Chapter: Motion in a Plane

1. Introduction

Definition: When an object changes its position with respect to time in a two-dimensional space (a plane), its motion is called Motion in a Plane or 2D Motion.
Examples: A projectile thrown into the air, a car driving around a circular track, a boat crossing a river.
To describe motion in a plane, we use vectors because direction becomes essential when dealing with multiple dimensions.

2. Scalars and Vectors

Scalar Quantities: Quantities that have only magnitude but no direction.
Examples: Mass, temperature, distance, speed, time, work.
Vector Quantities: Quantities that have both magnitude and a specific direction, and obey the laws of vector addition.
Examples: Displacement, velocity, acceleration, force, momentum.

Representation of a Vector

Geometrically, a vector is represented by a straight line with an arrowhead.
- The length of the line represents the magnitude.
- The arrowhead indicates the direction.
Symbolically, a vector is written with an arrow over its symbol (e.g., $\vec{A}$) or in bold type (A).

Important Types of Vectors

Position Vector ($\vec{r}$): A vector that specifies the position of a particle relative to an origin. For a point $P(x, y)$ in a plane:
$$\vec{r} = x\hat{i} + y\hat{j}$$
Displacement Vector ($\Delta\vec{r}$): The straight-line vector joining the initial position to the final position of a moving particle.
$$\Delta\vec{r} = \vec{r}_2 - \vec{r}_1 = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j}$$
Unit Vector: A vector of unit magnitude pointing in a specific direction. It is used to specify direction.
$$\hat{A} = \frac{\vec{A}}{|\vec{A}|}$$
where $\hat{i}$, $\hat{j}$, and $\hat{k}$ are unit vectors along the positive x, y, and z-axes respectively.
Equal Vectors: Vectors having the same magnitude and direction.
Null (Zero) Vector ($\vec{0}$): A vector with zero magnitude and an arbitrary direction.

3. Vector Addition and Subtraction

A. Triangle Law of Vector Addition

If two vectors are represented in magnitude and direction by the two sides of a triangle taken in the same order, then their resultant is represented in magnitude and direction by the third side of the triangle taken in the opposite order.

B. Parallelogram Law of Vector Addition

If two vectors acting simultaneously at a point are represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from a point, then their resultant is represented by the diagonal of the parallelogram passing through that same point.

Magnitude of the Resultant Vector ($R$):
$$R = |\vec{A} + \vec{B}| = \sqrt{A^2 + B^2 + 2AB\cos\theta}$$
(where $\theta$ is the angle between vectors $\vec{A}$ and $\vec{B}$ soul-to-tail)
Direction of the Resultant ($\alpha$ with vector $\vec{A}$):
$$\tan\alpha = \frac{B\sin\theta}{A + B\cos\theta}$$

C. Vector Subtraction

Subtracting a vector $\vec{B}$ from $\vec{A}$ is equivalent to adding the negative of $\vec{B}$ to $\vec{A}$:

$$\vec{A} - \vec{B} = \vec{A} + (-\vec{B})$$

Magnitude: $$R = \sqrt{A^2 + B^2 - 2AB\cos\theta}$$

4. Resolution of Vectors (Components of a Vector)

Splitting a single vector into two or more vectors along specific directions is called the resolution of a vector.

In a 2D Cartesian plane, a vector $\vec{A}$ making an angle $\theta$ with the x-axis can be resolved into:

Horizontal Component ($A_x$): $A_x = A\cos\theta$
Vertical Component ($A_y$): $A_y = A\sin\theta$

Vector form: $$\vec{A} = A_x\hat{i} + A_y\hat{j} = (A\cos\theta)\hat{i} + (A\sin\theta)\hat{j}$$

Magnitude: $$A = \sqrt{A_x^2 + A_y^2}$$
Direction: $$\tan\theta = \frac{A_y}{A_x}$$

5. Kinematics in a Plane (Position, Velocity, and Acceleration)

Any 2D motion can be treated as two separate, simultaneous 1D motions along mutually perpendicular axes (x and y).

Parameter	Vector Form	X-component	Y-component
Position	$\vec{r} = x\hat{i} + y\hat{j}$	$x$	$y$
Average Velocity	$\vec{v}_{avg} = \frac{\Delta\vec{r}}{\Delta t}$	$v_{x, avg} = \frac{\Delta x}{\Delta t}$	$v_{y, avg} = \frac{\Delta y}{\Delta t}$
Instantaneous Velocity	$\vec{v} = \frac{d\vec{r}}{dt} = v_x\hat{i} + v_y\hat{j}$	$v_x = \frac{dx}{dt}$	$v_y = \frac{dy}{dt}$
Instantaneous Acceleration	$\vec{a} = \frac{d\vec{v}}{dt} = a_x\hat{i} + a_y\hat{j}$	$a_x = \frac{dv_x}{dt} = \frac{d^2x}{dt^2}$	$a_y = \frac{dv_y}{dt} = \frac{d^2y}{dt^2}$

Equations of Motion for Uniform Acceleration in a Plane

$\vec{v} = \vec{u} + \vec{a}t \quad \implies \quad (v_x = u_x + a_xt \quad \text{and} \quad v_y = u_y + a_yt)$
$\vec{r} = \vec{u}t + \frac{1}{2}\vec{a}t^2 \quad \implies \quad (x = u_xt + \frac{1}{2}a_xt^2 \quad \text{and} \quad y = u_yt + \frac{1}{2}a_yt^2)$

6. Projectile Motion

When an object is thrown into space near the Earth's surface and moves along a curved path under the sole influence of gravity, it is called a projectile. Its path is called a trajectory.

Assumptions:

Air resistance is negligible.
Acceleration due to gravity ($g$) remains constant.

Analysis of Motion:

Let a body be projected with an initial velocity $u$ at an angle $\theta$ with the horizontal.

Horizontal component of velocity ($u_x$): $u\cos\theta$ (Remains constant throughout the flight because $a_x = 0$)
Vertical component of velocity ($u_y$): $u\sin\theta$ (Changes continuously because $a_y = -g$)

Key Formulas for Projectile Motion

Equation of Trajectory (Path of Projectile):
$$y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}$$
(Since this equation is quadratic in $x$, the path of a projectile is a parabola).
Time of Flight ($T$): Total time the projectile remains in the air.
$$T = \frac{2u\sin\theta}{g}$$
Maximum Height ($H$): The highest vertical point reached by the projectile.
$$H = \frac{u^2\sin^2\theta}{2g}$$
Horizontal Range ($R$): The total horizontal distance covered during the time of flight.
$$R = \frac{u^2\sin(2\theta)}{g}$$
- Maximum Horizontal Range ($R_{max}$): Occurs when $\theta = 45^\circ \implies R_{max} = \frac{u^2}{g}$.
- Note: For two complementary angles of projection ($\theta$ and $90^\circ - \theta$), the horizontal range $R$ is identical.

7. Uniform Circular Motion

When an object moves in a circular path with a constant speed, its motion is called Uniform Circular Motion.

Even though speed is constant, the velocity is constantly changing because its direction changes at every point. Hence, uniform circular motion is an accelerated motion.

Key Terms & Relations:

Angular Displacement ($\Delta\theta$): The angle swept out by the radius vector. Measured in radians ($rad$).
Angular Velocity ($\omega$): The rate of change of angular displacement.
$$\omega = \frac{d\theta}{dt} = \frac{2\pi}{T} = 2\pi\nu$$
(where $T$ = time period, $\nu$ = frequency)
Relation between Linear Velocity ($v$) and Angular Velocity ($\omega$):
$$v = \omega r$$
Centripetal Acceleration ($a_c$): The acceleration directed towards the center of the circular path.
$$a_c = \frac{v^2}{r} = \omega^2 r$$
Centripetal Force ($F_c$): The net force required to keep a body moving in a circular path.
$$F_c = \frac{mv^2}{r} = m\omega^2 r$$

8. Relative Velocity in Two Dimensions

If two objects $A$ and $B$ are moving with velocities $\vec{v}_A$ and $\vec{v}_B$ relative to the ground, the relative velocity of $A$ with respect to $B$ is given by:

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

Common Applications:

Rain-Umbrella Problem: If rain is falling vertically downward with velocity $\vec{v}_r$ and a person is walking horizontally with velocity $\vec{v}_m$, the relative velocity of rain with respect to the person is $\vec{v}_{rm} = \vec{v}_r - \vec{v}_m$. To protect themselves, the person must hold the umbrella at an angle $\theta$ with the vertical:
$$\tan\theta = \frac{v_m}{v_r}$$

        💡 Quick Facts:
        At the highest point of projectile motion, vertical velocity is zero ($v_y = 0$), but horizontal velocity is $u\cos\theta$ and acceleration is $g$ (downwards). The angle between the velocity and acceleration vectors at the highest point is exactly $90^\circ$.
Vector addition is commutative ($\vec{A} + \vec{B} = \vec{B} + \vec{A}$) and associative ($\vec{A} + (\vec{B} + \vec{C}) = (\vec{A} + \vec{B}) + \vec{C}$). Vector subtraction is neither.
In uniform circular motion, kinetic energy and speed remain constant, while velocity, momentum, and acceleration vectors change continuously.

    

Here are the NCERT textbook exercise solutions for Class 11 Physics Chapter 3 (traditionally Chapter 4, titled **Motion in a Plane**) formatted directly in HTML with structural KaTeX notation for easy copying, embedding, or previewing. ```html

NCERT Exercise Solutions: Motion in a Plane

Question 1: State, for each of the following physical quantities, if it is a scalar or a vector quantity:
Volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular momentum.

Answer:

Scalar Quantities: Volume, mass, speed, density, number of moles, angular frequency.
Reason: These quantities possess only magnitude and no specific direction. Even though angular frequency involves rotational behavior, it is treated as a scalar since it doesn’t follow the laws of vector addition.
Vector Quantities: Acceleration, velocity, displacement, angular momentum.
Reason: These quantities possess both magnitude and a defined direction, and they obey the rules of vector addition.

Question 2: Pick out the two scalar quantities from the following list:
force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity.

Answer:

The two scalar quantities are work and current.

Work: It is defined as the dot product of force and displacement ($\vec{F} \cdot \vec{d}$), which inherently yields a scalar value.
Current: Although electric current has a specified direction along a wire, it does not obey the laws of vector addition (currents at a junction add algebraically according to Kirchhoff's Current Law). Hence, it is a scalar quantity.

Question 3: Pick out the only vector quantity from the following list:
Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge.

Answer:

The only vector quantity is impulse.

Reason: Impulse is defined as the product of a large force and the short time interval for which it acts ($\vec{I} = \vec{F} \cdot \Delta t$). Since force is a vector, impulse points in the same direction as the applied force.

Question 4: State with reasons, whether the following algebraic operations with scalars and vectors are permissible:
(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.

Answer:

(a) Not permissible: We can only add two scalars if they represent the same physical quantity (e.g., you cannot add mass to temperature).
(b) Not permissible: A scalar cannot be added to a vector because a scalar lacks spatial direction.
(c) Permissible: Multiplying a vector by a scalar changes its magnitude (and unit, if the scalar has a unit), which is perfectly valid (e.g., $\vec{F} = m\vec{a}$).
(d) Permissible: Any two scalars can be multiplied regardless of their physical dimensions (e.g., $\text{mass} \times \text{acceleration} = \text{force}$).
(e) Not permissible: Two vectors can only be added if they represent the same physical quantity (e.g., you cannot add a velocity vector to a force vector).
(f) Not permissible: A component of a vector is a scalar quantity (or acts as one in absolute magnitude); adding a scalar component directly to its parental spatial vector is structurally invalid.

Question 5: 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.

Answer:

(a) True: The magnitude of a vector specifies its absolute size or length, which is a pure number carrying units but no spatial direction.
(b) False: While rectangular algebraic components ($A_x$, $A_y$) are treated as scalars, the actual component vectors ($A_x\hat{i}$, $A_y\hat{j}$) retain vector properties.
(c) False: Total path length is only equal to the magnitude of displacement if the particle moves in a straight line without reversing direction. For any curved path, path length is strictly greater than displacement.
(d) True: Since total path length $\ge$ magnitude of displacement, dividing both sides by the time interval $\Delta t$ ensures that $\text{average speed} \ge |\vec{v}_{\text{avg}}|$.
(e) True: To yield a null vector, the resultant of any two vectors must be equal and opposite to the third vector. This is only possible if all three vectors reside within the same geometric plane.

Question 6: Establish the following vector inequalities geometrically or otherwise:
(a) $|\vec{A}+\vec{B}| \le |\vec{A}| + |\vec{B}|$
(b) $|\vec{A}+\vec{B}| \ge ||\vec{A}| - |\vec{B}||$
(c) $|\vec{A}-\vec{B}| \le |\vec{A}| + |\vec{B}|$
(d) $|\vec{A}-\vec{B}| \ge ||\vec{A}| - |\vec{B}||$

Answer:

Let vectors $\vec{A}$ and $\vec{B}$ be represented by the sides $\vec{OP}$ and $\vec{PQ}$ of a triangle $OPQ$. According to the triangle law of vector addition, the side $\vec{OQ}$ represents the resultant vector $(\vec{A} + \vec{B})$.

(a) Proof for $|\vec{A}+\vec{B}| \le |\vec{A}| + |\vec{B}|$:

In any triangle $OPQ$, the length of any single side is always strictly less than the sum of the lengths of the other two sides:

$$OQ < OP + PQ \implies |\vec{A} + \vec{B}| < |\vec{A}| + |\vec{B}|$$

If the vectors are collinear and act in the same direction, then:

$$|\vec{A} + \vec{B}| = |\vec{A}| + |\vec{B}|$$

Combining both cases yields the requested inequality: **$|\vec{A}+\vec{B}| \le |\vec{A}| + |\vec{B}|$**.

(b) Proof for $|\vec{A}+\vec{B}| \ge ||\vec{A}| - |\vec{B}||$:

In triangle $OPQ$, the length of any single side is always greater than the absolute difference between the lengths of the remaining two sides:

$$OQ > |OP - PQ| \implies |\vec{A} + \vec{B}| > ||\vec{A}| - |\vec{B}||$$

If the vectors are collinear but act in opposite directions, equality holds. Thus: **$|\vec{A}+\vec{B}| \ge ||\vec{A}| - |\vec{B}||$**.

Note: Identities (c) and (d) can be evaluated identically by substituting $\vec{B}$ with $-\vec{B}$, noting that $|-\vec{B}| = |\vec{B}|$.

Question 7: Given that $\vec{A} + \vec{B} + \vec{C} + \vec{D} = 0$, which of the following statements are correct:
(a) $\vec{A}, \vec{B}, \vec{C},$ and $\vec{D}$ must each be a null vector,
(b) The magnitude of $(\vec{A} + \vec{C})$ equals the magnitude of $(\vec{B} + \vec{D})$,
(c) The magnitude of $\vec{A}$ can never be greater than the sum of magnitudes of $\vec{B}, \vec{C},$ and $\vec{D}$,
(d) $\vec{B} + \vec{C}$ must lie in the plane of $\vec{A}$ and $\vec{D}$ if $\vec{A}$ and $\vec{D}$ are not collinear, and along the line of $\vec{A}$ and $\vec{D}$ if they are collinear.

Answer:

(a) Incorrect: They do not have to be null vectors individually. There are infinite combinations of non-zero vectors whose vector sum can cancel out to zero.
(b) Correct: Given $\vec{A} + \vec{B} + \vec{C} + \vec{D} = 0$, rewriting the term gives:
$$\vec{A} + \vec{C} = -(\vec{B} + \vec{D})$$
Taking the magnitude on both sides: $|\vec{A} + \vec{C}| = |-(\vec{B} + \vec{D})| = |\vec{B} + \vec{D}|$.
(c) Correct: From the structural equation: $\vec{A} = -(\vec{B} + \vec{C} + \vec{D})$. By the triangle inequality extended to multiple vectors, the magnitude of any single side cannot exceed the combined sum of individual magnitudes of the remaining components:
$$|\vec{A}| \le |\vec{B}| + |\vec{C}| + |\vec{D}|$$
(d) Correct: Rearranging the expression gives $\vec{B} + \vec{C} = -(\vec{A} + \vec{D})$. Since the sum of $(\vec{A} + \vec{D})$ naturally sits in the plane containing $\vec{A}$ and $\vec{D}$, the vector $(\vec{B} + \vec{C})$ must lie within that exact same plane to successfully cancel it out.

Question 8: Three girls skating on a circular ice ground of radius $200\text{ m}$ 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 Fig. 4.20. What is the magnitude of the displacement vector for each? For which girl is this equal to the actual length of path skated?

Answer:

Displacement depends only on the initial position ($P$) and the final position ($Q$), independent of the path taken.

Since $P$ and $Q$ are diametrically opposite points on a circle of radius $r = 200\text{ m}$:

$$\text{Magnitude of Displacement} = \text{Diameter of Circle} = 2r = 2 \times 200 = \mathbf{400\text{ m}}$$

The magnitude of displacement for all three girls is $400\text{ m}$.

For Girl B, who skates directly along the straight diameter path $PQ$, the displacement magnitude is exactly equal to the actual path length skated.

Question 9: A cyclist starts from the centre $O$ of a circular park of radius $1\text{ km}$, reaches the edge $P$ of the park, then cycles along the circumference, and returns to the centre along $QO$ as shown in Fig. 4.21. If the round trip takes $10\text{ min}$, what is the (a) net displacement, (b) average velocity, and (c) average speed of the cyclist?

Answer:

Given: Radius ($r$) = $1\text{ km}$, Total Time ($t$) = $10\text{ min} = \frac{10}{60}\text{ h} = \frac{1}{6}\text{ h}$.

(a) Net Displacement:

Since the cyclist starts at the center $O$ and eventually returns to the exact starting point $O$, the net displacement is $0\text{ km}$.

(b) Average Velocity:

$$\text{Average Velocity} = \frac{\text{Net Displacement}}{\text{Total Time}} = \frac{0}{\frac{1}{6}} = \mathbf{0\text{ km/h}}$$

(c) Average Speed:

Total distance traveled = Path $OP$ + Arc $PQ$ + Path $QO$

$$\text{Total Distance} = r + \left(\frac{2\pi r}{4}\right) + r = 1 + \frac{\pi(1)}{2} + 1 = 2 + 1.57 = 3.57\text{ km}$$

$$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{3.57\text{ km}}{\frac{1}{6}\text{ h}} = 3.57 \times 6 = \mathbf{21.42\text{ km/h}}$$

Question 10: On an open ground, a motorist follows a track that turns to his left by an angle of $60^\circ$ after every $500\text{ m}$. Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of displacement with the total path length covered by the motorist in each case.

Answer:

A turn of $60^\circ$ after every equal segment implies the motorist is traveling along a regular hexagonal path. Let the side length $s = 500\text{ m}$.

1. At the Third Turn:

The motorist covers 3 sides of the hexagon. Total Path Length = $3 \times 500 = \mathbf{1500\text{ m}}$.
Displacement corresponds to the main diagonal connecting opposite vertices of a regular hexagon, which is equal to $2s$:
$\text{Magnitude of Displacement} = 2 \times 500 = \mathbf{1000\text{ m}}$.
Comparison Ratio: $\frac{\text{Displacement}}{\text{Path Length}} = \frac{1000}{1500} = \mathbf{\frac{2}{3}}$.

2. At the Sixth Turn:

The motorist completes a full lap around the hexagon. Total Path Length = $6 \times 500 = \mathbf{3000\text{ m}}$.
Since the final position coincides with the initial position, the **Magnitude of Displacement = $0\text{ m}$**.
Comparison Ratio: $\frac{0}{3000} = \mathbf{0}$.

3. At the Eighth Turn:

The motorist completes 1 full lap (6 turns) plus 2 more sides. Total Path Length = $8 \times 500 = \mathbf{4000\text{ m}}$.
The displacement is equivalent to the distance across 2 sides of a regular hexagon, forming an isosceles triangle with an interior angle of $120^\circ$:
$$\text{Displacement} = \sqrt{s^2 + s^2 + 2s^2\cos(60^\circ)} = s\sqrt{3} = 500 \times 1.732 = \mathbf{866\text{ m}}$$
Comparison Ratio: $\frac{\text{Displacement}}{\text{Path Length}} = \frac{866}{4000} \approx \mathbf{0.22}$.

Question 11: A passenger arriving in a new town wishes to go from the station to a hotel located $10 \text{ km}$ away on a straight road from the station. A dishonest cabman takes him along a circuitous path $23 \text{ km}$ long and reaches the hotel in $28 \text{ min}$. What is (a) the average speed of the taxi and (b) the magnitude of average velocity? Are the two equal?

Answer:

Given parameters:

Displacement magnitude ($|\Delta \vec{r}|$) = $10 \text{ km}$ (straight distance)
Total path length ($s$) = $23 \text{ km}$
Total time ($t$) = $28 \text{ min} = \frac{28}{60} \text{ h} = \frac{7}{15} \text{ h}$

(a) Average Speed:

$$\text{Average Speed} = \frac{\text{Total Path Length}}{\text{Total Time}} = \frac{23}{\frac{7}{15}} = \frac{23 \times 15}{7} = \frac{345}{7} \approx \mathbf{49.29 \text{ km/h}}$$

(b) Magnitude of Average Velocity:

$$|\vec{v}_{\text{avg}}| = \frac{\text{Magnitude of Displacement}}{\text{Total Time}} = \frac{10}{\frac{7}{15}} = \frac{150}{7} \approx \mathbf{21.43 \text{ km/h}}$$

Conclusion: No, the average speed and the magnitude of average velocity are not equal because the taxi travels along a curved path rather than a straight line.

Question 12: Rain is falling vertically with a speed of $30 \text{ m/s}$. A woman rides a bicycle with a speed of $10 \text{ m/s}$ in the north to south direction. What is the direction in which she should hold her umbrella?

Answer:

Let the vertically downward direction be along $-\hat{j}$ and the North-to-South direction be along $-\hat{i}$.

Velocity of rain: $\vec{v}_r = -30\hat{j} \text{ m/s}$
Velocity of the bicycle rider: $\vec{v}_w = -10\hat{i} \text{ m/s}$

The relative velocity of rain with respect to the woman ($\vec{v}_{rw}$) is:

$$\vec{v}_{rw} = \vec{v}_r - \vec{v}_w = -30\hat{j} - (-10\hat{i}) = 10\hat{i} - 30\hat{j}$$

Let $\theta$ be the angle that $\vec{v}_{rw}$ makes with the vertical downward line:

$$\tan\theta = \frac{|v_{rw, x}|}{|v_{rw, y}|} = \frac{10}{30} = \frac{1}{3}$$

$$\theta = \tan^{-1}\left(\frac{1}{3}\right) \approx \mathbf{18.43^\circ}$$

Conclusion: The woman should hold her umbrella at an angle of roughly $18.43^\circ$ leaning towards the South from the vertical line.

Question 13: A man can swim with a speed of $4.0 \text{ km/h}$ in still water. How long does he take to cross a river $1.0 \text{ km}$ wide if the river flows steadily at $3.0 \text{ km/h}$ and he makes his strokes normal to the river current? How far down the river does he go when he reaches the other bank?

Answer:

Given:

Speed of swimmer in still water ($v_m$) = $4.0 \text{ km/h}$
Width of the river ($d$) = $1.0 \text{ km}$
Speed of river flow ($v_r$) = $3.0 \text{ km/h}$

1. Time taken to cross the river ($t$):

Since his strokes are always normal (perpendicular) to the river current, only his swimming velocity handles the vertical crossing component:

$$t = \frac{\text{Width of river}}{\text{Speed perpendicular to bank}} = \frac{d}{v_m} = \frac{1.0 \text{ km}}{4.0 \text{ km/h}} = 0.25 \text{ h} = \mathbf{15 \text{ min}}$$

2. Distance drifted downstream ($x$):

The drift is caused entirely by the horizontal river current over that exact duration:

$$x = v_r \times t = 3.0 \text{ km/h} \times 0.25 \text{ h} = \mathbf{0.75 \text{ km}} = \mathbf{750 \text{ m}}$$

Question 14: In a harbour, wind is blowing at the speed of $72 \text{ km/h}$ and the flag on the mast of a boat anchored in the harbour flutters along the N-E direction. If the boat starts moving at a speed of $51 \text{ km/h}$ to the north, what is the direction of the flag on the mast of the boat?

Answer:

When the boat is static, the flag flutters purely along the direction of the wind vector ($\vec{v}_w$). When the boat moves with velocity $\vec{v}_b$, the flag flutters along the relative direction of the wind with respect to the moving boat ($\vec{v}_{wb}$):

$$\vec{v}_{wb} = \vec{v}_w - \vec{v}_b = \vec{v}_w + (-\vec{v}_b)$$

Magnitude of wind velocity $v_w = 72 \text{ km/h}$ at an angle of $45^\circ$ east of north.
Magnitude of boat velocity $v_b = 51 \text{ km/h}$ pointing due north. Therefore, $-\vec{v}_b$ points due south ($180^\circ$).

The angle between $\vec{v}_w$ and $-\vec{v}_b$ is $\theta = 180^\circ - 45^\circ = 135^\circ$. Let $\alpha$ be the angle of the resultant relative vector $\vec{v}_{wb}$ measured from the wind vector $\vec{v}_w$:

$$\tan\alpha = \frac{v_b\sin(135^\circ)}{v_w + v_b\cos(135^\circ)} = \frac{51 \times \frac{1}{\sqrt{2}}}{72 + 51 \times \left(-\frac{1}{\sqrt{2}}\right)} = \frac{36.06}{72 - 36.06} = \frac{36.06}{35.94} \approx 1$$

$$\alpha \approx 45^\circ$$

Since the wind was at $45^\circ$ East of North, shifting $45^\circ$ clockwise from that vector points the resultant exactly due **East**.

Question 15: The ceiling of a long hall is $25 \text{ m}$ high. What is the maximum horizontal distance that a ball thrown with a speed of $40 \text{ m/s}$ can go without hitting the ceiling of the hall?

Answer:

Given constraints:

Maximum allowable height ($H$) = $25 \text{ m}$
Initial speed ($u$) = $40 \text{ m/s}$
Acceleration due to gravity ($g$) = $9.8 \text{ m/s}^2$

Using the maximum height formula for a projectile:

$$H = \frac{u^2\sin^2\theta}{2g} \implies 25 = \frac{(40)^2\sin^2\theta}{2 \times 9.8}$$

$$\sin^2\theta = \frac{25 \times 19.6}{1600} = \frac{490}{1600} = 0.30625 \implies \sin\theta \approx 0.5534$$

From this, we evaluate $\cos^2\theta = 1 - \sin^2\theta = 1 - 0.30625 = 0.69375$. Now we compute the horizontal range $R$:

$$R = \frac{u^2 \sin(2\theta)}{g} = \frac{2u^2\sin\theta\cos\theta}{g}$$

$$R = \frac{2 \times 1600 \times 0.5534 \times \sqrt{0.69375}}{9.8} = \frac{3200 \times 0.5534 \times 0.8329}{9.8} \approx \mathbf{150.5 \text{ m}}$$

Question 16: A cricketer can throw a ball to a maximum horizontal distance of $100 \text{ m}$. How much high above the ground can the cricketer throw the same ball?

Answer:

The maximum horizontal range $R_{\text{max}}$ occurs at a launch angle of $\theta = 45^\circ$:

$$R_{\text{max}} = \frac{u^2}{g} = 100 \text{ m} \implies u^2 = 100g$$

To throw the ball to the maximum possible height ($H_{\text{max}}$), the cricketer must throw it straight up vertically ($\theta = 90^\circ$):

$$H_{\text{max}} = \frac{u^2}{2g} = \frac{100g}{2g} = \mathbf{50 \text{ m}}$$

Question 17: A stone tied to the end of a string $80 \text{ cm}$ long is whirled in a horizontal circle with a constant speed. If the stone makes $14$ revolutions in $25 \text{ s}$, what is the magnitude and direction of acceleration of the stone?

Answer:

Given properties:

Radius ($r$) = $80 \text{ cm} = 0.8 \text{ m}$
Frequency ($\nu$) = $\frac{\text{Revolutions}}{\text{Time}} = \frac{14}{25} \text{ Hz}$

Angular velocity ($\omega$) is given by:

$$\omega = 2\pi\nu = 2 \times \frac{22}{7} \times \frac{14}{25} = \frac{88}{25} = 3.52 \text{ rad/s}$$

For uniform circular motion, the acceleration is purely centripetal:

$$a_c = \omega^2 r = (3.52)^2 \times 0.8 = 12.39 \times 0.8 \approx \mathbf{9.91 \text{ m/s}^2}$$

Direction: The acceleration vector points radially inward towards the center of the circular orbit at every instant.

Question 18: An aircraft executes a horizontal loop of radius $1.00 \text{ km}$ with a steady speed of $900 \text{ km/h}$. Compare its centripetal acceleration with the acceleration due to gravity.

Answer:

Convert inputs to SI units:

Radius ($r$) = $1.00 \text{ km} = 1000 \text{ m}$
Speed ($v$) = $900 \text{ km/h} = 900 \times \frac{5}{18} = 250 \text{ m/s}$

Calculate centripetal acceleration ($a_c$):

$$a_c = \frac{v^2}{r} = \frac{(250)^2}{1000} = \frac{62500}{1000} = 62.5 \text{ m/s}^2$$

Comparing this value to gravity ($g = 9.8 \text{ m/s}^2$):

$$\frac{a_c}{g} = \frac{62.5}{9.8} \approx \mathbf{6.38}$$

Thus, the aircraft's centripetal acceleration is roughly $6.38$ times greater than the acceleration due to gravity.

Question 19: 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.

Answer:

(a) False: This holds true only for uniform circular motion. In non-uniform circular motion, there is an additional tangential component of acceleration ($a_t = \frac{dv}{dt}$), causing the net acceleration vector to slant away from the center.
(b) True: By definition, velocity represents the instantaneous direction of displacement. As the change in time approaches zero ($\Delta t \to 0$), the chord line merges into a tangent line at that specific point.
(c) True: In uniform circular motion, the acceleration vector has a constant magnitude but constantly rotates its direction uniformly over time. For every vector pointing inward at one instant, there is an equal and opposite vector pointing inward exactly half a cycle later, canceling out to zero over a full orbit.

Question 20: The position of a particle is given by $\vec{r} = 3.0t\hat{i} - 2.0t^2\hat{j} + 4.0\hat{k} \text{ m}$ where $t$ is in seconds. (a) Find the $\vec{v}$ and $\vec{a}$ of the particle. (b) What is the magnitude and direction of velocity of the particle at $t = 2.0 \text{ s}$?

Answer:

(a) Velocity ($\vec{v}$) and Acceleration ($\vec{a}$):

$$\vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{d}{dt}(3.0t)\hat{i} - \frac{d}{dt}(2.0t^2)\hat{j} + \frac{d}{dt}(4.0)\hat{k} = \mathbf{3.0\hat{i} - 4.0t\hat{j} \text{ m/s}}$$

$$\vec{a}(t) = \frac{d\vec{v}}{dt} = \frac{d}{dt}(3.0)\hat{i} - \frac{d}{dt}(4.0t)\hat{j} = \mathbf{-4.0\hat{j} \text{ m/s}^2}$$

(b) Velocity at $t = 2.0 \text{ s}$:

$$\vec{v}(2.0) = 3.0\hat{i} - 4.0(2.0)\hat{j} = 3.0\hat{i} - 8.0\hat{j} \text{ m/s}$$

Magnitude of velocity:

$$v = \sqrt{(3.0)^2 + (-8.0)^2} = \sqrt{9 + 64} = \sqrt{73} \approx \mathbf{8.54 \text{ m/s}}$$

Direction ($\theta$ with respect to the x-axis):

$$\tan\theta = \frac{v_y}{v_x} = \frac{-8.0}{3.0} = -2.667 \implies \theta = \tan^{-1}(-2.667) \approx \mathbf{-69.4^\circ}$$

This shows the velocity points at an angle of $69.4^\circ$ below the positive x-axis.

Question 21: A particle starts from the origin at $t = 0$ with a velocity of $10.0\hat{j} \text{ m/s}$ and moves in the $x\text{-}y$ plane with a constant acceleration of $(8.0\hat{i} + 2.0\hat{j}) \text{ m/s}^2$. (a) At what time is the $x$-coordinate of the particle $16 \text{ m}$? What is the $y$-coordinate of the particle at that time? (b) What is the speed of the particle at the time?

Answer:

Given terms: $\vec{u} = 10.0\hat{j} \text{ m/s}$ (meaning $u_x = 0, u_y = 10$), and $\vec{a} = 8.0\hat{i} + 2.0\hat{j} \text{ m/s}^2$ (meaning $a_x = 8, a_y = 2$).

(a) Finding time and $y$-coordinate when $x = 16 \text{ m}$:

$$x(t) = u_x t + \frac{1}{2}a_x t^2 \implies 16 = 0 + \frac{1}{2}(8)t^2 \implies 4t^2 = 16 \implies t = \mathbf{2.0 \text{ s}}$$

Substitute this time component to locate the position along the $y$-axis:

$$y(t) = u_y t + \frac{1}{2}a_y t^2 = (10)(2) + \frac{1}{2}(2)(2)^2 = 20 + 4 = \mathbf{24 \text{ m}}$$

(b) Calculating speed at $t = 2.0 \text{ s}$:

$$v_x = u_x + a_x t = 0 + (8)(2) = 16 \text{ m/s}$$

$$v_y = u_y + a_y t = 10 + (2)(2) = 14 \text{ m/s}$$

$$\text{Speed } v = \sqrt{v_x^2 + v_y^2} = \sqrt{16^2 + 14^2} = \sqrt{256 + 196} = \sqrt{452} \approx \mathbf{21.26 \text{ m/s}}$$

Question 22: $\hat{i}$ and $\hat{j}$ are unit vectors along $x$- and $y$-axes 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 $\vec{A} = 2\hat{i} + 3\hat{j}$ along the directions of $\hat{i} + \hat{j}$ and $\hat{i} - \hat{j}$?

Answer:

1. Evaluating $\hat{i} + \hat{j}$ and $\hat{i} - \hat{j}$:

Magnitude of $(\hat{i} + \hat{j}) = \sqrt{1^2 + 1^2} = \mathbf{\sqrt{2}}$. Direction: $\tan\theta = \frac{1}{1} \implies \mathbf{45^\circ}$ with the x-axis.
Magnitude of $(\hat{i} - \hat{j}) = \sqrt{1^2 + (-1)^2} = \mathbf{\sqrt{2}}$. Direction: $\tan\theta = \frac{-1}{1} \implies \mathbf{-45^\circ}$ with the x-axis.

2. Components of $\vec{A} = 2\hat{i} + 3\hat{j}$ along these directions:

The scalar component of a vector $\vec{A}$ along another vector $\vec{B}$ is computed via the projection formula: $\frac{\vec{A} \cdot \vec{B}}{|\vec{B}|}$.

Component along $(\hat{i} + \hat{j})$:
$$A_{\parallel (\hat{i}+\hat{j})} = \frac{(2\hat{i} + 3\hat{j}) \cdot (\hat{i} + \hat{j})}{\sqrt{2}} = \frac{2(1) + 3(1)}{\sqrt{2}} = \mathbf{\frac{5}{\sqrt{2}}}$$
Component along $(\hat{i} - \hat{j})$:
$$A_{\parallel (\hat{i}-\hat{j})} = \frac{(2\hat{i} + 3\hat{j}) \cdot (\hat{i} - \hat{j})}{\sqrt{2}} = \frac{2(1) + 3(-1)}{\sqrt{2}} = \mathbf{-\frac{1}{\sqrt{2}}}$$

Table of Contents

Reading Progress

Comments (0)