Chapter 3 Motion in a Plane

NCERT Solutions Class 11 Physics Chapter 3 – Motion in a Plane

NCERT Solutions for Class 11 Physics Chapter 3 (Motion in a Plane) extend the concepts of kinematics into two dimensions. While Chapter 2 dealt with linear paths, this chapter introduces Vectors, which are essential for describing motion in a curved path, such as a ball thrown at an angle or a car turning on a circular track.

The NCERT Solutions for Class 11 Physics Chapter 3 by ALLEN are prepared by expert faculty to build strong clarity in vector-based kinematics. The solutions use a systematic, step-by-step approach to help students confidently solve 2D motion problems for JEE and NEET.

This chapter is a base for JEE and NEET preparation. Projectile Motion and Uniform Circular Motion are common concepts tested. These solutions give a distinct mathematical approach to tackling complex 2D problems via vector addition, resolving vectors and the use of unit vectors (\hat{i} and \hat{j}).

1.0Class 11 Physics Chapter 3 : Key Concepts

This chapter focuses on the kinematics of particles moving in a two-dimensional coordinate system. Key lessons include:

Scalars and Vectors: Understanding magnitude and direction. Learning about unit vectors, null vectors, and equal vectors.
Vector Operations: * Addition and Subtraction: Using the Triangle Law and Parallelogram Law.
Resolution of Vectors: Splitting a vector into horizontal $(A_{x} = A cos θ) an d v er t i c a l (A_{y} = A sin θ)$ components.
Motion in a Plane with Constant Acceleration: Treating 2D motion as two independent 1D motions along the x and y axes.
Projectile Motion: Analyzing an object launched into the air under gravity.
Equation of Path (Trajectory): Proving the path is a parabola.

Time of Flight (T): $T = \frac{2 u s i n θ}{g}$
Maximum Height (H): $H = \frac{u ^{2} s i n ^{2} θ}{2 g}$
Horizontal Range (R): $R = \frac{u ^{2} s i n 2 θ}{g}$
Uniform Circular Motion: Motion in a circle at a constant speed.

Angular Velocity $(ω)$ and Angular Acceleration.

Centripetal Acceleration $(a_{c})$ : Understanding why an object at constant speed still accelerates toward the center: $a_{c} = \frac{v ^{2}}{R} = ω^{2} R$
Relative Velocity in 2D: Solving "Rain-Man" and "River-Boat" problems using vector subtraction: $v_{A B} = v_{A} - v_{B}$

2.0Detailed NCERT Textbook Solutions : Class 11 Physics Chapter 3

SOLVED EXAMPLES

Rain is falling vertically with a speed of $35 m s^{- 1}$ . Winds starts blowing after sometime with a speed of $12 m s^{- 1}$ in east to west direction. In which direction should a boy waiting at a bus stop hold his umbrella? Sol. The velocity of the rain and the wind are represented by the vectors $v_{r}$ and $v_{w}$ in Figure and are in the direction specified by the problem. Using the rule of vector addition, we see that the resultant of $v_{r}$ and $v_{w}$ is $R$ as shown in the figure. The magnitude of $R$ is

$R = v_{r}^{2} + v_{w}^{2} = 3 5^{2} + 1 2^{2} ms^{- 1} = 37 ms^{- 1}$

The direction $θ$ that $R$ makes with the vertical is given by

$tan θ = \frac{v _{w}}{v _{r}} = \frac{12}{35} = 0.343$

or, $θ = tan^{- 1} (0.343) = 1 9^{o}$ Therefore, the boy should hold his umbrella in the vertical plane at an angle of about 19° with the vertical towards the east.

2. Find the magnitude and direction of the resultant of two vectors $A$ and $B$ in terms of their magnitudes and angle $θ$ between them. Sol. Let OP and OQ represent the two vectors $A$ and $B$ making an angle $θ$ (Figure). Then, using the parallelogram method of vector addition, OS represents the resultant vector $R : R = A + B$

SN is normal to OP and PM is normal to OS. From the geometry of the figure,

$OS^{2} = ON^{2} + SN^{2}$

but $ON = OP + PN = A + B cos θ$

$SN = B sin θ$

$OS^{2} = (A + B cos θ)^{2} + (B sin θ)^{2}$ or, $R^{2} = A^{2} + B^{2} + 2 AB cos θ$

$R = A^{2} + B^{2} + 2 A B cos θ$

In $△ OSN, SN = OS sin α = R sin α$ , and in $Δ PSN, SN = PS sin θ = B sin θ$

$or, \frac{R}{s i n θ} = \frac{B}{s i n α}$

Similarly, $PM = A sin α = B sin β$

$or \frac{A}{s i n β} = \frac{B}{s i n α}$

Combining equation (ii) and (iii), we get

$\frac{R}{s i n θ} = \frac{A}{s i n β} = \frac{B}{s i n α}$

Using equation (iv), we get

$sin α = \frac{B}{R} sin θ$

where R is given by equation (i)

$or, tan α = \frac{SN}{OP + PN} = \frac{B s i n θ}{A + B c o s θ}$

Equation (i) gives the magnitude of the resultant and equation (v) and (vi) its direction. Equation (i) is known as the law of cosines and equation (iv) as the Law of sines.

A motorboat is racing towards north at 25 km/h and the water current in that region is 10 km/h in the direction of 60° east of south. Find the resultant velocity of the boat. Sol. The vector $v_{b}$ representing the velocity of the motorboat and the vector $v_{c}$ representing the water current are shown in Figure in directions specified by the problem. Using the parallelogram method of addition, the resultant $R$ is obtained in the direction shown in the figure.

We can obtain the magnitude of $R$ using the Law of cosine:

$R = v_{b}^{2} + v_{c}^{2} + 2 v_{b} v_{c} cos 12 0^{\circ} = 2 5^{2} + 1 0^{2} + 2 \times 25 \times 10 (- \frac{1}{2}) = 21.8 km / h$

To obtain the direction, we apply the Law of sines

$\frac{R}{sin θ} = \frac{v _{C}}{sin ϕ} or, sin ϕ = \frac{v _{C}}{R} sin θ = \frac{10 \times sin 12 0 ^{\circ}}{21.8} = \frac{10 3}{2 \times 21.8} = 0.397 = \frac{10 \times sin 12 0 ^{\circ}}{21.8} = \frac{10 3}{2 \times 21.8} = 0.397 ϕ = 23. 4^{\circ}$

The position of a particle is given by

$r = 3.0 t \hat{i} + 2.0 t^{2} \hat{j} + 5.0 \hat{k}$

where t is in seconds and the coefficients have the proper units for $r$ to be in metres. (a) Find $v (t)$ and $a (t)$ of the particle. (b) Find the magnitude and direction of $v (t)$ at $t = 1.0 s$ . Sol. $v (t) = \frac{dr}{dt} = \frac{d}{dt} (3.0 t \hat{i} + 2.0 t^{2} \hat{j} + 5.0 \hat{k})$

$v (t) = 3.0 \hat{i} + 4.0 t \hat{j} a (t) = \frac{d v}{d t} = + 4.0 \hat{j}$

$a = 4.0 ms^{- 2}$ along y-direction At $t = 1.0 s, v = 3.0 \hat{i} + 4.0 \hat{j}$ It's magnitude is $v = 3^{2} + 4^{2} = 5.0 ms^{- 1}$ and direction is $θ = tan^{- 1} (\frac{v _{y}}{v _{x}}) = tan^{- 1} (\frac{4}{3}) ≅ 5 3^{\circ}$ with x-axis.

A particle starts from origin at $t = 0$ with a velocity $5.0 \hat{i} m / s$ and moves in $x - y$ plane under action of a force which produces a constant acceleration of $(3.0 \hat{i} + 2.0 \hat{j}) m / s^{2}$ . (a) What is the y -coordinate of the particle at the instant its x-coordinate is 84 m ? (b) What is the speed of the particle at this time ? Sol. The position of the particle is given by

$r (t) = v_{0} t + \frac{1}{2} a t^{2} r (t) = 5.0 \hat{i} t + (\frac{1}{2}) (3.0 \hat{i} + 2.0 \hat{j}) t^{2} r (t) = (5.0 t + 1.5 t^{2}) \hat{i} + 1.0 t^{2} \hat{j}$

Therefore, $x (t) = 5.0 t + 1.5 t^{2}$

$y (t) = + 1.0 t^{2}$

Given $x (t) = 84 m, t =$ ?

$5.0 t + 1.5 t^{2} = 84 \Rightarrow t = 6 s$

at $t = 6 s, y = 1.0 (6)^{2} = 36.0 m$

Now, the velocity

$v = \frac{d r}{d t} = (5.0 + 3.0 t) \hat{i} + 2.0 t \hat{j} At t = 6 s, v = 23.0 \hat{i} + 12.0 \hat{j} Speed = ∣ v ∣ = 2 3^{2} + 1 2^{2} ≅ 26 ms^{- 1}$

Galileo, in his book Two new sciences, stated that "for elevations which exceed or fall short of 45° by equal amounts, the ranges are equal". Prove this statement. Sol. For a projectile launched with velocity $v_{0}$ , at an angle $θ_{0}$ , the range is given by

$R = \frac{v _{0}^{2} s i n 2 θ _{0}}{g}$

Now, for angles, $(4 5^{\circ} + α)$ and $(4 5^{\circ} - α), 2 θ_{0}$ is $(9 0^{o} + 2 α)$ and $(9 0^{o} - 2 α)$ , respectively. The values of $sin (9 0^{\circ} + 2 α)$ and $sin (9 0^{\circ} - 2 α)$ are the same, equal to that of $cos 2 α$ . Therefore, ranges are equal for elevations which exceed or fall short of 45° by equal amounts $α$ .

A hiker stands on the edge of a cliff 490 m above the ground and throws a stone horizontally with an initial speed of $15 ms^{- 1}$ . Neglecting air resistance, find the time taken by the stone to reach the ground, and the speed with which it hits the ground. (Take $g = 9.8 m$ $s^{- 2}$ ). Sol. We choose the origin of the x- and y-axis at the edge of the cliff and $t = 0 s$ at the instant the stone is thrown. Choose the positive direction of x-axis to be along the initial velocity and the positive direction of y-axis to be the vertically upward direction. The x- and y-components of the motion can be treated independently. The equations of motion are :

$x (t) = x_{0} + v_{o x} t y (t) = y_{o} + v_{oy} t + (\frac{1}{2}) a_{y} t^{2}$

Here, $x_{o} = y_{o} = 0, v_{oy} = 0, a_{y} = - g = - 9.8 ms^{- 2}$ , $v_{ox} = 15 m s^{- 1}$ .

The stone hits the ground when

$y (t) = - 490 m - 490 m = - (\frac{1}{2}) (9.8) t^{2} \Rightarrow t^{2} = \frac{9800}{98}$

This gives $t = 10 s$ . The velocity components are $v_{x} = v_{ox}$ and

$v_{y} = v_{oy} - gt$

so that when the stone hits the ground :

$v_{ox} v_{oy} = 15 ms^{- 1} = 0 - 9.8 \times 10 = - 98 m s^{- 1}$

Therefore, the speed of the stone is

$v_{x}^{2} + v_{y}^{2} = 1 5^{2} + 9 8^{2} = 99 ms^{- 1}$

A cricket ball is thrown at a speed of $28 ms^{- 1}$ in a direction 30° above the horizontal. Calculate (a) the maximum height, (b) the time taken by the ball to return to the same level, and (c) the distance from the thrower to the point where the ball returns to the same level.

Sol. (a) The maximum height is given by

$h_{m} = \frac{( v _{0} sin θ _{0} ) ^{2}}{2 g} = \frac{( 28 sin 3 0 ^{\circ} ) ^{2}}{2 ( 9.8 )} m = \frac{14 \times 14}{2 \times 9.8} = 10.0 m$

(b) The time taken to return to the same level is

$T_{f} = \frac{2 v _{0} sin θ _{0}}{g} = \frac{2 \times 28 sin 3 0 ^{\circ}}{9.8} = \frac{28}{9.8} s = 2.9 s$

$R = \frac{v _{0}^{2} s i n 2 θ _{0}}{g} = \frac{28 \times 28 \times s i n 6 0 ^{\circ}}{9.8} = 69 m$

An insect trapped in a circular groove of radius 12 cm moves along the groove steadily and completes 7 revolutions in 100 s. (a) What is the angular speed, and the linear speed of the motion? (b) Is the acceleration vector a constant vector ? What is its magnitude ? Sol. This is an example of uniform circular motion. Here $R = 12 cm$ . The angular speed $ω$ is given by

$ω = \frac{2 π}{T} = 2 π \times \frac{7}{100} = 0.44 rad / s$

The linear speed v is :

$v = ω R = 0.44 s^{- 1} \times 12 cm = 5.3 cm s^{- 1}$

The direction of velocity v is along the tangent to the circle at every point. The acceleration is directed towards the centre of the circle. Since this direction changes continuously, acceleration here is not a constant vector. However, the magnitude of acceleration is constant:

$a = ω^{2} R = (0.44 s^{- 1})^{2} (12 cm) = 2.3 cm s^{- 2}$

EXERCISE QUESTIONS WITH SOLUTIONS

State, for each of the following physical quantities, if it is a scalar or a vector : volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity. Sol. Scalar: Volume, mass, speed, density, number of moles, angular frequency. Vector: Acceleration, velocity, displacement, angular velocity. A scalar quantity is specified by its magnitude only. It does not have any direction associated with it. Volume, mass, speed, density, number of moles, and angular frequency are some of the scalar physical quantities. A vector quantity is specified by its magnitude as well as the direction associated with it. Acceleration, velocity, displacement, and angular velocity belong to this category.
Pick out the two scalar quantities in the following list : force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity. Sol. Work and current are scalar quantities. Work done is given by the dot product of force and displacement. Since the dot product of two quantities is always a scalar, work is a scalar physical quantity. Current is described only by its magnitude. Its direction is not taken into account. Hence, it is a scalar quantity.
Pick out the only vector quantity in the following list : Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge. Sol. Impulse is given by the product of force and time. Since force is a vector quantity, its product with time (a scalar quantity) gives a vector quantity.
State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful : (a) adding any two scalars, (b) adding a scalar to a vector of the same dimensions , (c) multiplying any vector by any scalar, (d) multiplying any two scalars, (e) adding any two vectors, (f) adding a component of a vector to the same vector. Sol. (a) Not Meaningful (b) Not Meaningful (c) Meaningful (d) Meaningful (e) Not Meaningful (f) Not Meaningful

Explanation:

(a) The addition of two scalar quantities is meaningful only if they both represent the same physical quantity. (b) The addition of a vector quantity with a scalar quantity is not meaningful. (c) A scalar can be multiplied with a vector. For example, force is multiplied with time to give impulse. (d) A scalar, irrespective of the physical quantity it represents, can be multiplied with another scalar having the same or different dimensions. (e) The addition of two vector quantities is meaningful only if they both represent the same physical quantity. (f) A component of a vector can be added to the same vector as they both have the same dimensions, but it is not needed.

Read each statement below carefully and state with reasons, if it is true or false: (a) The magnitude of a vector is always a scalar. (b) Each component of a vector is always a scalar. (c) The total path length is always equal to the magnitude of the displacement vector of a particle. (d) The average speed of a particle (defined as total path length divided by the time taken to cover the path) is either greater or equal to the magnitude of average velocity of the particle over the same interval of time, (e) Three vectors not lying in a plane can never add up to give a null vector.

Sol. (a) True

(b) False (c) False (d) True (e) True

Explanation:

(a) The magnitude of a vector is a number. Hence, it is a scalar. (b) Each component of a vector is also a vector. (c) Total path length is a scalar quantity, whereas displacement is a vector quantity. Hence, the total path length is always greater than the magnitude of displacement. It becomes equal to the magnitude of displacement only when a particle is moving in a straight line. (d) It is because of the fact that the total path length is always greater than or equal to the magnitude of displacement of a particle. (e) Three vectors, which do not lie in a plane, cannot be represented by the sides of a triangle taken in the same order.

Establish the following vector inequalities geometrically or otherwise : (a) $∣ a + b ∣ \leq ∣ a ∣ + ∣ b ∣$ (b) $∣ a + b ∣ \geq ∥ a ∣ - ∣ b ∥$ (c) $∣ a - b ∣ \leq ∣ a ∣ + ∣ b ∣$ (d) $∣ a - b ∣ \geq ∣∣ a ∣ - ∣ b ∣∣$

When does the equality sign above apply Sol. (a) Let two vectors $a$ and $b$ be represented by the adjacent sides of a parallelogram OMNP, as shown in the given figure

Here we can write

$∣ OM ∣ = ∣ a ∣ ∣ MN ∣ = ∣ OP ∣ = ∣ b ∣ ∣ ON ∣ = ∣ a + b ∣$

In a triangle, each side is smaller than If the two vectors $a$ and $b$ act along a the sum of the other two sides. straight line in the same direction, then Therefore, in $△ OMN$ , we have we can write:

$ON < (OM + MN) ∣ a + b ∣ < ∣ a ∣ + ∣ b ∣ .$

$∣ a + b ∣ = ∣∣ a ∣ - ∣ b ∣∣$

Combining equations (iv) and (v), we get If the two vectors $a$ and $b$ act along a

$∣ a + b ∣ \geq ∣∣ a ∣ - ∣ b ∣$

straight line in the same direction, then we can write:

$∣ a + b ∣ = ∣ a ∣ + ∣ b ∣$

Combining equations (iv) and (v), we get

$∣ a + b ∣ \leq ∣ a ∣ + ∣ b ∣$

(b) Let two vectors $a$ and $b$ be represented by the adjacent sides of a parallelogram OMNP, as shown in the given figure.

physics-chap-3-class-11-exercise-ques-6-(ii)

Here we can write

$∣ OM ∣ = ∣ a ∣ ∣ MN ∣ = ∣ OP ∣ = ∣ b ∣ ∣ ON ∣ = ∣ a + b ∣$

In a triangle, each side is smaller than the sum of the other two sides. Therefore, in $△ OMN$ , we have

$ON + MN > OM ON + OM > MN ∣ ON ∣ > ∣ OM - OP ∣ (∵ OP = MN) ∣ a + b ∣ > ∣∣ a ∣ - ∣ b ∣ \dots .. (iv)$

(c) Let two vectors $a$ and $b$ be represented by the adjacent sides of a parallelogram PORS, as shown in the given figure.

exercise-ques-6-(iii)-class-11-chap-3-physics

Here we have :

$∣ OR ∣ = ∣ PS ∣ = ∣ b ∣ ∣ OP ∣ = ∣ a ∣$

In triangle, each side is smaller than the sum of the other two sides. Therefore, in $Δ OPS$ , we have :

$OS < OP + PS ∣ a - b ∣ < ∣ a ∣ + ∣ - b ∣ ∣ a - b ∣ < ∣ a ∣ + ∣ b ∣$

If the two vectors act in a straight line but in opposite directions, then we can write:

$∣ a - b ∣ = ∣ a ∣ + ∣ b ∣$

Combining equation (iii) and (iv), we get

$∣ a - b ∣ \leq ∣ a ∣ + ∣ b ∣$

(d) Let two vectors $a$ and $b$ be represented by the adjacent sides of a parallelogram PORS, as shown in the given figure.

class-11-exercise-ques-6-(d)-physics-chap-3

The following relations can be written for the given parallelogram

$OS + PS > OP OS > OP - PS ∣ a - b ∣ > ∣ a ∣ - ∣ b ∣$

The quantity on the LHS is always positive and that on the RHS can be positive or negative. To make both quantities positive, we take modulus on both sides as:

$∥ a - b ∥ > ∣∣ a ∣ - ∣ b ∥ ∣ a - b ∣ > ∣ a ∣ - ∣ b ∣ ∣$

If the two vectors act in a straight line but in the opposite directions, then we can write:

$∣ a - b ∣ = ∣∣ a ∣ - ∣ b ∣∣$

Combining equations (iv) and (v), we get:

$∣ a - b ∣ \geq ∣∣ a ∣ - ∣ b ∣∣$

Given $a + b + c + d = 0$ , which of the following statements are correct: (a) $a, b, c$ and $d$ must each be a null vector. (b) The magnitude of $(a + c)$ equals the magnitude of $(b + d)$ . (c) The magnitude of $a$ can never be greater than the sum of the magnitudes of $b, c$ , and $d$ , (d) $b + c$ must lie in the plane of $a$ and $d$ if $a$ and $d$ are not collinear, and in the line of $a$ and $d$ , if they are collinear? Sol. (a) Incorrect In order to make $a + b + c + d = 0$ , it is not necessary to have all the four given vectors to be null vectors. There are many other combinations which can give the sum zero. (b) Correct

$a + b + c + d = 0 a + c = - (b + d)$

Taking modulus on both the sides, we get:

$∣ a + c ∣ = ∣ - (b + d) = ∣ b + d ∣$

Hence, the magnitude of $(a + c)$ is the same as the magnitude of $(b + d)$ . (c) Correct

$a + b + c + d = 0 a = (b + c + d)$

Taking modulus both sides, we get:

$∣ a ∣ = ∣ b + c + d ∣ ∣ a ∣ \leq ∣ b ∣ + ∣ c ∣ + ∣ d ∣$

Equation (i) shows that the magnitude of $a$ is equal to or less than the sum of the magnitudes of $b, c$ , and $d$ . Hence the magnitude of $a$ can never be greater than the sum of the magnitudes of $b, c$ , and $d$ . (d) Correct For $a + b + c + d = 0$

$a + (b + c) + d = 0$

The resultant sum of the three vectors $a$ , $(b + c)$ , and $d$ can be zero only if $(b + c)$ lie in a plane containing $a$ and $d$ , assuming that these three vectors are represented by the three sides of a triangle. If $a$ and $d$ are collinear, then it implies that the vector $(b + c)$ is in the line of $a$ and $d$ . This implication holds only then the vector sum of all the vectors will be zero.

Three girls skating on a circular ice ground of radius 200 m start from a point P on the edge of the ground and reach a point Q diametrically opposite to P following different paths as shown in Figure. What is the magnitude of the displacement vector for each ? For which girl is this equal to the actual length of path skated?

Sol. Displacement is given by the minimum distance between the initial and final positions of a particle. In the given case, all the girls start from point P and reach point Q. The magnitudes of their displacements will be equal to the diameter of the ground. Radius of the ground $= 200 m$ Diameter of the ground $= 2 \times 200 = 400 m$ Hence, the magnitude of the displacement for each girl is 400 m. This is equal to the actual length of the path skated by girl B

A cyclist starts from the centre O of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference, and returns to the centre along QO as shown in Figure. If the round trip takes 10 min, what is the (a) net displacement, (b) average velocity, and (c) average speed of the cyclist ?

Sol. Displacement is given by the minimum distance between the initial and final positions of a body. In the given case, the cyclist comes to the starting point after cycling for 10 minutes. Hence, his net displacement is zero. Average velocity is given by the relation: Average velocity $= \frac{Net displacement}{Total time}$ Since the net displacement of the cyclist is zero, his average velocity will also be zero. Average speed of the cyclist is given by the relation:

$Average speed = \frac{Total path length}{Total time}$

Total path length $= OP + PQ + QO$

$= 1 + \frac{1}{4} (2 π \times 1) + 1 = 2 + \frac{1}{2} π = 3.570 km$

Time taken $= 10 min = \frac{10}{60} = \frac{1}{6} h$

$∴ Average speed = \frac{3.570}{\frac{1}{6}} = 21.42 km / h$

On an open ground, a motorist follows a track that turns to his left by an angle of 60° after every 500 m. Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.

Sol. The path followed by the motorist is a regular hexagon with side 500 m, as shown in the given figure

class-11-physics-chap-3-exercise-ques-10

Let the motorist start from point P. The motorist takes the third turn at S. ∴ Magnitude of displacement

$= PS = PV + VS = 500 + 500 = 1000 m$

Total path length $= PQ + QR + RS$

$= 500 + 500 + 500 = 1500 m$

The motorist takes the sixth turn at point P, which is the starting point. ∴ Magnitude of displacement = 0 Total path length

$= PQ + QR + RS + ST + TU + UP = 500 + 500 + 500 + 500 + 500 + 500 = 3000 m$

The motorist takes the eight turn at point R ∴ Magnitude of displacement = PR

$= P Q^{2} + Q R^{2} + 2 (PQ) \cdot (QR) cos 6 0^{\circ} = 50 0^{2} + 50 0^{2} + (2 \times 500 \times 500 \times cos 6 0^{\circ}) = 250000 + 250000 + (500000 \times \frac{1}{2}) = 866.03 m β = tan^{- 1} (\frac{500 sin 6 0 ^{\circ}}{500 + 500 cos 6 0 ^{\circ}}) = 3 0^{\circ}$

Therefore, the magnitude of displacement is 866.03 m at an angle of 30 with PR.

Total path length = Circumference of the hexagon + PQ + QR

$= 6 \times 500 + 500 + 500 = 4000 m$

The magnitude of displacement and the total path length corresponding to the required turns is shown in the given table

Turn	Magnitude of displacement (m)	Total path length (m)
Third	1000	1500
Sixth	0	3000
Eighth	866.03 ; 30°	4000

11. A passenger arriving in a new town wishes to go from the station to a hotel located 10 km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23 km long and reaches the hotel in 28 min. What is (a) the average speed of the taxi, (b) the magnitude of average velocity? Are the two equal? Sol. Total distance travelled = 23km

$Total time taken = 28 min = \frac{28}{60} h$

∴ Average speed of the taxi

$= \frac{Total distance travelled}{Total time taken} = \frac{23}{( \frac{28}{60} )} = 49.29 km / h$

Distance between the hotel and the station $= 10 km =$ Displacement of the car ∴ Average velocity $= \frac{10}{\frac{28}{60}} = 21.43 km / h$ Therefore, the two physical quantities (average speed and average velocity) are not equal.

The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of $40 m s^{- 1}$ can go without hitting the ceiling of the hall? Sol. Speed of the ball, $u = 40 m / s$ Maximum height, $h = 25 m$ In projectile motion, the maximum height reached by a body projected at an angle $θ$ , is given by the relation:

$h = \frac{u ^{2} s i n ^{2} θ}{2 g}$

$25 = \frac{( 40 ) ^{2} s i n ^{2} θ}{2 \times 9.8}$

$sin^{2} θ = 0.30625$

$sin θ = 0.5534$

$∴ θ = sin^{- 1} (0.5534) = 33.6 0^{\circ}$

$Horizontal range, R = \frac{u ^{2} sin 2 θ}{g} = \frac{( 40 ) ^{2} \times sin 2 \times 33.6 0 ^{\circ}}{9.8} = \frac{1600 \times sin 67. 2 ^{\circ}}{9.8} = \frac{1600 \times 0.922}{9.8} = 150.53 m$

A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball? Sol. Maximum horizontal distance, $R = 100 m$ The cricketer will only be able to throw the ball to the maximum horizontal distance when the angle of projection is 45°,

$i.e., θ = 4 5^{\circ}$

The horizontal range for a projection velocity v , is given by the relation:

$R = \frac{u ^{2} sin 2 θ}{g} 100 = \frac{u ^{2}}{g} sin 9 0^{\circ} \frac{u ^{2}}{g} = 100$

The ball will achieve the maximum height when it is throws vertically upwards. For such motion, the final velocity v is zero at the maximum height H. Acceleration, $a = - g$ Using the third equation of motion

$v^{2} - u^{2} = - 2 gH H = \frac{1}{2} \times \frac{u ^{2}}{g} = \frac{1}{2} \times 100 = 50 m$

A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone ? Sol. Length of the string $ℓ = 80 cm = 0.8 m$

$Number of revolutions = 14 Time taken = 25 s Frequency, v = \frac{Number of revolutions}{Time taken} = \frac{14}{25} Hz$

Angular frequency, $ω = 2 π ν$

$= 2 \times \frac{22}{7} \times \frac{14}{25} = \frac{88}{25} rad s^{- 1}$

Centripetal acceleration $a_{c} = ω^{2} r$

$= (\frac{88}{25})^{2} \times 0.8 = 9.91 m / s^{2}$

The direction of centripetal acceleration is always directed along the string, towards the centre at all points.

An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity. Sol. Radius of the loop, $r = 1 km = 1000 m$ Speed of the aircraft,

$v = 900 km / h = 900 \times \frac{5}{18} = 250 m / s$

Centripetal acceleration, $a_{c} = \frac{v ^{2}}{r}$

$= \frac{( 250 ) ^{2}}{1000} = 62.5 m / s^{2}$

Acceleration due to gravity, $g = 9.8 m / s^{2}$

$\frac{a _{c}}{g} = \frac{62.5}{9.8} = 6.38 a_{c} = 6.38 g$

Read each statement below carefully and state, with reasons, if it is true or false: (a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre. (b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point. (c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector. Sol. (a) False The net acceleration of a particle in circular motion is not always directed along the radius of the circle toward the centre. It happens only in the case of uniform circular motion. (b) True At a point on a circular path, a particle appears to move tangentially to the circular path. Hence, the velocity vector of the particle is always along the tangent at a point. (c) True In uniform circular motion (UCM), the direction of the acceleration vector points toward the centre of the circle. However, it constantly changes with time. The average of these vectors over one cycle is a null vector.
The position of a particle is given by

$r = 3.0 t \hat{i} - 2.0 t^{2} \hat{j} + 4.0 \hat{k m}$

Where t is in seconds and the coefficients have the proper units for r to be in metres. Find the $v$ and $a$ of the particle? What is the magnitude and direction of velocity of the particle at $t = 2.0 s$ ? Sol. The position of the particle is given by :

$r = 3.0 \hat{i} - 2.0 t^{2} \hat{j} + 4.0 \hat{k}$

Velocity, $v$ of the particle is given as:

$v = \frac{d r}{d t} = \frac{d}{d t} (3.0 t \hat{i} - 2.0 t^{2} \hat{j} + 4.0 \hat{k}) ∴ v = 3.0 \hat{i} - 4.0 t \hat{j}$

Acceleration of the particle is given as:

$a = \frac{d v}{d t} = \frac{d}{d t} (3.0 \hat{i} - 4.0 t \hat{j}) ∴ a = - 4.0 \hat{j}$

We have velocity vector. $v = 3.0 \hat{i} - 4.0 tj$ At $t = 2.0 s : v = 3.0 \hat{i} - 8.0 \hat{j}$ The magnitude of velocity is given by :

$∣ v ∣ = 3^{2} + (- 8)^{2} = 75 = 8.54 m / s Direction, θ = tan^{- 1} (\frac{v _{y}}{v _{x}}) = tan^{- 1} (\frac{- 8}{3}) = - tan^{- 1} (2.667) = - 69.4 5^{\circ}$

The negative sign indicates that the direction of velocity is below the x-axis.

A particle starts from the origin at $t = 0 s$ with a velocity of $10.0 \hat{j}$ and moves in the x-y plane with a constant acceleration of $(8.0 \hat{i} + 2.0 \hat{j}) ms^{- 2}$ . At what time is the x-coordinate of the particle 16 m? What is the y-coordinate of the particle at that time? What is the speed of the particle at that time? Sol. Velocity of the particle, $v = 10.0 \hat{j} m / s$ Acceleration of the particle

$a = (8.0 \hat{i} + 2.0 \hat{j})$

Also,

$But, a = \frac{d v}{d t} = 8.0 \hat{i} + 2.0 \hat{j} d v = (8.0 \hat{i} + 2.0 \hat{j}) d t$

Integrating both sides:

$v (t) = 8.0 t \hat{i} + 2.0 t \hat{j} + u$

Where,

$u = Velocity vector of the particle at t = 0 v = Velocity vector of the particle at time t But = v = \frac{d r}{d t} d r = v d t = (8.0 t \hat{i} + 2.0 t \hat{j} + u) d t$

Integrating the equations with the conditions: at $t = 0; r = 0$ and at $t = t; r = r$

$r = u + \frac{1}{2} 8.0 t^{2} \hat{i} + \frac{1}{2} 2.0 t^{2} \hat{j} = u t + 4.0 t^{2} \hat{i} + t^{2} \hat{j} = (10.0 \hat{j}) t + 4.0 t^{2} \hat{i} + t^{2} \hat{j} x \hat{i} + y \hat{j} = 4.0 t^{2} \hat{i} + (10 t + t^{2}) \hat{j}$

Since the motion of the particle is confined to the x-y plane, on equating the coefficient of $\hat{i}$ and $\hat{j}$ , we get:

$x = 4 t^{2} t = (\frac{x}{4})^{\frac{1}{2}}$

And $y = 10 t + t^{2}$ When $x = 16 m$ :

$t = (\frac{16}{4})^{\frac{1}{2}} = 2 s ∴ y = 10 \times 2 + (2)^{2} = 24 m$

(b) Velocity of the particle is given by:

$v (t) = 8.0 t \hat{i} + 2.0 t \hat{j} + u at t = 2 s v (t) = 8.0 \times 2 \hat{i} + 2.0 \times 2 \hat{j} + 10 \hat{j} = 16 \hat{i} + 14 \hat{j}$

∴ Speed of the particle:

$∣ v ∣ = (16)^{2} + (14)^{2} = 256 + 196 = 452 = 21.26 m / s$

$\hat{i}$ and $\hat{j}$ are unit vectors along $x$ - and $y$ -axis respectively. What is the magnitude and direction of the vectors $\hat{i} + \hat{j}$ and $\hat{i} - \hat{j}$ ? What are the components of a vector $A = 2 \hat{i} + 3 \hat{j}$ along the directions of $\hat{i} + \hat{j}$ and $\hat{i} - \hat{j}$ ? [You may use graphical method] Sol. Consider a vectors $P$

$P = \hat{i} + \hat{j} P_{x} \hat{i} + P_{y} \hat{j} = \hat{i} + \hat{j}$

On comparing the components on both sides, we get:

$P_{x} = P_{y} = 1 ∣ P ∣ = P_{x}^{2} + P_{y}^{2} = 1^{2} + 1^{2} = 2$

Hence, the magnitude of the vector $\hat{i} + \hat{j}$ is $2$ Let $θ$ be the angle made by the vector $P$ , with the x-axis, as shown in the following figure.

exercise-ques-19-(i)-physics-class-11-chap-3

$∴ tan θ = (\frac{P _{y}}{P _{x}})$

$θ = tan^{- 1} (\frac{1}{1}) = 4 5^{\circ}$

Hence, the vector $\hat{i} + \hat{j}$ makes an angle of $4 5^{\circ}$ with the x-axis. Let

$Q = \hat{i} - \hat{j} Q_{x} \hat{i} - Q_{y} \hat{j} = \hat{i} - \hat{j} Q_{x} = Q_{y} = 1 ∣ Q ∣ = Q_{x}^{2} + Q_{y}^{2} = 2$

Hence, the magnitude of the vector $\hat{i} - \hat{j}$ is $2$

Let $θ$ be the angle made by the vector $Q$ , with the x-axis, as shown in the following figure.

physics-exercise-ques-19-(ii)-class-11-chap-3

$∴ tan θ = (\frac{Q _{y}}{Q _{x}}) θ = tan^{- 1} (- \frac{1}{1}) = - 4 5^{\circ}$

Hence, the vector $\hat{i} - \hat{j}$ makes and angle of -45º with the axis.

It is given that : On comparing the coefficients of $\hat{i}$ and $\hat{j}$ , we have

$A = 2 \hat{i} + 3 \hat{j} ∵ ∣ A ∣ = 2^{2} + 3^{2} = 13 A_{x} \hat{i} + A_{y} \hat{j} = 2 \hat{i} + 3 \hat{j}$

Let $A$ make an angle $θ$ with the x-axis, as shown in the following figure.

exercise-ques-19-(iii)-class-11-chap-3-physics

$∴ tan θ = (\frac{A _{x}}{A _{y}}) θ = tan^{- 1} (\frac{3}{2}) = tan^{- 1} (1.5) = 56.3 1^{\circ}$

Angle between the vectors $(2 \hat{i} + 3 \hat{j})$ and $(\hat{i} + \hat{j}), θ = 56.3 1^{o} - 4 5^{o} = 11.3 1^{o}$ $A$ , along the direction of $P$ making an angle $θ^{'}$ .

$= (A cos θ) P = (A cos 11.3 1^{\circ}) \frac{( i ^ + j ^ )}{2} = 13 \times \frac{0.9806}{2} (\hat{i} + \hat{j}) = 2.5 (\hat{i} + \hat{j}) = \frac{25}{10} \times 2 = \frac{5}{2} \dots\dots (v)$

Let $θ^{'}$ be the angle between the vectors $(2 \hat{i} + 3 \hat{j})$ and $(\hat{i} - \hat{j})$

$θ^{''} = 4 5^{\circ} + 56.3 1^{\circ} = 101.3 1^{\circ}$

Component of vector $A$ , along the direction of $Q$ , making an angle $θ^{''}$

$= (A cos θ) Q = (A cos θ) \frac{i ^ - j ^}{2} = 13 cos (101.3 1^{\circ}) \frac{( i ^ - j ^ )}{2} = - \frac{13}{2} sin 11.3 1^{\circ} (\hat{i} - \hat{j}) = - 2.550 \times 0.9161 (\hat{i} - \hat{j}) = - 0.5 (\hat{i} - \hat{j}) = - \frac{5}{10} \times 2 = - \frac{1}{2} \dots\dots (iv)$

For any arbitrary motion in space, which of the following relations are true: (a) $v_{average} = (\frac{1}{2}) [v (t_{1}) + v (t_{2})]$ (b) $v_{average} = \frac{[ r ( t _{2} ) - r ( t _{1} ) ]}{( t _{2} - t _{1} )}$ (c) $v (t) = v (0) + a t$ (d) $r (t) = r (0) + v (0) t + (\frac{1}{2}) a t^{2}$ (e) $v_{average} = \frac{[ v ( t _{2} ) - v ( t _{1} ) ]}{( t _{2} - t _{1} )}$

(The 'average' stands for average of the quantity over the time interval $t_{1}$ to $t_{2}$ )

Sol. (b) and (e)

(a) It is given that the motion of the particle is arbitrary. Therefore, the average velocity of the particle cannot be given by this equation. (b) The arbitrary motion of the particle can be represented by this equation. (c) The motion of the particle is arbitrary. The acceleration of the particle may also be non-uniform. Hence, this equation cannot represent the motion of the particle e in space. (d) The motion of the particle is arbitrary; acceleration of the particle may also be non-uniform. Hence, this equation cannot represent the motion of particle in space. (e) The arbitrary motion of the particle can be represented by this equation.

Read each statement below carefully and state, with reasons and examples, if it is true or false : A scalar quantity is one that (a) is conserved in a process (b) can never take negative values (c) must be dimensionless (d) does not vary from one point to another in space (e) has the same value for observers with different orientations of axes.

Sol. (a) False : Despite being a scalar quantity, energy is not conserved in inelastic collisions.

(b) False : Despite being a scalar quantity, temperature can take negative values. (c) False : Total path length is a scalar quantity. Yet it has the dimension of length. (d) False : A scalar quantity such as gravitational potential can vary from one point to another in space. (e) True : The value of a scalar does not vary for observers with different orientations of axes.

An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s apart is 30°, what is the speed of the aircraft ?

Sol. The positions of the observer and the aircraft are shown in the given figure.

20-exercise-ques-physics-chap-3-class-11

Height of the aircraft from ground $OR = 3400 m$ Angle subtended between the positions, $∠ POQ = 3 0^{\circ}$ Time $= 10 s$ In $△ PRO$ : $tan 1 5^{\circ} = \frac{PR}{OR}$ $PR = OR tan 1 5^{\circ} = 3400 \times tan 1 5^{\circ}$ $Δ PRO$ is similar to $Δ RQO$

$∴ PR = RQ PQ = PR + RQ = 2 PR = 2 \times 3400 tan 1 5^{\circ} = 6800 \times 0.268 = 1822.4 m$

$∴ Speed of the aircraft = \frac{1822.4}{10} = 182.24 m / s$

3.0NCERT Class 11 Physics – Chapter-Wise Solutions & Explanations

Get chapter-wise NCERT Solutions for Class 11 Physics with easy-to-understand explanations for every question. Understand textbook concepts, follow step-by-step solutions, and strengthen your Physics fundamentals across all chapters.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Units and Measurements
Chapter 2	Motion in a Straight Line
Chapter 4	Laws of Motion
Chapter 5	Work, Energy and Power
Chapter 6	System of Particles and Rotational Motion
Chapter 7	Gravitation
Chapter 8	Mechanical Properties of Solids
Chapter 9	Mechanical Properties of Fluids
Chapter 10	Thermal Properties of Matter
Chapter 11	Thermodynamics
Chapter 12	Kinetic Theory
Chapter 13	Oscillations
Chapter 14	Waves

4.0Key Features of NCERT Solutions for Class 11 Physics Chapter 3

Clear Understanding of Vectors: Learn vector addition, subtraction, resultant vectors, and vector components through simple explanations and step-by-step solutions.
Projectile Motion Explained: Understand the horizontal and vertical components of projectile motion and the effect of gravity on vertical motion.
Circular Motion Concepts: Understand the relationship between linear velocity and angular velocity using v=rωv = r\omega.
Relative Velocity and River-Crossing Problems: Learn to apply relative velocity to river-crossing problems and understand the conditions for the shortest path and shortest time.
Prepared by ALLEN Subject Experts: Get accurate and detailed NCERT Solutions for Class 11 Physics Chapter 3, prepared according to the latest NCERT syllabus.
Simple and Concept-Based Explanations: Understand Motion in a Plane, vectors, projectile motion, and circular motion through clear language, examples, and step-by-step methods.
Complete NCERT Exercise Coverage: Get detailed solutions to NCERT Class 11 Physics Chapter 3 examples and exercise questions, covering vectors and two-dimensional motion.

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