - c_{z} \end{array}

(3.23)

The vector lab lets you check this rule, and the graphical methods of Section 3.4, on any two vectors you like. Drag the arrow heads and compare the table of components with the drawing.

InteractiveVector addition lab

Fig. 3.8Drag the heads of

𝐀

and

𝐁

. Switch between the triangle method, the parallelogram method and the difference

𝐀 - 𝐁

; the table gives every component.

Example

Three legs of a delivery run

A delivery rider goes 4.0 km east, then 3.0 km in a direction 60° north of east, then 2.0 km towards the north-west. Find her overall displacement.

Try it yourself first. Then reveal the solution one step at a time.

Example 3.2NCERT Example 3.2 · p. 33

Magnitude and direction of a resultant

Two vectors $𝐀$ and $𝐁$ have magnitudes $A$ and $B$ , and the angle between them is θ. Find the magnitude $R$ of $𝐑 = 𝐀 + 𝐁$ and the angle α that $𝐑$ makes with $𝐀$ . Check your answer for $A = 3$ , $B = 4$ and $θ = 60 °$ .

Fig. 3.9

𝐑 = 𝐀 + 𝐁

drawn head to tail; SN is the height of the triangle.

A blue arrow A runs from O to P along a horizontal line. From P an orange arrow B rises to S at an angle theta above the extension of OP. A vermillion arrow R joins O to S, making an angle alpha with OP. A dashed line extends OP to N, directly below S, and a dashed vertical drops from S to N with a right-angle mark. Labels give PN = B cos theta and SN = B sin theta.

Try it yourself first. Then reveal the solution one step at a time.

These two results have names. With the angles of the triangle in Fig. 3.9 (β is the angle between $𝐑$ and $𝐁$ ), they are the law of cosines and the law of sines:

Law of cosines

\begin{aligned} R^{2} & = A^{2} + B^{2} \\ + 2 A B \cos θ \end{aligned}

(3.24a)

Law of sines

\frac{R}{\sin θ} = \frac{A}{\sin β} = \frac{B}{\sin α}

(3.24d)

\tan α = \frac{B \sin θ}{A + B \cos θ}

(3.24f)

If you know the cosine rule from trigonometry as $c^{2} = a^{2} + b^{2} - 2 a b \cos C$ , the plus sign here is no contradiction: the interior angle of the triangle at P is $180^{\circ} - θ$ , and $\cos (180^{\circ} - θ) = - \cos θ$ . Check the extremes: θ = 0 gives $R = A + B$ , θ = 180° gives $R = ∣ A - B ∣$ , and θ = 90° gives Pythagoras.

Example 3.3NCERT Example 3.3 · p. 34

An aircraft in a crosswind

A light aircraft heads due east with an airspeed of 240 km/h. A steady wind of 60 km/h blows from the south-east, that is, towards the north-west. Find the aircraft's velocity relative to the ground.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

𝐀 = 3 \hat{𝐢} - 2 \hat{𝐣}

and

𝐁 = - \hat{𝐢} + 6 \hat{𝐣}

, in units. What is the magnitude of

𝐀 + 𝐁

units

Show answer

Answer 4.47 units

𝐀 + 𝐁 = 2 \hat{𝐢} + 4 \hat{𝐣}

, so the magnitude is

\sqrt{4 + 16} = \sqrt{20} \approx 4.47

3.7 Motion in a plane

With vectors in hand we can describe motion that is not confined to a line: a carrom striker gliding across the board, a pigeon circling a temple dome, a scooter taking a bend. The three ideas of Chapter 2, displacement, velocity and acceleration, carry over, with vectors in place of signed numbers.

3.7.1 Position vector and displacement

Set up $x$ – $y$ axes with the origin at any convenient point. A particle at P with coordinates $(x, y)$ has position vector $𝐫 = x \hat{𝐢} + y \hat{𝐣}$ ; the coordinates of a point are simply the components of its position vector. If the particle is at P at time $t$ and at P′ at a later time $t^{'}$ , its displacement is

Δ 𝐫 = 𝐫^{'} - 𝐫

(3.25)

Subtracting component by component,

Δ 𝐫 = Δ x \hat{𝐢} + Δ y \hat{𝐣}

(3.26)

where

$Δ x$: $x^{'} - x$ , the change in the $x$ -coordinate
$Δ y$: $y^{'} - y$ , the change in the $y$ -coordinate

In the simulation, O is fixed while P and Q can be dragged. The segments OP and OQ have the lengths of the two position vectors, PQ has the length of the displacement from P to Q, and the right-triangle legs on each segment show its $x$ - and $y$ -components.

SimulationPosition vectors and displacement

A square coordinate grid from −10 to 10 on both axes with a fixed origin O and two draggable points, P at (−4, 2) and Q at (4, 8). Segments join O to P, O to Q and P to Q, each labelled with its length, and right-triangle legs show the x- and y-components of each segment. Dragging a point updates its coordinates and every length.

Fig. 3.10Drag P and Q. OP and OQ are the position vectors; PQ is the displacement, with legs Δx and Δy.NCERT Fig. 3.12 · p. 35

Try this

Drag Q until the displacement PQ is 5 units long with Δx = 3. How many grid points work?
Move P and Q by the same amount in the same direction. Which lengths change, and which one stays fixed?
Make OQ shorter than OP while PQ stays 10 units long.

Velocity

The average velocity over the interval $Δ t = t^{'} - t$ is the displacement divided by the time taken:

\begin{aligned} \overset{ˉ}{𝐯} & = \frac{Δ 𝐫}{Δ t} \\ = \frac{Δ x}{Δ t} \hat{𝐢} + \frac{Δ y}{Δ t} \hat{𝐣} \end{aligned}

(3.27)

Dividing by a positive scalar does not change a direction, so $\overset{ˉ}{𝐯}$ points along $Δ 𝐫$ , the chord from P to P′. The velocity (instantaneous velocity) is the limit of the average as the interval shrinks to zero:

𝐯 = \lim_{Δ t \to 0} \frac{Δ 𝐫}{Δ t} = \frac{d 𝐫}{d t}

(3.28)

What does the limit do to the direction? Fig. 3.11 keeps P fixed and lets the later point slide back along the path towards it. The chord swings round, and in the limit it lies along the tangent to the path at P. So at every point of a path the velocity is tangent to the path and points the way the object is moving. Sparks flying off a grinding wheel leave along tangents for exactly this reason.

4/4In the limit the direction is the tangent at P: this is the direction of $𝐯$ .

Fig. 3.11As Δt → 0 the chord PP′ turns into the tangent at P.NCERT Fig. 3.13 · p. 36

A curved path rises from the lower left, arches over and comes down on the right, with a fixed point P near its lower left end. Step 1: a blue arrow, the displacement, runs from P to a point P1 far along the path, on the far side of the arch. Step 2: the first chord fades and a second chord runs to a nearer point P2 near the top; it is tilted more steeply. Step 3: a third chord runs to a point closer still. Step 4: the chords are faint and a thick vermillion arrow v leaves P along the tangent to the path, the direction the chords were approaching.

In components, take the limit of each term separately:

v_{x} = \frac{d x}{d t}, v_{y} = \frac{d y}{d t}

(3.30a)

So if you know $x (t)$ and $y (t)$ , differentiating gives the velocity. Its magnitude, the speed, and its angle θ with the $x$ -axis follow as for any vector:

\begin{array}{c} v = \sqrt{v_{x}^{2} + v_{y}^{2}} \\ \tan θ = \frac{v_{y}}{v_{x}} \end{array}

(3.30b)

Acceleration

Acceleration is built from velocity exactly as velocity was built from position. Over an interval $Δ t$ the average acceleration is

\begin{aligned} \overset{ˉ}{𝐚} & = \frac{Δ 𝐯}{Δ t} \\ = \frac{Δ v_{x}}{Δ t} \hat{𝐢} + \frac{Δ v_{y}}{Δ t} \hat{𝐣} \end{aligned}

(3.31)

and the acceleration (instantaneous acceleration) is its limit:

\begin{array}{c} 𝐚 = \lim_{Δ t \to 0} \frac{Δ 𝐯}{Δ t} = \frac{d 𝐯}{d t} \\ a_{x} = \frac{d v_{x}}{d t} = \frac{d^{2} x}{d t^{2}} \\ a_{y} = \frac{d v_{y}}{d t} = \frac{d^{2} y}{d t^{2}} \end{array}

(3.32)

The average acceleration points along $Δ 𝐯$ , which you find with the triangle rule (Fig. 3.12). This need not be the direction of motion, and that is new. Along a line, $𝐯$ and $𝐚$ can only point the same way (speeding up) or opposite ways (slowing down). In a plane they can make any angle at all between 0° and 180°. When it is exactly 90°, the speed holds steady while the direction turns, as in Section 3.10.

4/4 $\overset{ˉ}{𝐚} = Δ 𝐯 / Δ t$ points along $Δ 𝐯$ , here up and to the left of the path.

Fig. 3.12The acceleration follows Δv, not the direction of motion.NCERT Fig. 3.15 · p. 37

A curved path bends upwards from the lower left. Step 1: at point P a blue arrow v points along the path. Step 2: at a later point P-prime a longer green arrow v-prime points along the path in a new, steeper direction. Step 3: on the right the two velocities are redrawn from one common tail, and a vermillion arrow delta v joins the head of v to the head of v-prime. Step 4: a purple arrow, the average acceleration, is drawn on the path between P and P-prime, pointing the same way as delta v, up and to the left, not along the path.

Example 3.4NCERT Example 3.4 · p. 37

Velocity and acceleration from a position vector

A particle moves so that $𝐫 = (2.0 t^{2} \hat{𝐢} + 6.0 t \hat{𝐣} + 4.0 \hat{𝐤}) m$ , with $t$ in seconds and each coefficient carrying the unit needed. (a) Find $𝐯 (t)$ and $𝐚 (t)$ . (b) Find the speed and the direction of motion at $t = 2.0 s$ . (c) What is the angle between $𝐯$ and $𝐚$ at that instant?

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

A car goes round a curve. At any instant its velocity points

towards the centre of the curvealong the chord to the end of the curvealong the line from the start of the curvealong the tangent to the curve, in the direction of motion

Show answer

Answer (D) along the tangent to the curve, in the direction of motion

Velocity is the limit of Δr/Δt, and in the limit the chord turns into the tangent. The centre of the curve is where the acceleration (not the velocity) points when the speed is steady.

3.8 Motion in a plane with constant acceleration

Often the acceleration stays fixed in size and direction for a while: a ball in flight accelerates at $𝐠$ , a puck on a tilted air table feels a steady push down the slope. Suppose $𝐚$ is constant and the particle has velocity $𝐯_{0}$ and position $𝐫_{0}$ at $t = 0$ .

DerivationEquations for constant acceleration

With a constant acceleration, the average acceleration over any interval equals $𝐚$ . Between $0$ and $t$ , $𝐚 = (𝐯 - 𝐯_{0}) / t$ , so

𝐯 = 𝐯_{0} + 𝐚 t

(3.33a)

The velocity changes linearly with time, so its average over the interval is the mean of its end values, $(𝐯_{0} + 𝐯) / 2$ . The displacement $Δ 𝐫 = 𝐫 - 𝐫_{0}$ is this average multiplied by $t$ :

\begin{aligned} Δ 𝐫 & = \frac{𝐯_{0} + 𝐯}{2} t \\ = \frac{2 𝐯_{0} + 𝐚 t}{2} t \end{aligned}

𝐫 = 𝐫_{0} + 𝐯_{0} t + \frac{1}{2} 𝐚 t^{2}

(3.34a)

Check: differentiating gives $d 𝐫 / d t = 𝐯_{0} + 𝐚 t$ , which is (3.33a), and at $t = 0$ it gives $𝐫 = 𝐫_{0}$ .

Written in components, the two vector equations become four ordinary ones:

\begin{array}{c} v_{x} = v_{0 x} + a_{x} t \\ v_{y} = v_{0 y} + a_{y} t \end{array}

(3.33b)

\begin{array}{c} x = x_{0} + v_{0 x} t + \frac{1}{2} a_{x} t^{2} \\ y = y_{0} + v_{0 y} t + \frac{1}{2} a_{y} t^{2} \end{array}

(3.34b)

Notice what is missing. The $x$ -equations contain only $x$ -quantities and the $y$ -equations only $y$ -quantities. Motion in a plane with constant acceleration is two independent straight-line motions, along perpendicular axes, happening at the same time, each obeying the equations of Chapter 2. Fig. 3.13 shows this with “shadows”: the projection of the particle on each axis moves like a separate one-dimensional object. In three dimensions a third axis adds a third independent motion.

4/4Seen on its own, each shadow obeys a one-dimensional equation; together they trace the curve.

Fig. 3.13Two straight-line motions, one on each axis, combine into the curved path.NCERT p. 38

Axes with origin at the lower left and a dotted curve showing the path of a particle that starts at the origin moving along x and curves upwards. A vermillion dot marks the particle at t = 0, 1, 2 and 3 s, one per step, with earlier positions left faint. Dashed lines drop from the particle to each axis, where a blue dot marks its x-shadow and a green dot its y-shadow. Step by step the x-shadow moves along the x-axis by larger and larger equal-time steps, and the y-shadow climbs the y-axis by growing steps, each like a separate straight-line motion with constant acceleration.

The vector forms (3.33a) and (3.34a) make no reference to any axes at all. You choose axes only when you want numbers, and you are free to pick whichever pair makes the algebra simplest. For a projectile that will be horizontal and vertical.

Example 3.5NCERT Example 3.5 · p. 38

A model boat with a steady push

A model boat on a pond starts from the origin at $t = 0$ with velocity $6.0 \hat{𝐢} m / s$ . Its motor and a steady breeze together give it a constant acceleration $(2.0 \hat{𝐢} + 4.0 \hat{𝐣}) m / s^{2}$ . (a) Where is the boat when its $y$ -coordinate is 18 m? (b) What is its velocity then?

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

A puck has a constant acceleration along

+ x

and starts with a velocity along

+ y

. Which statement is true?

v_{x}

and

v_{y}

both grow

v_{y}

stays constant while

v_{x}

grows

v_{y}

falls steadily to zeroThe puck moves in a straight line

Show answer

Answer (B) $v_{y}$ stays constant while $v_{x}$ grows

There is no acceleration along

y

, so

v_{y}

never changes;

v_{x}

grows from zero. The path curves, because the velocity keeps turning towards

+ x

3.9 Projectile motion

Throw a ball, kick a football, or watch water arc from a garden hose: once released, each is a projectile (प्रक्षेप्य), an object in flight acted on only by gravity. We neglect air resistance throughout, which is reasonable for compact, fairly slow objects over short distances. The acceleration is then the constant $𝐠$ , straight down, so Section 3.8 applies directly. Take $x$ horizontal and $y$ vertically up. The horizontal motion has no acceleration at all, while the vertical motion is ordinary free fall, and the two run side by side without affecting each other.

Try this

One click or two?

You’ll need

two similar coins
a 30 cm ruler
a table on a hard floor

Predict first

One coin is knocked off the table horizontally at the same moment as the other drops straight down. Which reaches the floor first?

The coin that drops straight downThe coin that is knocked sidewaysBoth at the same time

Place a coin flat at the very edge of a table.
Put a ruler flat on the table beside it, with one end sticking out past the edge, and lay a second coin on that end.
Holding the ruler near its middle as a pivot, flick the inner end sideways quickly so that the ruler knocks the edge coin horizontally off the table while the coin on the ruler drops almost straight down.
Listen carefully, then repeat with a harder flick.

Launch the projectile from the origin with speed $v_{0}$ at an angle $θ_{0}$ above the horizontal. Its acceleration and the components of its initial velocity are

a_{x} = 0, a_{y} = - g

(3.35)

\begin{array}{c} v_{0 x} = v_{0} \cos θ_{0} \\ v_{0 y} = v_{0} \sin θ_{0} \end{array}

(3.36)

Putting these into the constant-acceleration equations with $x_{0} = y_{0} = 0$ gives the position and velocity at any time $t$ :

\begin{array}{c} x = (v_{0} \cos θ_{0}) t \\ y = (v_{0} \sin θ_{0}) t - \frac{1}{2} g t^{2} \end{array}

(3.37)

\begin{array}{c} v_{x} = v_{0} \cos θ_{0} \\ v_{y} = v_{0} \sin θ_{0} - g t \end{array}

(3.38)

The horizontal velocity never changes. Only the vertical velocity does, dropping by $g$ every second: it shrinks to zero at the top and then grows again pointing down (Fig. 3.14). At the highest point the velocity is horizontal, not zero, because the ball is still moving forward at $v_{0} \cos θ_{0}$ .

3/5At the top $v_{y} = 0$ and the velocity is horizontal; the height here is $h_{m}$ .

Fig. 3.14The horizontal velocity stays the same; the vertical velocity changes steadily.NCERT Fig. 3.17 · p. 39

A dashed parabola rises from the launch point at the left of a level ground line and lands at the right. Each step shows a yellow ball at one point of the path with a blue arrow for the horizontal velocity, always the same length, a green arrow for the vertical velocity and a dashed vermillion arrow for the total velocity. Step 1, launch: a long green arrow points up and the total velocity v0 slants up at the launch angle. Step 2, rising: the green arrow is half as long. Step 3, top: there is no green arrow, only the blue one, and a dashed line drops to the ground marked h_m. Step 4, falling: the green arrow points down, half length. Step 5, landing: the green arrow points down with the same length as at launch, and a double-headed arrow along the ground marks the range R.

Equation of path of a projectile

DerivationThe path is a parabola

Eliminate the time. From the first of (3.37), $t = x / (v_{0} \cos θ_{0})$ . Substitute this into the second:

\begin{aligned} y & = x \tan θ_{0} \\ - \frac{g x^{2}}{2 v_{0}^{2} \cos^{2} θ_{0}} \end{aligned}

(3.39)

For a given launch, $g$ , $v_{0}$ and $θ_{0}$ are constants, so this has the form $y = a x - b x^{2}$ with $b > 0$ : a parabola through the origin that opens downwards.

So the path, or trajectory, of every projectile is a parabola, as long as air resistance is negligible.

Time of maximum height and time of flight

DerivationHow long the projectile rises and flies

At the top $v_{y} = 0$ . From (3.38), $v_{0} \sin θ_{0} - g t_{m} = 0$ , so

t_{m} = \frac{v_{0} \sin θ_{0}}{g}

(3.40a)

The projectile returns to its launch height when $y = 0$ again. Setting $y = 0$ in (3.37) gives $t (v_{0} \sin θ_{0} - \frac{1}{2} g t) = 0$ , so either $t = 0$ (the launch) or

T_{f} = \frac{2 v_{0} \sin θ_{0}}{g}

(3.40b)

This time of flight is exactly $2 t_{m}$ : over level ground the descent mirrors the climb.

Maximum height

Put $t = t_{m}$ into $y (t)$ , or use $v_{y}^{2} = v_{0 y}^{2} - 2 g y$ with $v_{y} = 0$ at the top. Either way the greatest height reached is

h_{m} = \frac{(v_{0} \sin θ_{0})^{2}}{2 g}

(3.41)

Horizontal range

The horizontal range $R$ is the horizontal distance covered by the time the projectile comes back to its launch height. Multiply the steady horizontal velocity by the time of flight, and use $2 \sin θ_{0} \cos θ_{0} = \sin 2 θ_{0}$ :

\begin{aligned} R & = (v_{0} \cos θ_{0}) T_{f} \\ = \frac{v_{0}^{2} \sin 2 θ_{0}}{g} \end{aligned}

(3.42a)

For a fixed launch speed, $R$ is largest when $\sin 2 θ_{0} = 1$ , that is when $θ_{0} = 45^{\circ}$ :

R_{m} = \frac{v_{0}^{2}}{g}

(3.42b)

Angles equally far above and below 45° give the same range, as Fig. 3.15 shows and Example 3.6 proves. The steeper launch climbs higher and stays up longer. You can test all of this in the projectile lab.

3/3At 45° the range is the largest possible, $v_{0}^{2} / g$ .

Fig. 3.15Complementary angles share a range; 45° goes furthest.NCERT p. 40

Parabolic paths launched from one point at the left of a level ground line, all at the same speed. Step 1: a solid blue path at 30 degrees and a dashed blue path at 60 degrees; the 60 degree path climbs three times as high, yet both land on the same marked point, labelled same range. Step 2: those paths fade and an orange pair at 15 and 75 degrees lands on a nearer common point. Step 3: a thick vermillion path at 45 degrees lands furthest of all, labelled maximum range, with the other pairs faint for comparison.

InteractiveProjectile lab

A side view of level ground with a launcher at the left on an adjustable platform. Sliders set the launch speed from 5 to 30 m/s, the launch angle from 0 to 90 degrees and the launch height from 0 to 30 m. Before playing, the predicted parabola is drawn dashed; pressing Play sends a ball along it, leaving dots every tenth of a second, with arrows for the horizontal and vertical velocity. Readouts show the time of flight, the maximum height, the range and the landing speed. An optional twin ball launched at 90 degrees minus the angle shows that complementary angles share a range when the launch height is zero.

Fig. 3.16Set the launch speed, angle and height, then press Play. Turn on the twin at 90° − θ to compare complementary angles.

Example 3.6NCERT Example 3.6 · p. 40

Complementary angles, equal ranges

(a) Show that launch angles of $45^{\circ} + α$ and $45^{\circ} - α$ give the same horizontal range for the same launch speed. (b) A ball is thrown at 25 m/s, once at 30° and once at 60°. Compare the ranges, greatest heights and flight times. Take $g = 9.8 m / s^{2}$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 3.7NCERT Example 3.7 · p. 40

A parcel thrown from a bridge

From a bridge 44.1 m above a river, a boy throws a parcel horizontally at 12 m/s to a friend in a boat. Neglect air resistance and take $g = 9.8 m / s^{2}$ . (a) How long is the parcel in the air? (b) How far from the foot of the bridge, horizontally, does it land? (c) With what velocity does it hit the water?

Try it yourself first. Then reveal the solution one step at a time.

Example 3.8NCERT Example 3.8 · p. 40

A football kicked at an angle

A footballer kicks a ball from the ground at 22 m/s, 37° above the horizontal. Find (a) the greatest height the ball reaches, (b) the time until it lands on level ground, and (c) how far away it lands. Take $g = 9.8 m / s^{2}$ and ignore air resistance.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

At the highest point of its path, a projectile has

a horizontal velocity, and acceleration g downwardszero velocity, and acceleration g downwardszero velocity and zero accelerationa horizontal velocity and zero acceleration

Show answer

Answer (A) a horizontal velocity, and acceleration g downwards

Only

v_{y}

is zero at the top;

v_{x} = v_{0} \cos θ_{0}

carries on. Gravity never switches off, so the acceleration is

g

downwards throughout the flight.

3.10 Uniform circular motion

A stone whirled on a string, the tip of a ceiling-fan blade, a satellite in a circular orbit, a car circling a roundabout with a steady speedometer reading: each moves on a circle at constant speed. This is uniform circular motion. Only the speed is steady: the direction of motion changes from instant to instant, so the velocity changes, and the object must be accelerating even though its speed never does. Which way does this acceleration point, and how large is it?

Let the object move with speed $v$ on a circle of radius $R$ about the centre C, and let it be at P at time $t$ and at P′ a short time $Δ t$ later. Its velocities $𝐯$ and $𝐯^{'}$ are tangents, so each is perpendicular to the radius drawn to its point. Both have length $v$ , so $Δ 𝐯 = 𝐯^{'} - 𝐯$ is the base of an isosceles triangle. Fig. 3.17 draws this triangle next to the circle. When P and P′ are placed symmetrically, $Δ 𝐯$ points straight inwards, along the line that bisects the angle PCP′, so the average acceleration $Δ 𝐯 / Δ t$ also points inwards. As $Δ t$ shrinks, P′ closes in on P, and in the limit the acceleration at P points exactly at the centre.

3/3In the limit the acceleration at P points at C, with magnitude $v^{2} / R$ .

Fig. 3.17However small the interval, Δv points towards the centre.NCERT Fig. 3.18 · p. 41

A circle with centre C on the left. Step 1: points P and P-prime lie 60 degrees apart on the right of the circle, joined to C by radii with the angle delta theta marked; a blue velocity arrow v at P and a green arrow v-prime at P-prime are drawn along the tangents. On the right the two arrows are redrawn from one common tail, and a vermillion arrow delta v joins their heads, pointing left, the direction from the arc towards C. Step 2: the same construction with the points only 30 degrees apart; delta v is shorter and still points towards the centre. Step 3: in the limit there is a single point P with a blue tangent velocity v and a thick vermillion acceleration arrow a pointing straight at C, labelled a_c = v squared over R.

DerivationMagnitude of the centripetal acceleration

The radii CP and CP′, each of length $R$ , meet at an angle $Δ θ$ . Each velocity is turned 90° from its radius, so the angle between $𝐯$ and $𝐯^{'}$ is also $Δ θ$ . The triangle of position vectors (sides $R$ , $R$ , $∣ Δ 𝐫 ∣$ ) and the triangle of velocities (sides $v$ , $v$ , $∣ Δ 𝐯 ∣$ ) are therefore similar isosceles triangles, and

\frac{∣ Δ 𝐯 ∣}{v} = \frac{∣ Δ 𝐫 ∣}{R}

Divide by $Δ t$ and take the limit. For a very short interval the chord $∣ Δ 𝐫 ∣$ equals the arc $v Δ t$ , so $∣ Δ 𝐫 ∣ / Δ t \to v$ :

\begin{aligned} a & = \lim_{Δ t \to 0} \frac{∣ Δ 𝐯 ∣}{Δ t} \\ = \frac{v}{R} \lim_{Δ t \to 0} \frac{∣ Δ 𝐫 ∣}{Δ t} \\ = \frac{v}{R} \cdot v \end{aligned}

a_{c} = \frac{v^{2}}{R}

(3.43)

This centripetal acceleration (अभिकेंद्र त्वरण) keeps the same size, $v^{2} / R$ , all the way round, but it always aims at the centre: downwards when the object is at the top of the circle, upwards when it is at the bottom. So $𝐚_{c}$ is not a constant vector, and the constant-acceleration equations of Section 3.8 can never be used for circular motion.

Angular speed

Circular motion is often easier to describe by angle than by distance. As the object goes from P to P′, the radius sweeps through the angle $Δ θ$ , its angular displacement. The angular speed ω is the rate at which this angle changes:

ω = \frac{Δ θ}{Δ t}

(3.44)

It is measured in radians per second. The arc covered in $Δ t$ is $Δ s = R Δ θ$ (with θ in radians), so the speed and the centripetal acceleration are

v = \frac{Δ s}{Δ t} = R ω

(3.45)

a_{c} = \frac{v^{2}}{R} = ω^{2} R

(3.46)

One full revolution takes the period $T$ and covers a distance $2 π R$ . The frequency $ν = 1 / T$ is the number of revolutions per second. Hence

v = \frac{2 π R}{T} = 2 π R ν

(3.47)

\begin{array}{c} ω = 2 π ν \\ a_{c} = 4 π^{2} ν^{2} R \end{array}

(3.48)

Try this

Let go!

You’ll need

a soft ball or a knotted cloth bundle
about 1 m of strong string
an open playground

Predict first

The ball whirls anticlockwise above your head. You let go when it is straight in front of you, moving to your left. Which way does it fly?

Straight ahead, away from youTo your left, along its direction of motion at that instantBack towards you

Tie the ball firmly to the string.
In an open space, whirl the ball slowly in a horizontal circle above your head.
Release the string at the moment the ball is straight in front of you, and watch which way it flies.
Repeat a few times, releasing at different points of the circle.

Example 3.9NCERT Example 3.9 · p. 42

A toy train on a circular track

A toy train runs steadily round a circular track of radius 0.75 m and completes 12 laps in 60 s. (a) Find its angular speed and its speed. (b) Is its acceleration a constant vector? Find its magnitude.

Try it yourself first. Then reveal the solution one step at a time.

Recap

Motion in a Plane: key points and summary

Scalars have magnitude only and follow ordinary algebra; vectors have magnitude and direction and add by the triangle law (Section 3.2).
Displacement depends only on the end points; its magnitude never exceeds the path length (Section 3.2.1).
Equal vectors have the same magnitude and direction; $λ 𝐀$ has magnitude $∣ λ ∣ A$ and is reversed when λ < 0 (Section 3.2.2, Section 3.3).
Vectors add head to tail or by the parallelogram rule; addition is commutative and associative, and $𝐀 - 𝐁 = 𝐀 + (- 𝐁)$ (Section 3.4).
The null vector has zero magnitude and no direction: the displacement of a round trip ((3.3)).
Any vector in a plane splits uniquely along two non-parallel directions: $𝐀 = λ 𝐚 + μ 𝐛$ (Section 3.5).
$𝐀 = A_{x} \hat{𝐢} + A_{y} \hat{𝐣}$ with $A_{x} = A \cos θ$ , $A_{y} = A \sin θ$ , $A = \sqrt{A_{x}^{2} + A_{y}^{2}}$ and $\tan θ = A_{y} / A_{x}$ (Rectangular components).
Components of a sum are sums of components; the magnitude of $𝐀 + 𝐁$ is $\sqrt{A^{2} + B^{2} + 2 A B \cos θ}$ (Section 3.6).
$𝐯 = d 𝐫 / d t$ is tangent to the path; $𝐚 = d 𝐯 / d t$ points along $Δ 𝐯$ and can make any angle with $𝐯$ (Section 3.7).
With constant $𝐚$ : $𝐯 = 𝐯_{0} + 𝐚 t$ and $𝐫 = 𝐫_{0} + 𝐯_{0} t + \frac{1}{2} 𝐚 t^{2}$ , two independent 1D motions (Section 3.8).
A projectile moves uniformly sideways while it falls freely; its path is a parabola (Section 3.9).
$t_{m} = v_{0} \sin θ_{0} / g$ , $T_{f} = 2 t_{m}$ , $h_{m} = (v_{0} \sin θ_{0})^{2} / 2 g$ , $R = v_{0}^{2} \sin 2 θ_{0} / g$ , largest at 45° (Horizontal range).
In uniform circular motion $a_{c} = v^{2} / R = ω^{2} R$ , always towards the centre, so it is not a constant vector (Section 3.10).
$v = R ω$ , $ω = 2 π ν = 2 π / T$ and $a_{c} = 4 π^{2} ν^{2} R$ (Angular speed).

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Quick check: MCQs on Motion in a Plane

3.2.2 Equality of vectors4 questions
3.3 Multiplication of vectors by real numbers2 questions
3.4 Addition and subtraction of vectors — graphical method4 questions
Vectors in three dimensions3 questions
3.6 Vector addition — analytical method3 questions
Acceleration3 questions
3.8 Motion in a plane with constant acceleration3 questions
Horizontal range3 questions
Angular speed4 questions

Glossary · 15

Key terms (English–Hindi)

angular speedकोणीय चालkoniya chaal: The rate of change of the angle swept by the radius, $ω = Δ θ / Δ t$ , in rad/s.First seen in the chapter →

centripetal accelerationअभिकेंद्र त्वरणabhikendra tvaran: The acceleration $v^{2} / R$ of uniform circular motion, always pointing towards the centre.First seen in the chapter →
componentघटकghatak: The part of a vector along a chosen axis; $A_{x} = A \cos θ$ can be positive, negative or zero.First seen in the chapter →

displacementविस्थापनvisthapan: The change in position, $Δ 𝐫 = 𝐫^{'} - 𝐫$ , from the initial to the final point, whatever the route.First seen in the chapter →

horizontal rangeक्षैतिज परासkshaitij paras: The horizontal distance a projectile covers before returning to its launch height, $v_{0}^{2} \sin 2 θ_{0} / g$ .First seen in the chapter →

null vectorशून्य सदिशshunya sadish: The vector of zero magnitude, $𝟎$ , with no definite direction.First seen in the chapter →

position vectorस्थिति सदिशsthiti sadish: The arrow from a chosen origin to the point where an object is: $𝐫 = x \hat{𝐢} + y \hat{𝐣}$ .First seen in the chapter →
projectileप्रक्षेप्यprakshepya: An object in flight after being thrown or launched, moving under gravity alone.First seen in the chapter →

resultantपरिणामीparinami: The single vector equal to the sum of two or more vectors.First seen in the chapter →

scalarअदिशadish: A quantity fully described by a number and a unit, such as mass, time or temperature.First seen in the chapter →

time of flightउड्डयन कालuddayan kaal: The time a projectile spends in the air before returning to its launch height, $2 v_{0} \sin θ_{0} / g$ .First seen in the chapter →
trajectoryप्रक्षेप पथprakshep path: The path followed by a moving object; for a projectile without air drag, a parabola.First seen in the chapter →

uniform circular motionएकसमान वृत्तीय गतिeksaman vrittiya gati: Motion along a circle at constant speed; the velocity keeps changing direction.First seen in the chapter →
unit vectorएकांक सदिशekank sadish: A dimensionless vector of magnitude 1 that only shows a direction, such as $\hat{𝐢}$ , $\hat{𝐣}$ , $\hat{𝐤}$ .First seen in the chapter →

vectorसदिशsadish: A quantity with magnitude and direction that adds by the triangle law, such as displacement or force.First seen in the chapter →

Motion in a Plane: Class 11 Physics · Chapter 3

What you will learn in Motion in a Plane

3.1 Introduction

3.2 Scalars and vectors

3.2.1 Position and displacement vectors

3.2.2 Equality of vectors

3.3 Multiplication of vectors by real numbers

3.4 Addition and subtraction of vectors — graphical method

3.5 Resolution of vectors

Unit vectors

Rectangular components

Vectors in three dimensions

3.6 Vector addition — analytical method

3.7 Motion in a plane

3.7.1 Position vector and displacement

Velocity

Acceleration

3.8 Motion in a plane with constant acceleration

3.9 Projectile motion

Equation of path of a projectile

Time of maximum height and time of flight

Maximum height

Horizontal range

3.10 Uniform circular motion

Angular speed

Motion in a Plane: key points and summary

Practice questions with answers (NCERT “Exercises”)

Find your NCERT exercise question

Quick check: MCQs on Motion in a Plane

Key terms (English–Hindi)

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