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Chapter 4 Motion In A Plane
Introduction
In the previous chapter, we described motion along a straight line (one dimension). The directional aspect was handled using positive (+) and negative (-) signs. However, to describe motion in two dimensions (a plane) or three dimensions (space), we need a more robust mathematical tool. This tool is the vector.
This chapter introduces the language of vectors, including their properties and operations like addition, subtraction, and multiplication. We will then use vectors to define physical quantities such as position, displacement, velocity, and acceleration in a plane. We will apply these concepts to analyze two important types of motion in a plane: projectile motion and uniform circular motion.
Scalars and Vectors
Physical quantities can be broadly classified into two categories: scalars and vectors.
Scalars
A scalar quantity is a quantity that has only magnitude and no direction. It is completely specified by a single number along with its proper unit. Scalars are combined using the rules of ordinary algebra.
Examples: Distance, mass, temperature, time, speed, volume, density.
Vectors
A vector quantity is a quantity that has both magnitude and direction and obeys the triangle law or parallelogram law of vector addition.
Examples: Displacement, velocity, acceleration, force, momentum.
Vector Notation
- In print, vectors are represented by boldface letters (e.g., v, A, r).
- When written by hand, an arrow is placed over the letter (e.g., $\vec{v}, \vec{A}, \vec{r}$).
- The magnitude (or absolute value) of a vector A is a scalar and is written as |A| or simply A.
Position and Displacement Vectors
Position Vector
To specify the position of an object P in a plane, we choose an origin O. The straight line vector OP, with its tail at the origin O and its head at the point P, is called the position vector of the object. It is usually denoted by r.
Displacement Vector
If an object moves from an initial position P (with position vector r) to a final position P' (with position vector r'), the displacement vector is the vector PP'. It is the change in the position vector.
$\Delta \textbf{r} = \textbf{r'} - \textbf{r}$
The displacement vector is the straight line connecting the initial and final points and its direction points from the initial to the final point. Its magnitude is always less than or equal to the path length.
Equality of Vectors
Two vectors A and B are considered equal if and only if they have the same magnitude and the same direction. A vector remains unchanged if it is displaced parallel to itself.
Multiplication of Vectors by Real Numbers
Multiplying a vector A by a real number (scalar) $\lambda$ results in a new vector, $\lambda\textbf{A}$.
- The magnitude of the new vector is $|\lambda|$ times the magnitude of the original vector: $|\lambda\textbf{A}| = |\lambda| |\textbf{A}|$.
- If $\lambda$ is positive, the direction of $\lambda\textbf{A}$ is the same as the direction of A.
- If $\lambda$ is negative, the direction of $\lambda\textbf{A}$ is opposite to the direction of A.
For example, the vector $2\textbf{A}$ has twice the magnitude of A and the same direction. The vector $-\textbf{A}$ has the same magnitude as A but the opposite direction.
If the scalar $\lambda$ has dimensions, the resulting vector will have new dimensions. For example, multiplying velocity (a vector) by time (a scalar) gives displacement (a vector).
Addition and Subtraction of Vectors — Graphical Method
Vectors do not add like ordinary numbers; they add geometrically according to specific laws.
Triangle Law of Vector Addition (Head-to-Tail Method)
To add two vectors A and B graphically:
- Draw vector A.
- Place the tail of vector B at the head of vector A.
- The resultant vector R = A + B is the vector drawn from the tail of A to the head of B.
Properties of Vector Addition
- Commutative Law: The order of addition does not matter. A + B = B + A.
- Associative Law: When adding three or more vectors, the grouping does not matter. (A + B) + C = A + (B + C).
Null Vector (Zero Vector)
A null vector, denoted by 0, is a vector with zero magnitude. Since its magnitude is zero, its direction is indeterminate.
- Adding a vector and its negative gives a null vector: $\textbf{A} + (-\textbf{A}) = \textbf{0}$.
- Adding a null vector to any vector leaves it unchanged: $\textbf{A} + \textbf{0} = \textbf{A}$.
- Multiplying a vector by the number 0 gives a null vector: $0 \times \textbf{A} = \textbf{0}$.
Subtraction of Vectors
The subtraction of vector B from vector A is defined as the addition of vector A and the negative of vector B.
$\textbf{A} - \textbf{B} = \textbf{A} + (-\textbf{B})$
Parallelogram Law of Vector Addition
This is an equivalent method to the triangle law for adding two vectors.
- Draw vectors A and B with their tails at a common origin O.
- Complete the parallelogram using A and B as adjacent sides.
- The resultant vector R = A + B is the diagonal of the parallelogram starting from the common origin O.
Resolution of Vectors
Resolution of a vector is the process of splitting a single vector into two or more vectors in different directions, which when added together give the original vector. The split vectors are called the components of the original vector.
Unit Vectors
A unit vector is a vector of unit magnitude (magnitude = 1) and points in a specific direction. It is dimensionless and unitless, used only to specify direction.
- Unit vectors along the x, y, and z-axes of a rectangular coordinate system are denoted by $\hat{\textbf{i}}$, $\hat{\textbf{j}}$, and $\hat{\textbf{k}}$, respectively.
- $|\hat{\textbf{i}}| = |\hat{\textbf{j}}| = |\hat{\textbf{k}}| = 1$.
- Any vector A can be written as the product of its magnitude and a unit vector $\hat{\textbf{n}}$ in its direction: $\textbf{A} = |\textbf{A}|\hat{\textbf{n}}$.
Rectangular Components of a Vector
It is convenient to resolve a vector along the axes of a rectangular coordinate system. A vector A in the x-y plane can be resolved into two perpendicular components, one along the x-axis ($\textbf{A}_x$) and one along the y-axis ($\textbf{A}_y$).
$\textbf{A} = \textbf{A}_x + \textbf{A}_y$
In terms of unit vectors, this is written as:
$\textbf{A} = A_x\hat{\textbf{i}} + A_y\hat{\textbf{j}}$
Here, $A_x$ and $A_y$ are scalar quantities called the x and y-components of vector A.
Finding Components from Magnitude and Direction
If a vector A has magnitude A and makes an angle $\theta$ with the positive x-axis:
$ A_x = A \cos \theta $
$ A_y = A \sin \theta $
Finding Magnitude and Direction from Components
If the components $A_x$ and $A_y$ are known:
Magnitude: $ A = \sqrt{A_x^2 + A_y^2} $
Direction: $ \theta = \tan^{-1}\left(\frac{A_y}{A_x}\right) $
This concept can be extended to three dimensions, where a vector A is expressed as $\textbf{A} = A_x\hat{\textbf{i}} + A_y\hat{\textbf{j}} + A_z\hat{\textbf{k}}$, and its magnitude is $A = \sqrt{A_x^2 + A_y^2 + A_z^2}$.
Vector Addition — Analytical Method
Adding vectors graphically can be tedious. A more precise and easier method is the analytical method, which involves adding the components of the vectors.
Consider two vectors $\textbf{A} = A_x\hat{\textbf{i}} + A_y\hat{\textbf{j}}$ and $\textbf{B} = B_x\hat{\textbf{i}} + B_y\hat{\textbf{j}}$. Their sum, the resultant vector $\textbf{R} = R_x\hat{\textbf{i}} + R_y\hat{\textbf{j}}$, is found by adding their corresponding components:
$ \textbf{R} = \textbf{A} + \textbf{B} = (A_x + B_x)\hat{\textbf{i}} + (A_y + B_y)\hat{\textbf{j}} $
So, the components of the resultant are:
$ R_x = A_x + B_x $
$ R_y = A_y + B_y $
The magnitude and direction of the resultant R can then be found using $R = \sqrt{R_x^2 + R_y^2}$ and $\theta = \tan^{-1}(R_y/R_x)$.
Law of Cosines and Law of Sines
For two vectors A and B with an angle $\theta$ between them, the magnitude and direction of their resultant R can be found without resolving into components.
Law of Cosines (Magnitude of Resultant)
The magnitude $R$ of the resultant vector $\textbf{R} = \textbf{A} + \textbf{B}$ is given by:
$ R = \sqrt{A^2 + B^2 + 2AB \cos \theta} $
Law of Sines (Direction of Resultant)
If the resultant R makes an angle $\alpha$ with vector A, then:
$ \frac{R}{\sin \theta} = \frac{A}{\sin \beta} = \frac{B}{\sin \alpha} $
Where $\beta$ is the angle between R and B. The direction $\alpha$ can also be found using:
$ \tan \alpha = \frac{B \sin \theta}{A + B \cos \theta} $
Example 1. A motorboat is racing towards north at 25 km/h and the water current is 10 km/h in the direction of 60° east of south. Find the resultant velocity of the boat.
Answer:
Let $\textbf{v}_b$ be the velocity of the boat and $\textbf{v}_c$ be the velocity of the current.
Magnitude of boat's velocity, $v_b = 25$ km/h (North).
Magnitude of current's velocity, $v_c = 10$ km/h (60° East of South).
The angle between the North direction and the South direction is 180°. The angle between North and 60° East of South is $180^\circ - 60^\circ = 120^\circ$. So, the angle $\theta$ between $\textbf{v}_b$ and $\textbf{v}_c$ is $120^\circ$.
Using the Law of Cosines to find the magnitude of the resultant velocity $R$:
$R = \sqrt{v_b^2 + v_c^2 + 2v_b v_c \cos \theta}$
$R = \sqrt{25^2 + 10^2 + 2(25)(10) \cos 120^\circ}$
$R = \sqrt{625 + 100 + 500(-0.5)} = \sqrt{725 - 250} = \sqrt{475} \approx 21.8 \text{ km/h}$
To find the direction, let $\phi$ be the angle the resultant makes with the North direction. Using the Law of Sines:
$\frac{\sin \phi}{v_c} = \frac{\sin \theta}{R} \implies \sin \phi = \frac{v_c \sin \theta}{R} = \frac{10 \sin 120^\circ}{21.8} = \frac{10(\sqrt{3}/2)}{21.8} \approx 0.397$
$\phi = \sin^{-1}(0.397) \approx 23.4^\circ$
The resultant velocity of the boat is approximately 21.8 km/h in a direction 23.4° East of North.
Motion in a Plane
We can now use vectors to describe motion in two dimensions (a plane).
Position Vector and Displacement
The position of a particle P in an x-y plane is described by its position vector r:
$\textbf{r} = x\hat{\textbf{i}} + y\hat{\textbf{j}}$
The displacement $\Delta \textbf{r}$ of the particle as it moves from P to P' is:
$\Delta \textbf{r} = \textbf{r'} - \textbf{r} = (x' - x)\hat{\textbf{i}} + (y' - y)\hat{\textbf{j}} = \Delta x\hat{\textbf{i}} + \Delta y\hat{\textbf{j}}$
Velocity
The average velocity $\bar{\textbf{v}}$ is the displacement divided by the time interval:
$\bar{\textbf{v}} = \frac{\Delta \textbf{r}}{\Delta t} = \frac{\Delta x}{\Delta t}\hat{\textbf{i}} + \frac{\Delta y}{\Delta t}\hat{\textbf{j}} = \bar{v}_x\hat{\textbf{i}} + \bar{v}_y\hat{\textbf{j}}$
The instantaneous velocity v is the time derivative of the position vector. Its direction is always tangent to the path of the particle.
$\textbf{v} = \lim_{\Delta t \to 0} \frac{\Delta \textbf{r}}{\Delta t} = \frac{d\textbf{r}}{dt}$
In component form:
$\textbf{v} = \frac{dx}{dt}\hat{\textbf{i}} + \frac{dy}{dt}\hat{\textbf{j}} = v_x\hat{\textbf{i}} + v_y\hat{\textbf{j}}$
Acceleration
The average acceleration $\bar{\textbf{a}}$ is the change in velocity divided by the time interval:
$\bar{\textbf{a}} = \frac{\Delta \textbf{v}}{\Delta t} = \frac{\Delta v_x}{\Delta t}\hat{\textbf{i}} + \frac{\Delta v_y}{\Delta t}\hat{\textbf{j}} = \bar{a}_x\hat{\textbf{i}} + \bar{a}_y\hat{\textbf{j}}$
The instantaneous acceleration a is the time derivative of the velocity vector.
$\textbf{a} = \lim_{\Delta t \to 0} \frac{\Delta \textbf{v}}{\Delta t} = \frac{d\textbf{v}}{dt}$
In component form:
$\textbf{a} = \frac{dv_x}{dt}\hat{\textbf{i}} + \frac{dv_y}{dt}\hat{\textbf{j}} = a_x\hat{\textbf{i}} + a_y\hat{\textbf{j}}$
An important point for 2D/3D motion is that the acceleration vector can have any angle with the velocity vector. For example, in circular motion, they are perpendicular.
Motion in a Plane with Constant Acceleration
If an object moves in a plane with a constant acceleration a, we can derive kinematic equations similar to those for one-dimensional motion.
Let the initial velocity at $t=0$ be $\textbf{v}_0$ and the velocity at time $t$ be v.
From the definition of constant acceleration, $\textbf{a} = (\textbf{v} - \textbf{v}_0)/t$, we get the first equation:
$\textbf{v} = \textbf{v}_0 + \textbf{a}t$
Let the initial position be $\textbf{r}_0$ and the position at time $t$ be r. The displacement $\textbf{r} - \textbf{r}_0$ is the average velocity multiplied by time.
$\textbf{r} - \textbf{r}_0 = \left(\frac{\textbf{v}_0 + \textbf{v}}{2}\right)t = \left(\frac{\textbf{v}_0 + (\textbf{v}_0 + \textbf{a}t)}{2}\right)t$
This simplifies to the second equation:
$\textbf{r} = \textbf{r}_0 + \textbf{v}_0t + \frac{1}{2}\textbf{a}t^2$
A key takeaway from these vector equations is that motion in a plane can be treated as two separate, simultaneous one-dimensional motions along perpendicular axes. The x-component of motion is independent of the y-component of motion.
Relative Velocity in Two Dimensions
The concept of relative velocity extends to two dimensions. If object A moves with velocity $\textbf{v}_A$ and object B moves with velocity $\textbf{v}_B$ (both with respect to a common reference frame, like the ground), then:
The velocity of object A relative to object B is:
$\textbf{v}_{AB} = \textbf{v}_A - \textbf{v}_B$
The velocity of object B relative to object A is:
$\textbf{v}_{BA} = \textbf{v}_B - \textbf{v}_A$
Note that $\textbf{v}_{AB} = -\textbf{v}_{BA}$.
Example 2. Rain is falling vertically with a speed of 35 m s$^{-1}$. A woman rides a bicycle with a speed of 12 m s$^{-1}$ in the east to west direction. What is the direction in which she should hold her umbrella?
Answer:
Let's set up a coordinate system. Let the positive y-axis be downward (direction of rain) and the positive x-axis be west (direction of bicycle).
Velocity of rain with respect to ground, $\textbf{v}_r = 35\hat{\textbf{j}} \text{ m/s}$.
Velocity of bicycle with respect to ground, $\textbf{v}_b = 12\hat{\textbf{i}} \text{ m/s}$.
The woman needs to protect herself from the velocity of the rain relative to her (the bicycle). This is $\textbf{v}_{rb}$.
$\textbf{v}_{rb} = \textbf{v}_r - \textbf{v}_b = (35\hat{\textbf{j}}) - (12\hat{\textbf{i}}) = -12\hat{\textbf{i}} + 35\hat{\textbf{j}}$.
This resultant vector points downwards and to the east. To protect herself, the woman must hold her umbrella in the opposite direction, i.e., upwards and towards the west.
Let's find the angle $\theta$ this relative velocity vector makes with the vertical (downward y-axis).
$\tan \theta = \frac{|\text{horizontal component}|}{|\text{vertical component}|} = \frac{|-12|}{35} = \frac{12}{35} \approx 0.343$
$\theta = \tan^{-1}(0.343) \approx 19^\circ$
She should hold her umbrella at an angle of about 19° with the vertical, towards the west (the direction she is moving).
Projectile Motion
An object that is thrown or projected into the air and is then subject only to the acceleration of gravity is called a projectile. We will assume that air resistance is negligible.
Projectile motion is a classic example of 2D motion with constant acceleration. It can be analyzed as the result of two independent motions:
- Horizontal Motion: Uniform velocity (zero acceleration).
- Vertical Motion: Constant downward acceleration ($g$).
Equations of Projectile Motion
Consider a projectile launched from the origin with an initial velocity $v_0$ at an angle $\theta_0$ with the horizontal.
Initial velocity components:
$v_{0x} = v_0 \cos \theta_0$
$v_{0y} = v_0 \sin \theta_0$
Acceleration components:
$a_x = 0$
$a_y = -g$
Velocity at time t
$v_x(t) = v_{0x} = v_0 \cos \theta_0$ (remains constant)
$v_y(t) = v_{0y} + a_y t = v_0 \sin \theta_0 - gt$
Position at time t
$x(t) = v_{0x}t = (v_0 \cos \theta_0)t$
$y(t) = v_{0y}t + \frac{1}{2}a_y t^2 = (v_0 \sin \theta_0)t - \frac{1}{2}gt^2$
Key Characteristics of Projectile Motion
Equation of Path (Trajectory)
By eliminating time $t$ from the position equations, we get the equation for the path:
$y = (\tan \theta_0)x - \left(\frac{g}{2v_0^2 \cos^2 \theta_0}\right)x^2$
This is the equation of a parabola.
Time of Maximum Height ($t_m$)
At the maximum height, the vertical component of velocity is zero ($v_y = 0$).
$0 = v_0 \sin \theta_0 - gt_m$
$t_m = \frac{v_0 \sin \theta_0}{g}$
Maximum Height ($h_m$)
Substitute $t_m$ into the equation for y(t).
$h_m = \frac{(v_0 \sin \theta_0)^2}{2g}$
Time of Flight ($T_f$)
This is the total time the projectile is in the air. Due to symmetry, $T_f = 2t_m$.
$T_f = \frac{2v_0 \sin \theta_0}{g}$
Horizontal Range (R)
This is the horizontal distance covered during the time of flight.
$R = v_x T_f = (v_0 \cos \theta_0)\left(\frac{2v_0 \sin \theta_0}{g}\right) = \frac{v_0^2(2 \sin \theta_0 \cos \theta_0)}{g}$
Using the identity $\sin 2\theta = 2\sin\theta\cos\theta$:
$R = \frac{v_0^2 \sin(2\theta_0)}{g}$
The range is maximum when $\sin(2\theta_0)$ is maximum (i.e., 1), which occurs when $2\theta_0 = 90^\circ$, or $\theta_0 = 45^\circ$.
Uniform Circular Motion
When an object moves along a circular path at a constant speed, its motion is called uniform circular motion. Although the speed is constant, the velocity is continuously changing because its direction is changing at every point. Since the velocity changes, the object is accelerating.
Centripetal Acceleration ($a_c$)
For an object in uniform circular motion, the acceleration is always directed towards the center of the circle. This acceleration is called centripetal acceleration (meaning "center-seeking").
Derivation of Magnitude
Consider an object moving with speed $v$ in a circle of radius $R$. In a small time interval $\Delta t$, it moves from position P to P' and its position vector rotates by an angle $\Delta \theta$. Its velocity vector also rotates by the same angle $\Delta \theta$.
Using similar triangles formed by the position vectors ($\textbf{r}, \textbf{r'}$) and the velocity vectors ($\textbf{v}, \textbf{v'}$), we can show:
$\frac{|\Delta \textbf{v}|}{v} = \frac{|\Delta \textbf{r}|}{R}$
Since $|\Delta \textbf{v}| = a \Delta t$ and for small intervals $|\Delta \textbf{r}| \approx v \Delta t$, we have:
$\frac{a \Delta t}{v} = \frac{v \Delta t}{R}$
This gives the magnitude of the centripetal acceleration as:
$a_c = \frac{v^2}{R}$
Angular Speed, Period, and Frequency
Angular Speed ($\omega$): The rate of change of angular position. $\omega = \frac{\Delta \theta}{\Delta t}$. Its unit is rad/s.
Relation between Linear and Angular Speed:
$v = R\omega$
Using this, the centripetal acceleration can also be expressed as:
$a_c = \frac{(R\omega)^2}{R} = \omega^2 R$
Time Period (T): The time taken to complete one revolution. The distance covered is $2\pi R$.
$v = \frac{2\pi R}{T}$
Frequency (n or f): The number of revolutions made per second. $n = \frac{1}{T}$.
Example 3. 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?
Answer:
Given: Radius $R = 12 \text{ cm} = 0.12 \text{ m}$.
The frequency of revolution is $n = \frac{7 \text{ rev}}{100 \text{ s}} = 0.07 \text{ rev/s}$.
(a) Angular and Linear Speed
Angular speed $\omega = 2\pi n = 2\pi(0.07) \text{ rad/s} \approx 0.44 \text{ rad/s}$.
Linear speed $v = R\omega = (12 \text{ cm})(0.44 \text{ rad/s}) \approx 5.3 \text{ cm/s}$.
(b) Acceleration Vector
The acceleration vector is not constant because its direction is continuously changing, always pointing towards the center of the groove.
However, its magnitude is constant and is given by:
$a_c = \omega^2 R = (0.44 \text{ rad/s})^2 (12 \text{ cm}) \approx 2.3 \text{ cm/s}^2$.
Exercises
Question 4.1. 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.
Answer:
Question 4.2. 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.
Answer:
Question 4.3. 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.
Answer:
Question 4.4. 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.
Answer:
Question 4.5. 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:
Question 4.6. Establish the following vector inequalities geometrically or otherwise :
(a) $|\textbf{a}+\textbf{b}| \le |\textbf{a}| + |\textbf{b}|$
(b) $|\textbf{a}+\textbf{b}| \ge ||\textbf{a}| - |\textbf{b}||$
(c) $|\textbf{a}-\textbf{b}| \le |\textbf{a}| + |\textbf{b}|$
(d) $|\textbf{a}-\textbf{b}| \ge ||\textbf{a}| - |\textbf{b}||$
When does the equality sign above apply?
Answer:
Question 4.7. Given $\textbf{a} + \textbf{b} + \textbf{c} + \textbf{d} = \textbf{0}$, which of the following statements are correct :
(a) $\textbf{a}, \textbf{b}, \textbf{c},$ and $\textbf{d}$ must each be a null vector,
(b) The magnitude of ($\textbf{a} + \textbf{c}$) equals the magnitude of ($\textbf{b} + \textbf{d}$),
(c) The magnitude of $\textbf{a}$ can never be greater than the sum of the magnitudes of $\textbf{b}, \textbf{c},$ and $\textbf{d}$,
(d) $\textbf{b} + \textbf{c}$ must lie in the plane of $\textbf{a}$ and $\textbf{d}$ if $\textbf{a}$ and $\textbf{d}$ are not collinear, and in the line of $\textbf{a}$ and $\textbf{d}$, if they are collinear ?
Answer:
Question 4.8. 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 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 skate ?
Answer:
Question 4.9. 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 Fig. 4.21. If the round trip takes 10 min, what is the (a) net displacement, (b) average velocity, and (c) average speed of the cyclist ?
Answer:
Question 4.10. On an open ground, a motorist follows a track that turns to his left by an angle of 60$^o$ 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.
Answer:
Question 4.11. 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 ?
Answer:
Question 4.12. Rain is falling vertically with a speed of 30 m s$^{-1}$. A woman rides a bicycle with a speed of 10 m s$^{-1}$ in the north to south direction. What is the direction in which she should hold her umbrella ?
Answer:
Question 4.13. A man can swim with a speed of 4.0 km/h in still water. How long does he take to cross a river 1.0 km wide if the river flows steadily at 3.0 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:
Question 4.14. In a harbour, wind is blowing at the speed of 72 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 km/h to the north, what is the direction of the flag on the mast of the boat ?
Answer:
Question 4.15. 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 ?
Answer:
Question 4.16. 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 ?
Answer:
Question 4.17. 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 ?
Answer:
Question 4.18. 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.
Answer:
Question 4.19. 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:
Question 4.20. The position of a particle is given by
$\textbf{r} = 3.0t \hat{\textbf{i}} - 2.0t^2 \hat{\textbf{j}} + 4.0 \hat{\textbf{k}}$ m
where t is in seconds and the coefficients have the proper units for r to be in metres.
(a) Find the $\textbf{v}$ and $\textbf{a}$ of the particle? (b) What is the magnitude and direction of velocity of the particle at t = 2.0 s ?
Answer:
Question 4.21. A particle starts from the origin at t = 0 s with a velocity of 10.0 $\hat{\textbf{j}}$ m/s and moves in the x-y plane with a constant acceleration of (8.0 $\hat{\textbf{i}}$ + 2.0 $\hat{\textbf{j}}$) m s$^{-2}$. (a) At what time is the x- coordinate of the particle 16 m? What is the y-coordinate of the particle at that time? (b) What is the speed of the particle at the time ?
Answer:
Question 4.22. $\hat{\textbf{i}}$ and $\hat{\textbf{j}}$ are unit vectors along x- and y- axis respectively. What is the magnitude and direction of the vectors $\hat{\textbf{i}} + \hat{\textbf{j}}$, and $\hat{\textbf{i}} - \hat{\textbf{j}}$ ? What are the components of a vector $\textbf{A}= 2\hat{\textbf{i}} + 3\hat{\textbf{j}}$ along the directions of $\hat{\textbf{i}} + \hat{\textbf{j}}$ and $\hat{\textbf{i}} - \hat{\textbf{j}}$? [You may use graphical method]
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Question 4.23. For any arbitrary motion in space, which of the following relations are true :
(a) $\textbf{v}_{average} = (1/2) (\textbf{v}(t_1) + \textbf{v}(t_2))$
(b) $\textbf{v}_{average} = [\textbf{r}(t_2) - \textbf{r}(t_1) ] /(t_2 – t_1)$
(c) $\textbf{v}(t) = \textbf{v}(0) + \textbf{a}t$
(d) $\textbf{r}(t) = \textbf{r}(0) + \textbf{v}(0)t + (1/2)\textbf{a}t^2$
(e) $\textbf{a}_{average} = [\textbf{v}(t_2) - \textbf{v}(t_1) ] /(t_2 – t_1)$
(The ‘average’ stands for average of the quantity over the time interval $t_1$ to $t_2$)
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Question 4.24. 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.
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Question 4.25. 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 ?
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Additional Exercises
Question 4.26. A vector has magnitude and direction. Does it have a location in space ? Can it vary with time ? Will two equal vectors a and b at different locations in space necessarily have identical physical effects ? Give examples in support of your answer.
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Question 4.27. A vector has both magnitude and direction. Does it mean that anything that has magnitude and direction is necessarily a vector ? The rotation of a body can be specified by the direction of the axis of rotation, and the angle of rotation about the axis. Does that make any rotation a vector ?
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Question 4.28. Can you associate vectors with (a) the length of a wire bent into a loop, (b) a plane area, (c) a sphere ? Explain.
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Question 4.29. A bullet fired at an angle of 30° with the horizontal hits the ground 3.0 km away. By adjusting its angle of projection, can one hope to hit a target 5.0 km away ? Assume the muzzle speed to be fixed, and neglect air resistance.
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Question 4.30. A fighter plane flying horizontally at an altitude of 1.5 km with speed 720 km/h passes directly overhead an anti-aircraft gun. At what angle from the vertical should the gun be fired for the shell with muzzle speed 600 m s$^{-1}$ to hit the plane ? At what minimum altitude should the pilot fly the plane to avoid being hit ? (Take g = 10 m s$^{-2}$ ).
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Question 4.31. A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.50 m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn ?
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Question 4.32. (a) Show that for a projectile the angle between the velocity and the x-axis as a function of time is given by
$\theta(t) = \tan^{-1}\left(\frac{v_{0y} - gt}{v_{ox}}\right)$
(b) Shows that the projection angle $\theta_0$ for a projectile launched from the origin is given by
$\theta_0 = \tan^{-1}\left(\frac{4h_m}{R}\right)$
where the symbols have their usual meaning.
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