Class 11 Physics (Physics Part-I) Chapter 3 Motion In a Plane

Key Concept: Null Vector Concept

c) $0$

[Solution Description] Adding a vector $A$ to its opposite $-A$ results in the null vector. The equation for this is $A + (-A) = 0$. Here, $0$ represents the null vector, which has zero magnitude and no direction.

Key Concept: Vector Magnitude

b) 5

[Solution Description] The magnitude of a vector $r = x\hat{i} + y\hat{j}$ is calculated using the formula $\sqrt{x^2 + y^2}$. For $r = 4\hat{i} + 3\hat{j}$, we have:

$\text{Magnitude} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$

Thus, the magnitude of the vector is 5.

Key Concept: Unit Vectors

b) $\hat{i}$

[Solution Description] A unit vector has a magnitude of 1. Among the options given, $\hat{i}$ is one of the standard unit vectors with a magnitude of 1.

Key Concept: Negative Scaling

d) Direction reverses and magnitude becomes 4 times

[Solution Description] When a vector is multiplied by a negative scalar, the direction of the resultant vector reverses, and the magnitude becomes four times the original magnitude.

Key Concept: Basic Concept

c) Towards the center of the circle

[Solution Description] In uniform circular motion, centripetal acceleration always points towards the center of the circular path. This inward acceleration ensures that the object continues to move in a circular path rather than moving in a straight line.

Key Concept: Position Analysis

b) $9 \, \text{m}$

[Solution Description] Use the position equation for motion along the y-axis:

$y = y_0 + v_{oy} t + \frac{1}{2} a_y t^2$

Since the particle starts from rest, $v_{oy} = 0$. Assume $y_0 = 0$ for simplicity.

Substituting the given values:

$y = 0 + 0 \times 3 + \frac{1}{2} \times 2 \times (3)^2$

$y = 0 + \frac{1}{2} \times 2 \times 9$

$y = 9 \, \text{m}$

Therefore, the position is $9 \, \text{m}$.

Key Concept: Maximum Height Calculation

c) 63.78 m

[Solution Description] The maximum height is calculated using the formula:

$h_m = \frac{v_o^2 \sin^2 \theta_o}{2g}$

Given $v_o = 50\, \text{m/s}$, $\theta_o = 45^\circ$, and $g = 9.8\, \text{m/s}^2$.

$h_m = \frac{(50)^2 (\sin(45^\circ))^2}{2 \times 9.8}$

$h_m = \frac{2500 \times \left(\frac{\sqrt{2}}{2}\right)^2}{19.6}$

$h_m = \frac{2500 \times \frac{1}{2}}{19.6}$

$h_m = \frac{1250}{19.6} \approx 63.78\, \text{m}$

The maximum height is approximately $63.78 \, \text{m}$.

Key Concept: Vector Basics

b) A quantity with both magnitude and direction

[Solution Description] A vector is defined as a quantity that has both magnitude and direction.

Key Concept: Displacement Vector

a) (3, -4)

[Solution Description]

The displacement vector $\mathbf{AB}$ can be calculated by subtracting the initial position vector from the final position vector. Here, A(2, 3) and B(5, -1):

$\mathbf{AB} = (x_2 – x_1, y_2 – y_1) = (5 – 2, -1 – 3) = (3, -4)$

Hence, the displacement vector is $(3, -4)$.

Key Concept: Commutative Law

b) $B + A$

[Solution Description] According to the commutative law of vector addition, the order of vectors does not matter. So, $A + B = B + A$.

Key Concept: Unit Vector Identification

c) $\hat{k}$

[Solution Description] The unit vector along the z-axis is denoted by $\hat{k}$. It has a magnitude of 1 and points in the positive z-direction.

Key Concept: Basic Definitions

d) Velocity

[Solution Description]

A vector quantity has both magnitude and direction. The options given are evaluated: temperature is scalar, speed is scalar, mass is scalar, while velocity has both magnitude and direction.

Key Concept: Calculation of Vector Components

a) 2.5

[Solution Description] The component $Cx$ of vector $\mathbf{C}$ along the x-axis can be calculated using trigonometry:

$Cx = C \cos \theta = 5 \cos 60^\circ$

$Cx = 5 \times \frac{1}{2} = 2.5$

Thus, the component of the vector along the x-axis is 2.5 units.

Key Concept: Instantaneous Velocity

c) 13 m/s

[Solution Description] The instantaneous velocity can be found by differentiating the position function with respect to time and then substituting $t = 2$.

The derivative of $x(t)$ is:

$v(t) = \frac{dx}{dt} = \frac{d}{dt}(4t^2 – 3t + 2) = 8t – 3.$

Now substitute $t = 2$:

$v(2) = 8(2) – 3 = 16 – 3 = 13 \text{ m/s}.$

Therefore, the instantaneous velocity at $t = 2$ s is 13 m/s.

Key Concept: Position Vector

a) OP

[Solution Description]

A position vector from point O to point P is represented by OP, denoted as $\mathbf{r}$ or $\overrightarrow{OP}$.

Key Concept: Deriving Relationships, Complex Problem Solving

a) $v = \frac{2πR}{T}$
[Solution Description] We know that the frequency $ν$ is related to the time period $T$ as $ν = 1/T$. The linear speed $v$ can be expressed as $v = 2πRν$.
Substitute $ν = 1/T$ into the equation for linear speed:
$$v = 2πR\left(\frac{1}{T}\right) = \frac{2πR}{T}.$$
Therefore, the derived formula for linear speed in terms of time period and radius is $v = \frac{2πR}{T}$.

Key Concept: Zero Vector Definition

c) It has zero magnitude.

[Solution Description] A null vector or zero vector is defined as a vector that has zero magnitude. It does not have any specific direction.

Key Concept: Frequency and Acceleration, Complex Calculations

c) 72π² m/s²

[Solution Description]
Centripetal acceleration $a_c$ in terms of frequency $\nu$ is expressed as:
$a_c = 4\pi^2\nu^2R$
Given that $\nu = 3 \text{ Hz}$ and $R = 2 \text{ m}$, substituting these into the equation gives:
$a_c = 4\pi^2(3)^2(2)$
Calculate each component step by step:
$a_c = 4 \times \pi^2 \times 9 \times 2$
$a_c = 72\pi^2 \text{ m/s}^2$
Hence, the centripetal acceleration is $72\pi^2$ m/s².

Key Concept: Derivation-Based, Multi-Step Calculations

c) 16 radians

[Solution Description]

To find the total angle covered, we use the formula for angular displacement with constant angular acceleration: $\theta = \omega_1 \Delta t + \frac{1}{2} \alpha (\Delta t)^2$. Here, $\alpha = \frac{\omega_2 – \omega_1}{\Delta t} = \frac{12 – 4}{2} = 4$ rad/s². Then, $\theta = 4 \times 2 + \frac{1}{2} \times 4 \times (2)^2 = 8 + 8 = 16$ radians.

Key Concept: Velocity at Time t

c) $7 \hat{i} + 9 \hat{j}$ m/s

[Solution Description] The velocity of the particle at time $t$ can be found using the formula $v = v_0 + at$. Here, $v_0 = 4 \hat{i} + 3 \hat{j}$ m/s, $a = 1 \hat{i} + 2 \hat{j}$ m/s², and $t = 3$ s.

First, calculate each component separately:

$v_x  = v_{0x} + a_x t = 4 + 1 \times 3 = 7 \text{ m/s},$

$v_y  = v_{0y} + a_y t = 3 + 2 \times 3 = 9 \text{ m/s}.$

Therefore, the velocity vector $v$ at $t = 3$ s is $7 \hat{i} + 9 \hat{j}$ m/s.

Key Concept: Commutative Property

d) $A + B = B + A$

[Solution Description]

The commutative property for vector addition states that the order in which two vectors are added does not affect the resultant vector. Therefore, $A + B = B + A$.

Key Concept: Position Calculation

c) $(15, 9)$

[Solution Description] To find the position coordinates, we use the equations of motion for each direction independently. For the x-direction:

$x  = x_0 + v_{ox}t + \frac{1}{2}a_xt^2$

$= 0 + 5(3) + \frac{1}{2}(0)(3)^2$

$= 15 \, \text{m}$

For the y-direction:

$y = y_0 + v_{oy}t + \frac{1}{2}a_yt^2$

$= 0 + 0 + \frac{1}{2}(2)(3)^2$

$= 9 \, \text{m}$

Therefore, the position coordinates are $(15, 9)$.

Key Concept: Displacement Calculation

c) $30 \, \text{m}$
[Solution Description]
The displacement can be found using the equation $\mathbf{r} = \mathbf{r}_0 + \mathbf{v}_0t + \frac{1}{2}\mathbf{a}t^2$. Assuming $\mathbf{r}_0 = 0$, we substitute $v_0 = 2 \, \text{m/s}$, $a = 1 \, \text{m/s}^2$, and $t = 6 \, \text{s}$:
$\mathbf{r} = (2 \, \text{m/s})(6 \, \text{s}) + \frac{1}{2}(1 \, \text{m/s}^2)(6 \, \text{s})^2$
Calculate:
$\mathbf{r} = 12 \, \text{m} + \frac{1}{2} \times 1 \, \text{m/s}^2 \times 36 \, \text{s}^2 = 12 \, \text{m} + 18 \, \text{m} = 30 \, \text{m}$
Therefore, the displacement is $30 \, \text{m}$.

Key Concept: Negative Scalar Multiplication

d) The magnitude is 24 and direction reversed

[Solution Description] When vector $B$ is multiplied by a negative scalar -3, the direction of the vector reverses and its magnitude becomes $\lambda |B|$, where $\lambda = 3$ and $|B| = 8$. The calculation is as follows:

$\text{Magnitude} = 3 \times 8 = 24$

Hence, the magnitude of the resultant vector is 24, and its direction is opposite to that of $B$.

Key Concept: Commutative Property

b) $A + B = B + A$

[Solution Description] The commutative property states that changing the order of vectors in addition does not change the resultant vector, i.e., $A + B = B + A$.

Key Concept: Range Calculation with Time

c) 91.84 meters
[Solution Description]
The formula for horizontal range $R$ is given by:
$R = \frac{v_0^2 \sin 2\theta_0}{g}$
Substituting the values:
$v_0   = 30 \, \text{m/s},$
$\theta_0  = 45^\circ,$
$g   = 9.8 \, \text{m/s}^2$
We have:
$R = \frac{30^2 \times \sin(90^\circ)}{9.8}$
FSince $\sin 90^\circ = 1$, it simplifies to:
$R = \frac{900}{9.8} \approx 91.84 \, \text{meters}$
Thus, the horizontal range is approximately $91.84$ meters.

Key Concept: Vector Magnitude

c) 10

[Solution Description] The magnitude of a vector $B$ with components $B_x$ and $B_y$ is given by:

$|B| = \sqrt{B_x^2 + B_y^2}$

Substituting the given values:

$|B| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100}$

Therefore, the magnitude of the vector is:

$|B| = 10$

Key Concept: Velocity Components Over Time

b) -1.116 m/s

[Solution Description]

The vertical component of velocity as a function of time is given by:

$v_y = v_{oy} – g t$

First, we calculate the initial vertical component $v_{oy}$:

$v_{oy} = v_o \sin \theta_o = 40 \sin 45^\circ = 40 \times 0.7071 = 28.284 \, \text{m/s}$

Then, substitute into the equation for $v_y$:

$v_y = 28.284 – 9.8 \times 3$

$v_y = 28.284 – 29.4$

$v_y = -1.116 \, \text{m/s}$

Key Concept: Vector Addition

a) $5 \hat{i} + 3 \hat{j}$

[Solution Description] The resultant vector is found by adding the corresponding components of vectors $C$ and $D$:

$C + D = (2 \hat{i} + 5 \hat{j}) + (3 \hat{i} – 2 \hat{j})$

Simplifying gives:

$C + D = (2 + 3) \hat{i} + (5 – 2) \hat{j} = 5 \hat{i} + 3 \hat{j}$

Key Concept: Time to Reach Ground

b) 6.06 s

[Solution Description] The time taken for the ball to reach the ground depends only on the vertical motion. Using $t = \sqrt{\frac{2h}{g}}$, substitute $h = 180 \, \text{m}$ and $g = 9.8 \, \text{m/s}^2$:
$t = \sqrt{\frac{2 \times 180}{9.8}} = \sqrt{\frac{360}{9.8}} = \sqrt{36.73} \approx 6.06 \, \text{s}.$

Key Concept: Complex Vector Operations, Resultant Magnitude and Direction

c) 11.83

[Solution Description] The vector subtraction is calculated as follows: $\mathbf{A} – \mathbf{B} = (3 + 2)\hat{i} + (4 – 1)\hat{j} + (5 – 6)\hat{k} = 5\hat{i} + 3\hat{j} – \hat{k}$. Now scale this by $k$: $2(5\hat{i} + 3\hat{j} – \hat{k}) = 10\hat{i} + 6\hat{j} – 2\hat{k}$. The magnitude of $\mathbf{R}$ is given by $\sqrt{10^2 + 6^2 + (-2)^2} = \sqrt{100 + 36 + 4} = \sqrt{140}$. Simplifying gives approximately $11.83$.

Key Concept: Maximum Height

a) $\frac{v_0^2 \sin^2 \theta_0}{2g}$

[Solution Description] The maximum height reached by a projectile is calculated using the formula $h_m = \frac{v_0^2 \sin^2 \theta_0}{2g}$. In this equation, $v_0$ represents the initial velocity, $\theta_0$ is the angle of projection, and $g$ is the acceleration due to gravity. Substituting these parameters into the formula gives us the peak height the projectile achieves during its flight.

Key Concept: Theoretical Analysis, Complex Problem Solving

b) Doubled

[Solution Description] The centripetal force $F_c$ for an object in uniform circular motion is given by:
$F_c = \frac{mv^2}{R}$
Initially, let the mass of the object be $m$. Therefore, the initial centripetal force $F_{c1}$ is:
$F_{c1} = \frac{mv^2}{R}$
If the mass is doubled, i.e., the new mass becomes $2m$, the new centripetal force $F_{c2}$ becomes:
$F_{c2} = \frac{2mv^2}{R} = 2\left(\frac{mv^2}{R}\right) = 2F_{c1}$
Hence, the required centripetal force doubles.

Key Concept: Magnitude from Components

b) 10 units

[Solution Description] The magnitude of a vector from its components can be calculated using the formula $A = \sqrt{A_x^2 + A_y^2}$.

Plugging in values:

$A  = \sqrt{6^2 + 8^2}$

$= \sqrt{36 + 64}$

$= \sqrt{100}$

$= 10$

Hence, the magnitude is 10 units.

Key Concept: Vector Components

b) $(2, -3, 1)$
[Solution Description]
The vector $\mathbf{A} = 2\hat{i} – 3\hat{j} + \hat{k}$ can be directly separated into its components as follows:
– The x-component is: $A_x = 2$
– The y-component is: $A_y = -3$
– The z-component is: $A_z = 1$
Therefore, the x, y, and z components are $(2, -3, 1)$.

Key Concept: Component Form

b) $3\hat{i} + 3\hat{j}$

[Solution Description] The displacement vector $\Delta r$ is calculated as follows:

$\Delta r = r’ – r = (6\hat{i} + 7\hat{j}) – (3\hat{i} + 4\hat{j}) = (6-3)\hat{i} + (7-4)\hat{j} = 3\hat{i} + 3\hat{j}.$

Therefore, the displacement vector is $3\hat{i} + 3\hat{j}$.

Key Concept: Negative Scaling

a) South

[Solution Description] Multiplying a vector by a negative scalar reverses its direction. Therefore, if vector $B$ originally points north, multiplying it by -2 results in a direction opposite to north, which is south.

Key Concept: Null Vector Concept

c) $0$

[Solution Description] Adding a vector $A$ to its negative counterpart $-A$ results in a null vector, meaning their sum is zero: $A + (-A) = 0$.

Key Concept: Basic Definition

c) $r = x \hat{i} + y \hat{j}$

[Solution Description]

The position vector $r$ in a plane is defined by the equation $r = x \hat{i} + y \hat{j}$, where $x$ and $y$ are the coordinates of the point.

Key Concept: Vector Addition

b) $2\sqrt{10}$ m/s

[Solution Description] To find the resultant velocity, we add the two given velocity vectors:

$v_{\text{resultant}} = (3 \hat{i} + 4 \hat{j}) + (-1 \hat{i} + 2 \hat{j}) = (3 – 1) \hat{i} + (4 + 2) \hat{j} = 2 \hat{i} + 6 \hat{j}.$

The magnitude of the resultant velocity is found using the Pythagorean theorem:

$|v_{\text{resultant}}| = \sqrt{(2)^2 + (6)^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10} \, \text{m/s}.$

Thus, the magnitude of the resultant velocity is $2\sqrt{10} m/s$.

Key Concept: Magnitude Calculation

c) 15

[Solution Description]

The magnitude of a vector given its components is calculated by:

$C = \sqrt{C_x^2 + C_y^2} = \sqrt{9^2 + 12^2}$

Simplifying inside the square root:

$= \sqrt{81 + 144} = \sqrt{225} = 15$

Key Concept: Maximum Height Calculation

c) 11.48 m

[Solution Description] To calculate the maximum height, we use the formula:
$H = \frac{(v_{oy})^2}{2g}$
where $v_{oy} = 15 \, \text{m/s}$ and $g = 9.8 \, \text{m/s}^2$.
Substitute the given values:
$H = \frac{(15)^2}{2 \times 9.8}$
$H = \frac{225}{19.6}$
$H = 11.48 \, \text{m}$
So, the maximum height reached by the projectile is approximately 11.48 m.

Key Concept: Position Change

b) $14\hat{i} + 8\hat{j}$
[Solution Description] The new position $r$ can be found using the formula $r = r_0 + v \cdot t$. Substituting the given values:
$r = (4\hat{i} + 3\hat{j}) + (2\hat{i} + 1\hat{j}) \times 5$
$r = 4\hat{i} + 3\hat{j} + 10\hat{i} + 5\hat{j}$
$r = 14\hat{i} + 8\hat{j}$
Therefore, the new position is $r = 14\hat{i} + 8\hat{j}$.

Key Concept: Max Height Calculation

b) 5.1 m

[Solution Description] To find the maximum height $(h_m)$, use the formula:$h_m = \frac{{v_0}^2 \sin^2 \theta_0}{2g}.$
Substitute $v_0 = 20 \, \text{m/s}$, $\theta_0 = 30^\circ$, and $g = 9.8 \, \text{m/s}^2$:
$h_m = \frac{{(20)^2} (\sin 30^\circ)^2}{2 \times 9.8}.$
Since $\sin 30^\circ = 0.5$: $h_m = \frac{(400)(0.25)}{19.6} = \frac{100}{19.6} \approx 5.10 \, \text{m}.$

Key Concept: Magnitude Calculation

a) $2\sqrt{2}$

[Solution Description] First, calculate the change in velocity $\Delta v = v_2 – v_1$:

$\Delta v = (13\hat{i} + 10\hat{j}) – (5\hat{i} + 2\hat{j})$

$= (13 – 5)\hat{i} + (10 – 2)\hat{j}$

$= 8\hat{i} + 8\hat{j}.$

The acceleration $a = \frac{\Delta v}{\Delta t}$ where $\Delta t = 4\, \text{s}$:

$a = \frac{8\hat{i} + 8\hat{j}}{4}$

$= 2\hat{i} + 2\hat{j}.$

Now, calculate the magnitude $|a|$:

$|a| = \sqrt{(2)^2 + (2)^2}$

$= \sqrt{4 + 4}$

$= \sqrt{8}$

$= 2\sqrt{2}.$

The magnitude of the acceleration is $2\sqrt{2}$.

Key Concept: Vector Multiplication

d) Its direction reverses and magnitude triples

[Solution Description]
When a vector is multiplied by a negative scalar, its magnitude becomes the absolute value of the product of the original magnitude and the scalar, while the direction of the vector reverses.
Let the original magnitude of vector $\mathbf{B}$ be $|\mathbf{B}|$. Then, after multiplication:
$|\mathbf{B’}| = |-3| \cdot |\mathbf{B}| = 3 \cdot |\mathbf{B}|$
However, due to the negative scalar, the direction of $\mathbf{B}$ reverses. So, multiplying $\mathbf{B}$ by $-3$ results in a vector that is three times as long and pointing in the opposite direction.

Key Concept: Vector Addition

c) Dot product method

[Solution Description] To find the resultant vector of two vectors, either the head-to-tail method or the parallelogram method can be used graphically. Both methods involve arranging the vectors such that they form a shape or sequence which helps in determining the resultant.

Key Concept: Vector Components Analysis

a) Yes

[Solution Description] The vectors are equal if both their x-components and y-components are equal. Here, $A_x = B_x = 5$ and $A_y = B_y = -3$. Thus, the vectors are equal.

Key Concept: Vector Representation

a) $B = 5 \hat{i} + 8 \hat{j}$

[Solution Description]

A vector can be represented in terms of its components by using unit vectors $\hat{i}$ and $\hat{j}$. For vector $B$, with components $B_x = 5$ and $B_y = 8$, it can be represented as:

$B = B_x \hat{i} + B_y \hat{j} = 5 \hat{i} + 8 \hat{j}$

Thus, $B$ is expressed as $5 \hat{i} + 8 \hat{j}$.

Key Concept: Time of Flight

b) $T_f = \frac{2 v_o \sin \theta_o}{g}$
[Solution Description] The total time of flight for a projectile can be calculated using the formula $T_f = \frac{2 v_o \sin \theta_o}{g}$. This formula arises from the symmetry of projectile motion and accounts for the upward and downward paths of the projectile.

Key Concept: Projectile Path

a) $y = x \sqrt{3} – \frac{49x^2}{2250}$

[Solution Description] We know that the general form of the parabolic path of a projectile is given by:

$y = x \tan \theta_0 – \frac{gx^2}{2(v_0 \cos \theta_0)^2}$

With: Launch angle: $\theta_0 = 60^\circ$

Initial speed: $v_0 = 30$ m/s

Gravitational acceleration: $g = 9.8$ m/s²

Solving for components:

$\tan \theta_0 = \sqrt{3}$

$v_0 \cos \theta_0 = 30 \cdot \frac{1}{2} = 15 \text{ m/s}$

Putting values in equation:

$y = x \cdot \sqrt{3} – \frac{9.8x^2}{2(15)^2}$

Simplifying the second term:

$\frac{9.8}{450}x^2 = \frac{49}{2250}x^2$

Hence, $y = x\sqrt{3} – \frac{49x^2}{2250}$

Key Concept: Basic Components

c) 4

[Solution Description] The z-component of a vector is simply the coefficient of $\hat{k}$ in its expression. For vector $A = 6\hat{i} + 3\hat{j} + 4\hat{k}$, the z-component is 4.

Key Concept: Symmetry in Motion

b) $6 \, \text{seconds}$

[Solution Description] The time to reach maximum height ($t_m$) is half of the total time of flight ($T_f$). Thus, $T_f = 2 \times t_m$.

Step 1: Given $t_m = 3 \, \text{seconds}$.

Step 2: Calculate the total time of flight:

$T_f = 2 \times 3 = 6 \, \text{seconds}$

Hence, the total time of flight is $6$ seconds.

Key Concept: Graphical Representation

c) No, because they have different directions

[Solution Description] Vectors are equal only if they have the same magnitude and direction. Although $\vec{A}$ and $\vec{B}$ have the same magnitude, their directions are opposite. Therefore, they are not equal.

Key Concept: Vector Basics

c) Velocity

[Solution Description] A vector quantity has both magnitude and direction, while a scalar only has magnitude. Among the given options, velocity is a vector as it includes both speed (magnitude) and a specified direction.

Key Concept: Horizontal Range Calculation

b) $R = \frac{v_0^2 \sin 2\theta_0}{g}$

[Solution Description] The horizontal range $R$ can be calculated using the formula $R = \frac{v_0^2 \sin 2\theta_0}{g}$. By recognizing that the range depends on the square of the initial velocity and the sine of twice the launch angle, one can compute $R$ when given specific values for $v_0$, $\theta_0$, and $g$.

Key Concept: Vector Addition

b) $2\hat{i} + 8\hat{j}$

[Solution Description]

To find the resultant vector, we add the corresponding components of $\mathbf{A}$ and $\mathbf{B}$.

The x-component of the resultant vector is:

$A_x + B_x = 4 + (-2) = 2$

The y-component of the resultant vector is:

$A_y + B_y = 3 + 5 = 8$

Therefore, the resultant vector is $2\hat{i} + 8\hat{j}$.

Key Concept: Angular Speed and Frequency Relationship, Multi-step Solutions

d) $80\pi m/s$
[Solution Description] Given: $\nu = 5$ Hz and $R = 8$ m.
First, calculate the angular speed using the relationship: $\omega = 2 \pi \nu = 2 \pi \times 5 = 10 \pi \text{ rad/s}$
Next, find the linear speed using the relation between linear speed and angular speed: $v = \omega R = 10 \pi \times 8 = 80 \pi \text{ m/s}$
So, the linear speed $v$ is $80\pi m/s$.

Key Concept: Vector Subtraction, Real Number Multiplication

c) $-6\hat{i} + 4\hat{j}$

[Solution Description] Compute $\mathbf{A}-\mathbf{B}$ first:

$\mathbf{A} – \mathbf{B} = (5\hat{i} – \hat{j}) – (2\hat{i} + \hat{j}) = (5 – 2)\hat{i} + (-1 – 1)\hat{j} = 3\hat{i} – 2\hat{j}$

Now calculate $-2(\mathbf{A}-\mathbf{B})$:

$-2(\mathbf{A}-\mathbf{B}) = -2(3\hat{i} – 2\hat{j}) = -6\hat{i} + 4\hat{j}$

Therefore, $-2(\mathbf{A}-\mathbf{B}) = -6\hat{i} + 4\hat{j}$.

Key Concept: Path Independence

b) It is the shortest distance between two points

[Solution Description]

Displacement is defined as the straight line joining the initial and final positions. It is independent of the path taken. Therefore, the correct statement is that displacement is the shortest distance between two points.

Key Concept: Associative Law

a) $(A + B) + C = A + (B + C)$

[Solution Description] The associative law of vector addition indicates that when three vectors are added, the grouping of the vectors does not change the result. Thus, $(A + B) + C = A + (B + C)$ is an illustration of this law.

Key Concept: Unit Identification

a) Kilogram

[Solution Description] The standard unit of mass in the International System of Units (SI) is the kilogram (kg).