Motion in a Plane

Given: A list of physical quantities.

Concept: Scalar quantities have only magnitude; vector quantities have both magnitude and direction.

| Physical Quantity | Type |
|---|---|
| Volume | Scalar |
| Mass | Scalar |
| Speed | Scalar |
| Acceleration | Vector |
| Density | Scalar |
| Number of moles | Scalar |
| Velocity | Vector |
| Angular frequency | Scalar |
| Displacement | Vector |
| Angular velocity | Vector |

Explanation:
- Scalars: Volume, mass, speed, density, number of moles, and angular frequency are completely described by their magnitude alone.
- Vectors: Acceleration, velocity, displacement, and angular velocity require both magnitude and direction for complete description.

Given: A list of physical quantities.

Concept: Scalars have only magnitude; vectors have magnitude and direction.

Examining each quantity:
- Force → Vector
- Angular momentum → Vector
- Work → Scalar (dot product of two vectors gives a scalar)
- Current → Scalar (though current has a direction of flow, it does not obey vector addition laws and is treated as a scalar)
- Linear momentum → Vector
- Electric field → Vector
- Average velocity → Vector
- Magnetic moment → Vector
- Relative velocity → Vector

Answer: The two scalar quantities are work and current.

Given: A list of physical quantities.

Concept: A vector quantity has both magnitude and direction and obeys vector addition laws.

Examining each quantity:
- Temperature → Scalar
- Pressure → Scalar (it is force per unit area, but acts equally in all directions — treated as scalar)
- Impulse → Vector (Impulse = Force × time; since force is a vector, impulse is also a vector)
- Time → Scalar
- Power → Scalar
- Total path length → Scalar
- Energy → Scalar
- Gravitational potential → Scalar
- Coefficient of friction → Scalar
- Charge → Scalar

Answer: The only vector quantity is impulse.

Concept: Only quantities of the same nature (both scalars or both vectors) and same dimensions can be added. Multiplication of a vector by a scalar is always meaningful.

(a) Adding any two scalars:
Meaningful — only if they represent the same physical quantity (same dimensions). For example, adding two masses or two lengths is meaningful. Adding mass and temperature is not meaningful.

(b) Adding a scalar to a vector of the same dimensions:
Not meaningful — A scalar and a vector are fundamentally different mathematical objects. Even if they have the same dimensions, they cannot be added (e.g., you cannot add speed (scalar) to velocity (vector)).

(c) Multiplying any vector by any scalar:
Meaningful — The product of a vector and a scalar is always a vector. For example, force $\mathbf{F} = m\mathbf{a}$ (mass × acceleration). The result has magnitude equal to the product of the magnitudes and direction same as (or opposite to) the original vector.

(d) Multiplying any two scalars:
Meaningful — The product of two scalars is always a scalar. For example, work = pressure × volume.

(e) Adding any two vectors:
Meaningful — only if they represent the same physical quantity (same dimensions). For example, two displacement vectors or two force vectors can be added using vector addition rules. Adding a force vector and a velocity vector is not meaningful.

(f) Adding a component of a vector to the same vector:
Not meaningful — A component of a vector is itself a vector, but it lies along one axis. Adding it to the original vector would be like adding two vectors of the same type, which is mathematically possible, but physically this operation is not meaningful because the component is already a part of the vector.

(a) The magnitude of a vector is always a scalar.
True. The magnitude of a vector is a positive real number with no direction. For example, $|\mathbf{v}| = v$ is a scalar (speed). Magnitude is always a non-negative scalar quantity.

(b) Each component of a vector is always a scalar.
False. The components of a vector (e.g., $A_x\hat{\mathbf{i}}$, $A_y\hat{\mathbf{j}}$) are themselves vectors directed along the respective axes. The scalar parts $A_x$ and $A_y$ are scalars, but the components $A_x\hat{\mathbf{i}}$ and $A_y\hat{\mathbf{j}}$ are vectors.

(c) The total path length is always equal to the magnitude of the displacement vector of a particle.
False. The total path length is the actual distance travelled along the path, while displacement is the shortest straight-line distance between the initial and final positions. They are equal only when the particle moves in a straight line without reversing direction. In all other cases, path length $\geq$ |displacement|.

(d) The average speed of a particle is either greater or equal to the magnitude of average velocity of the particle over the same interval of time.
True.
$$\text{Average speed} = \frac{\text{Total path length}}{\Delta t} \geq \frac{|\Delta \mathbf{r}|}{\Delta t} = |\text{Average velocity}|$$
Since total path length $\geq$ magnitude of displacement, average speed $\geq$ magnitude of average velocity. Equality holds when the particle moves in a straight line without reversing direction.

(e) Three vectors not lying in a plane can never add up to give a null vector.
True. For three vectors to add up to a null vector, they must form a closed triangle (head-to-tail). A triangle is always a planar figure. Therefore, three vectors that do not lie in the same plane cannot form a closed triangle and hence cannot give a null (zero) vector.

Concept: Using the parallelogram law of vector addition and properties of triangles.

Let $\theta$ be the angle between vectors $\mathbf{a}$ and $\mathbf{b}$.

By the parallelogram law:
$$|\mathbf{a} + \mathbf{b}|^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 + 2|\mathbf{a}||\mathbf{b}|\cos\theta$$

(a) $|\mathbf{a} + \mathbf{b}| \leq |\mathbf{a}| + |\mathbf{b}|$

Since $\cos\theta \leq 1$:
$$|\mathbf{a} + \mathbf{b}|^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 + 2|\mathbf{a}||\mathbf{b}|\cos\theta \leq |\mathbf{a}|^2 + |\mathbf{b}|^2 + 2|\mathbf{a}||\mathbf{b}|$$
$$|\mathbf{a} + \mathbf{b}|^2 \leq (|\mathbf{a}| + |\mathbf{b}|)^2$$
$$\therefore |\mathbf{a} + \mathbf{b}| \leq |\mathbf{a}| + |\mathbf{b}|$$

Equality holds when $\cos\theta = 1$, i.e., $\theta = 0°$ — both vectors are parallel and in the same direction.

(b) $|\mathbf{a} + \mathbf{b}| \geq ||\mathbf{a}| - |\mathbf{b}||$

Since $\cos\theta \geq -1$:
$$|\mathbf{a} + \mathbf{b}|^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 + 2|\mathbf{a}||\mathbf{b}|\cos\theta \geq |\mathbf{a}|^2 + |\mathbf{b}|^2 - 2|\mathbf{a}||\mathbf{b}|$$
$$|\mathbf{a} + \mathbf{b}|^2 \geq (|\mathbf{a}| - |\mathbf{b}|)^2 = \big||\mathbf{a}| - |\mathbf{b}|\big|^2$$
$$\therefore |\mathbf{a} + \mathbf{b}| \geq \big||\mathbf{a}| - |\mathbf{b}|\big|$$

Equality holds when $\cos\theta = -1$, i.e., $\theta = 180°$ — both vectors are antiparallel.

(c) $|\mathbf{a} - \mathbf{b}| \leq |\mathbf{a}| + |\mathbf{b}|$

Write $\mathbf{a} - \mathbf{b} = \mathbf{a} + (-\mathbf{b})$. The magnitude of $-\mathbf{b}$ is $|\mathbf{b}|$.

Applying result (a) with $-\mathbf{b}$ in place of $\mathbf{b}$:
$$|\mathbf{a} + (-\mathbf{b})| \leq |\mathbf{a}| + |-\mathbf{b}| = |\mathbf{a}| + |\mathbf{b}|$$
$$\therefore |\mathbf{a} - \mathbf{b}| \leq |\mathbf{a}| + |\mathbf{b}|$$

Equality holds when $\mathbf{a}$ and $\mathbf{b}$ are antiparallel (i.e., $\theta = 180°$), so $\mathbf{a}$ and $-\mathbf{b}$ are parallel.

(d) $|\mathbf{a} - \mathbf{b}| \geq ||\mathbf{a}| - |\mathbf{b}||$

Applying result (b) with $-\mathbf{b}$ in place of $\mathbf{b}$:
$$|\mathbf{a} + (-\mathbf{b})| \geq \big||\mathbf{a}| - |-\mathbf{b}|\big| = \big||\mathbf{a}| - |\mathbf{b}|\big|$$
$$\therefore |\mathbf{a} - \mathbf{b}| \geq \big||\mathbf{a}| - |\mathbf{b}|\big|$$

Equality holds when $\mathbf{a}$ and $\mathbf{b}$ are parallel (same direction, $\theta = 0°$).

Given: $\mathbf{a} + \mathbf{b} + \mathbf{c} + \mathbf{d} = \mathbf{0}$

(a) a, b, c, and d must each be a null vector.
Incorrect. The sum of four vectors can be zero without each being a null vector. For example, if $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$, $\mathbf{d}$ are four sides of a closed quadrilateral (taken in order), their vector sum is zero, yet none need be zero.

(b) The magnitude of $(\mathbf{a} + \mathbf{c})$ equals the magnitude of $(\mathbf{b} + \mathbf{d})$.
Correct. From the given condition:
$$\mathbf{a} + \mathbf{b} + \mathbf{c} + \mathbf{d} = \mathbf{0}$$
$$(\mathbf{a} + \mathbf{c}) + (\mathbf{b} + \mathbf{d}) = \mathbf{0}$$
$$(\mathbf{a} + \mathbf{c}) = -(\mathbf{b} + \mathbf{d})$$
Taking magnitudes: $|\mathbf{a} + \mathbf{c}| = |\mathbf{b} + \mathbf{d}|$. ✓

(c) The magnitude of $\mathbf{a}$ can never be greater than the sum of the magnitudes of $\mathbf{b}$, $\mathbf{c}$, and $\mathbf{d}$.
Correct. From the given condition:
$$\mathbf{a} = -(\mathbf{b} + \mathbf{c} + \mathbf{d})$$
$$|\mathbf{a}| = |\mathbf{b} + \mathbf{c} + \mathbf{d}| \leq |\mathbf{b}| + |\mathbf{c}| + |\mathbf{d}|$$
(by triangle inequality applied repeatedly). So $|\mathbf{a}|$ can never exceed $|\mathbf{b}| + |\mathbf{c}| + |\mathbf{d}|$. ✓

(d) $\mathbf{b} + \mathbf{c}$ must lie in the plane of $\mathbf{a}$ and $\mathbf{d}$ if $\mathbf{a}$ and $\mathbf{d}$ are not collinear, and in the line of $\mathbf{a}$ and $\mathbf{d}$, if they are collinear.
Correct. From the given condition:
$$\mathbf{a} + (\mathbf{b} + \mathbf{c}) + \mathbf{d} = \mathbf{0}$$
$$\mathbf{b} + \mathbf{c} = -(\mathbf{a} + \mathbf{d})$$
The vector $-(\mathbf{a} + \mathbf{d})$ lies in the plane defined by $\mathbf{a}$ and $\mathbf{d}$ (if they are not collinear), or along their common line (if they are collinear). Hence $\mathbf{b} + \mathbf{c}$ must lie in the plane of $\mathbf{a}$ and $\mathbf{d}$ when they are not collinear, and along the line of $\mathbf{a}$ and $\mathbf{d}$ when they are collinear. ✓

Correct statements: (b), (c), and (d).

Given:
- Radius of circular ground, $r = 200$ m
- Starting point: $P$ (on the edge)
- Ending point: $Q$ (diametrically opposite to $P$)

Displacement:
Displacement is the straight-line distance from the initial position $P$ to the final position $Q$.

Since $P$ and $Q$ are diametrically opposite:
$$|\text{Displacement}| = \text{Diameter} = 2r = 2 \times 200 = 400 \text{ m}$$

The magnitude of displacement for each girl is 400 m, regardless of the path taken, because displacement depends only on the initial and final positions.

For which girl is path length = displacement?

Path length equals displacement only when the girl moves along a straight line from $P$ to $Q$ without changing direction. This is the girl who skates along the diameter (straight line path).

From the figure description, the girl who follows the straight-line path (diameter) has her path length equal to the displacement of 400 m.

Answer: Displacement = 400 m for all three girls. The displacement equals the actual path length only for the girl who skates along the straight-line diameter from $P$ to $Q$.

Given:
- Radius of circular park, $r = 1$ km $= 1000$ m
- The cyclist goes: $O \to P$ (radius), then along arc $P$ to $Q$ (quarter circumference), then $Q \to O$ (radius)
- Total time, $t = 10$ min $= 10 \times 60 = 600$ s

Path length calculation:
- $O$ to $P$: distance $= r = 1$ km
- Arc $P$ to $Q$ (quarter of circumference): distance $= \frac{1}{4} \times 2\pi r = \frac{\pi r}{2} = \frac{\pi \times 1}{2} = \frac{\pi}{2}$ km
- $Q$ to $O$: distance $= r = 1$ km

Total path length:
$$L = 1 + \frac{\pi}{2} + 1 = 2 + \frac{\pi}{2} \approx 2 + 1.571 = 3.571 \text{ km}$$

(a) Net Displacement:
The cyclist starts at $O$ and returns to $O$.
$$\text{Net displacement} = \mathbf{0} \text{ (zero)}$$

(b) Average Velocity:
$$\mathbf{v}_{\text{avg}} = \frac{\text{Net displacement}}{\text{Total time}} = \frac{0}{600} = \mathbf{0} \text{ m/s}$$

(c) Average Speed:
$$v_{\text{avg}} = \frac{\text{Total path length}}{\text{Total time}} = \frac{3.571 \times 1000}{600} \approx \frac{3571}{600} \approx 5.95 \text{ m/s}$$

Answers:
- (a) Net displacement = zero
- (b) Average velocity = zero
- (c) Average speed $\approx$ 5.95 m/s (approximately 6 m/s)

Given:
- The motorist turns left by $60°$ after every $500$ m.
- Each straight segment has length $500$ m.

Understanding the geometry:
At each turn, the motorist turns left by $60°$. The exterior angle is $60°$, so the interior angle of the polygon traced is $180° - 60° = 120°$. This means the motorist traces a regular hexagon (since $360°/60° = 6$ turns complete a full cycle).

Let the starting point be $A$, and let the motorist move along directions making angles $0°, 60°, 120°, 180°, 240°, 300°$ with the initial direction (turning left = counterclockwise).

Let each segment be along unit vectors. Taking initial direction as $+x$:
- Segment 1: direction $0°$
- Segment 2: direction $60°$
- Segment 3: direction $120°$
- Segment 4: direction $180°$
- Segment 5: direction $240°$
- Segment 6: direction $300°$
- Segment 7: direction $0°$ (same as segment 1, cycle repeats)
- Segment 8: direction $60°$

At the 3rd turn (after 3 segments, i.e., 1500 m of path):

Displacement = sum of 3 vectors of magnitude 500 m at $0°$, $60°$, $120°$:

$x$-component: $500(\cos 0° + \cos 60° + \cos 120°) = 500(1 + 0.5 - 0.5) = 500$ m

$y$-component: $500(\sin 0° + \sin 60° + \sin 120°) = 500(0 + \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}) = 500\sqrt{3}$ m

$$|\mathbf{d}_3| = \sqrt{500^2 + (500\sqrt{3})^2} = \sqrt{250000 + 750000} = \sqrt{1000000} = 1000 \text{ m}$$

Direction: $\theta = \tan^{-1}\left(\frac{500\sqrt{3}}{500}\right) = \tan^{-1}(\sqrt{3}) = 60°$ from initial direction.

Total path length at 3rd turn $= 3 \times 500 = 1500$ m

$$\frac{|\mathbf{d}_3|}{L_3} = \frac{1000}{1500} = \frac{2}{3}$$

At the 6th turn (after 6 segments, i.e., 3000 m of path):

The 6 segments complete a full regular hexagon, returning to the starting point.

$x$-component: $500(\cos 0° + \cos 60° + \cos 120° + \cos 180° + \cos 240° + \cos 300°)$
$= 500(1 + 0.5 - 0.5 - 1 - 0.5 + 0.5) = 500 \times 0 = 0$

$y$-component: $500(\sin 0° + \sin 60° + \sin 120° + \sin 180° + \sin 240° + \sin 300°)$
$= 500(0 + \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} + 0 - \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2}) = 0$

$$|\mathbf{d}_6| = 0 \text{ m}$$

Total path length at 6th turn $= 6 \times 500 = 3000$ m

The displacement is zero at the 6th turn.

At the 8th turn (after 8 segments, i.e., 4000 m of path):

Segments 7 and 8 repeat directions $0°$ and $60°$ (same as segments 1 and 2).

Net displacement = displacement after 6 turns + displacement due to segments 7 and 8
$= 0 + 500(\cos 0° + \cos 60°)\hat{\mathbf{i}} + 500(\sin 0° + \sin 60°)\hat{\mathbf{j}}$
$= 500(1 + 0.5)\hat{\mathbf{i}} + 500(0 + \frac{\sqrt{3}}{2})\hat{\mathbf{j}}$
$= 750\hat{\mathbf{i}} + 250\sqrt{3}\hat{\mathbf{j}}$ m

$$|\mathbf{d}_8| = \sqrt{750^2 + (250\sqrt{3})^2} = \sqrt{562500 + 187500} = \sqrt{750000} = 500\sqrt{3} \approx 866 \text{ m}$$

Direction: $\theta = \tan^{-1}\left(\frac{250\sqrt{3}}{750}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = 30°$ from initial direction.

Total path length at 8th turn $= 8 \times 500 = 4000$ m

$$\frac{|\mathbf{d}_8|}{L_8} = \frac{500\sqrt{3}}{4000} = \frac{\sqrt{3}}{8} \approx \frac{866}{4000} \approx 0.217$$

Summary:
| Turn | Displacement | Path Length | Ratio |
|---|---|---|---|
| 3rd | 1000 m at 60° | 1500 m | 2/3 |
| 6th | 0 m | 3000 m | 0 |
| 8th | $500\sqrt{3}$ m ≈ 866 m at 30° | 4000 m | $\sqrt{3}/8 \approx 0.22$ |

Given:
- Displacement (straight-line distance from station to hotel) $= 10$ km
- Total path length (circuitous route) $= 23$ km
- Time taken $= 28$ min $= \frac{28}{60}$ h $= \frac{7}{15}$ h

(a) Average speed of the taxi:
$$v_{\text{avg}} = \frac{\text{Total path length}}{\text{Time}} = \frac{23 \text{ km}}{\frac{28}{60} \text{ h}} = \frac{23 \times 60}{28} \text{ km/h}$$
$$v_{\text{avg}} = \frac{1380}{28} \approx 49.3 \text{ km/h}$$

Converting to m/s: $49.3 \times \frac{1000}{3600} \approx 13.7$ m/s

(b) Magnitude of average velocity:
$$|\mathbf{v}_{\text{avg}}| = \frac{|\text{Displacement}|}{\text{Time}} = \frac{10 \text{ km}}{\frac{28}{60} \text{ h}} = \frac{10 \times 60}{28} \text{ km/h}$$
$$|\mathbf{v}_{\text{avg}}| = \frac{600}{28} \approx 21.4 \text{ km/h}$$

Converting to m/s: $21.4 \times \frac{1000}{3600} \approx 5.95$ m/s

Are the two equal?
No, the average speed ($\approx 49.3$ km/h) is not equal to the magnitude of average velocity ($\approx 21.4$ km/h). They would be equal only if the path length equals the displacement, which is not the case here since the cabman took a circuitous route.

Given:
- Maximum height allowed (ceiling height), $h = 25$ m
- Initial speed of ball, $v_0 = 40$ m/s
- $g = 10$ m/s²

Concept: For projectile motion, the maximum height reached is:
$$h_m = \frac{(v_0 \sin\theta)^2}{2g}$$

For the ball not to hit the ceiling:
$$h_m \leq 25 \text{ m}$$
$$\frac{(v_0 \sin\theta)^2}{2g} = 25$$
$$(40 \sin\theta)^2 = 2 \times 10 \times 25 = 500$$
$$\sin^2\theta = \frac{500}{1600} = \frac{5}{16}$$
$$\sin\theta = \sqrt{\frac{5}{16}} = \frac{\sqrt{5}}{4}$$
$$\cos\theta = \sqrt{1 - \frac{5}{16}} = \sqrt{\frac{11}{16}} = \frac{\sqrt{11}}{4}$$

Horizontal range:
$$R = \frac{v_0^2 \sin 2\theta}{g} = \frac{v_0^2 \cdot 2\sin\theta\cos\theta}{g}$$
$$R = \frac{(40)^2 \times 2 \times \frac{\sqrt{5}}{4} \times \frac{\sqrt{11}}{4}}{10}$$
$$R = \frac{1600 \times 2 \times \frac{\sqrt{55}}{16}}{10}$$
$$R = \frac{1600 \times \frac{\sqrt{55}}{8}}{10} = \frac{200\sqrt{55}}{10} = 20\sqrt{55}$$
$$R = 20 \times 7.416 \approx 148.3 \text{ m}$$

The maximum horizontal distance the ball can travel without hitting the ceiling is approximately $\boxed{148.3}$ m.

Given:
- Maximum horizontal range, $R_{\max} = 100$ m
- $g = 10$ m/s²

Step 1: Find the initial speed.

Maximum range occurs at $\theta = 45°$:
$$R_{\max} = \frac{v_0^2}{g}$$
$$v_0^2 = R_{\max} \times g = 100 \times 10 = 1000 \text{ m}^2/\text{s}^2$$

Step 2: Find maximum height.

Maximum height is achieved when the ball is thrown vertically upward ($\theta = 90°$):
$$h_{\max} = \frac{v_0^2}{2g} = \frac{1000}{2 \times 10} = \frac{1000}{20} = 50 \text{ m}$$

The cricketer can throw the ball to a maximum height of $\boxed{50}$ m above the ground.

*Note:* The maximum height is half the maximum range, i.e., $h_{\max} = \frac{R_{\max}}{2}$.

Given:
- Length of string (radius), $r = 80$ cm $= 0.80$ m
- Number of revolutions $= 14$ in $25$ s
- Frequency, $\nu = \frac{14}{25}$ rev/s

Step 1: Find angular speed.
$$\omega = 2\pi\nu = 2\pi \times \frac{14}{25} = \frac{28\pi}{25} \text{ rad/s}$$

Step 2: Find centripetal acceleration.
$$a_c = \omega^2 r = \left(\frac{28\pi}{25}\right)^2 \times 0.80$$
$$a_c = \frac{784\pi^2}{625} \times 0.80$$
$$a_c = \frac{784 \times 9.87}{625} \times 0.80$$
$$a_c = \frac{7738.1}{625} \times 0.80 = 12.38 \times 0.80 \approx 9.91 \text{ m/s}^2$$

More precisely:
$$a_c = \frac{784 \times \pi^2 \times 0.8}{625} = \frac{784 \times 9.8696 \times 0.8}{625} \approx \frac{6187.6}{625} \approx 9.90 \text{ m/s}^2$$

Magnitude of acceleration $\approx 9.9$ m/s²

Direction: The acceleration (centripetal acceleration) is always directed towards the centre of the circular path, i.e., along the string towards the point of suspension, directed inward (centripetal direction).

Given:
- Radius of loop, $r = 1.00$ km $= 1000$ m
- Speed, $v = 900$ km/h $= 900 \times \frac{1000}{3600} = 250$ m/s
- $g = 9.8$ m/s²

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

Comparison with $g$:
$$\frac{a_c}{g} = \frac{62.5}{9.8} \approx 6.38$$

$$a_c \approx 6.38 \, g$$

The centripetal acceleration of the aircraft is approximately 6.38 times the acceleration due to gravity.

(a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre.

False. This is true only for uniform circular motion (constant speed), where the only acceleration is the centripetal acceleration directed towards the centre. In non-uniform circular motion, there is also a tangential component of acceleration (due to changing speed). The net acceleration is then the vector sum of centripetal and tangential accelerations, which is not directed purely 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.

True. The instantaneous velocity is defined as $\mathbf{v} = \frac{d\mathbf{r}}{dt}$, which gives the direction of motion at that instant. For any curved path, the direction of motion at any point is along the tangent to the curve at that point. This is a general result valid for any type of motion.

(c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector.

True. In uniform circular motion, the centripetal acceleration vector continuously changes direction (always pointing towards the centre, but the centre is fixed while the particle revolves). Over one complete cycle, the acceleration vectors are symmetrically distributed in all directions. By symmetry, their vector sum (and hence the average) over one complete revolution is zero:
$$\mathbf{a}_{\text{avg}} = \frac{1}{T}\int_0^T \mathbf{a}\, dt = \frac{\Delta\mathbf{v}}{T} = \frac{\mathbf{v}(T) - \mathbf{v}(0)}{T} = \frac{\mathbf{0}}{T} = \mathbf{0}$$
(since after one complete cycle, the velocity returns to its initial value). Hence the average acceleration is a null vector.

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

(a) Finding velocity $\mathbf{v}$ and acceleration $\mathbf{a}$:

Velocity is the time derivative of position:
$$\mathbf{v} = \frac{d\mathbf{r}}{dt} = \frac{d}{dt}(3.0t\hat{\mathbf{i}} - 2.0t^2\hat{\mathbf{j}} + 4.0\hat{\mathbf{k}})$$
$$\boxed{\mathbf{v} = 3.0\hat{\mathbf{i}} - 4.0t\hat{\mathbf{j}} \text{ m/s}}$$

Acceleration is the time derivative of velocity:
$$\mathbf{a} = \frac{d\mathbf{v}}{dt} = \frac{d}{dt}(3.0\hat{\mathbf{i}} - 4.0t\hat{\mathbf{j}})$$
$$\boxed{\mathbf{a} = -4.0\hat{\mathbf{j}} \text{ m/s}^2}$$

The acceleration is constant and directed along the negative $y$-axis.

(b) Magnitude and direction of velocity at $t = 2.0$ s:

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

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

Direction:
Let $\theta$ be the angle with the positive $x$-axis:
$$\tan\theta = \frac{v_y}{v_x} = \frac{-8.0}{3.0} = -2.667$$
$$\theta = \tan^{-1}(-2.667) \approx -69.4°$$

The velocity makes an angle of approximately 69.4° below the positive $x$-axis (i.e., $69.4°$ with the $x$-axis in the fourth quadrant, or equivalently $-69.4°$ from the $+x$ direction).

Given:
- Initial position: origin, $\mathbf{r}_0 = \mathbf{0}$
- Initial velocity: $\mathbf{v}_0 = 10.0\hat{\mathbf{j}}$ m/s (so $v_{0x} = 0$, $v_{0y} = 10.0$ m/s)
- Constant acceleration: $\mathbf{a} = 8.0\hat{\mathbf{i}} + 2.0\hat{\mathbf{j}}$ m/s² (so $a_x = 8.0$ m/s², $a_y = 2.0$ m/s²)

Equations of motion:
$$x = v_{0x}t + \frac{1}{2}a_x t^2 = 0 + \frac{1}{2}(8.0)t^2 = 4.0t^2$$
$$y = v_{0y}t + \frac{1}{2}a_y t^2 = 10.0t + \frac{1}{2}(2.0)t^2 = 10.0t + t^2$$

(a) Time when $x = 16$ m:
$$4.0t^2 = 16$$
$$t^2 = 4$$
$$t = 2 \text{ s}$$

$y$-coordinate at $t = 2$ s:
$$y = 10.0(2) + (2)^2 = 20 + 4 = 24 \text{ m}$$

(b) Speed at $t = 2$ s:

Velocity components:
$$v_x = v_{0x} + a_x t = 0 + 8.0 \times 2 = 16 \text{ m/s}$$
$$v_y = v_{0y} + a_y t = 10.0 + 2.0 \times 2 = 14 \text{ m/s}$$

Speed:
$$v = \sqrt{v_x^2 + v_y^2} = \sqrt{(16)^2 + (14)^2} = \sqrt{256 + 196} = \sqrt{452} \approx 21.3 \text{ m/s}$$

Answers:
- (a) $t = 2$ s; $y$-coordinate $= 24$ m
- (b) Speed $\approx 21.3$ m/s

Part 1: Magnitude and direction of $\hat{\mathbf{i}} + \hat{\mathbf{j}}$

$$|\hat{\mathbf{i}} + \hat{\mathbf{j}}| = \sqrt{1^2 + 1^2} = \sqrt{2}$$

Direction: $\theta = \tan^{-1}\left(\frac{1}{1}\right) = 45°$ with the positive $x$-axis.

Part 2: Magnitude and direction of $\hat{\mathbf{i}} - \hat{\mathbf{j}}$

$$|\hat{\mathbf{i}} - \hat{\mathbf{j}}| = \sqrt{1^2 + (-1)^2} = \sqrt{2}$$

Direction: $\theta = \tan^{-1}\left(\frac{-1}{1}\right) = -45°$ with the positive $x$-axis (i.e., $45°$ below the $x$-axis).

Part 3: Components of $\mathbf{A} = 2\hat{\mathbf{i}} + 3\hat{\mathbf{j}}$ along $\hat{\mathbf{i}} + \hat{\mathbf{j}}$ and $\hat{\mathbf{i}} - \hat{\mathbf{j}}$

Unit vector along $\hat{\mathbf{i}} + \hat{\mathbf{j}}$:
$$\hat{n}_1 = \frac{\hat{\mathbf{i}} + \hat{\mathbf{j}}}{\sqrt{2}}$$

Unit vector along $\hat{\mathbf{i}} - \hat{\mathbf{j}}$:
$$\hat{n}_2 = \frac{\hat{\mathbf{i}} - \hat{\mathbf{j}}}{\sqrt{2}}$$

Component of $\mathbf{A}$ along $\hat{\mathbf{i}} + \hat{\mathbf{j}}$:
$$A_1 = \mathbf{A} \cdot \hat{n}_1 = (2\hat{\mathbf{i}} + 3\hat{\mathbf{j}}) \cdot \frac{(\hat{\mathbf{i}} + \hat{\mathbf{j}})}{\sqrt{2}} = \frac{2 \times 1 + 3 \times 1}{\sqrt{2}} = \frac{5}{\sqrt{2}} = \frac{5\sqrt{2}}{2} \approx 3.54$$

Component of $\mathbf{A}$ along $\hat{\mathbf{i}} - \hat{\mathbf{j}}$:
$$A_2 = \mathbf{A} \cdot \hat{n}_2 = (2\hat{\mathbf{i}} + 3\hat{\mathbf{j}}) \cdot \frac{(\hat{\mathbf{i}} - \hat{\mathbf{j}})}{\sqrt{2}} = \frac{2 \times 1 + 3 \times (-1)}{\sqrt{2}} = \frac{2 - 3}{\sqrt{2}} = \frac{-1}{\sqrt{2}} = -\frac{\sqrt{2}}{2} \approx -0.71$$

Summary:
- $|\hat{\mathbf{i}} + \hat{\mathbf{j}}| = \sqrt{2}$, directed at $45°$ to $x$-axis
- $|\hat{\mathbf{i}} - \hat{\mathbf{j}}| = \sqrt{2}$, directed at $-45°$ to $x$-axis
- Component of $\mathbf{A}$ along $(\hat{\mathbf{i}} + \hat{\mathbf{j}})$ direction $= \dfrac{5}{\sqrt{2}}$
- Component of $\mathbf{A}$ along $(\hat{\mathbf{i}} - \hat{\mathbf{j}})$ direction $= \dfrac{-1}{\sqrt{2}}$

Context: The question asks which relations hold for any arbitrary motion (not necessarily uniform acceleration).

(a) $\mathbf{v}_{\text{average}} = \frac{1}{2}(\mathbf{v}(t_1) + \mathbf{v}(t_2))$

False (in general). This relation holds only for uniformly accelerated motion (constant acceleration), where velocity changes linearly with time. For arbitrary motion, the average velocity is not simply the arithmetic mean of initial and final velocities.

(b) $\mathbf{v}_{\text{average}} = \frac{\mathbf{r}(t_2) - \mathbf{r}(t_1)}{t_2 - t_1}$

True. This is the definition of average velocity — it is the ratio of displacement to the time interval. This holds for any type of motion, arbitrary or otherwise.

(c) $\mathbf{v}(t) = \mathbf{v}(0) + \mathbf{a}t$

False (in general). This relation is valid only for uniformly accelerated motion (constant $\mathbf{a}$). For arbitrary motion, acceleration may vary with time, so this equation does not hold in general.

(d) $\mathbf{r}(t) = \mathbf{r}(0) + \mathbf{v}(0)t + \frac{1}{2}\mathbf{a}t^2$

False (in general). This kinematic equation is valid only for constant acceleration. For arbitrary motion with variable acceleration, this equation does not apply.

(e) $\mathbf{a}_{\text{average}} = \frac{\mathbf{v}(t_2) - \mathbf{v}(t_1)}{t_2 - t_1}$

True. This is the definition of average acceleration — the change in velocity divided by the time interval. This holds for any type of motion.

Correct relations: (b) and (e)

(a) A scalar quantity is one that is conserved in a process.

False. Being a scalar has nothing to do with conservation. A scalar is defined by having only magnitude (no direction). Many scalars are not conserved (e.g., kinetic energy is a scalar but is not conserved in an inelastic collision). Conversely, some vector quantities (like momentum) are conserved.

(b) A scalar quantity can never take negative values.

False. Scalars can be negative. For example, temperature can be negative (e.g., $-10°C$), potential energy can be negative, and work done can be negative. The sign of a scalar indicates whether it is above or below a reference value.

(c) A scalar quantity must be dimensionless.

False. Scalars can have dimensions. For example, mass has dimensions $[M]$, temperature has dimensions $[\Theta]$, and energy has dimensions $[ML^2T^{-2}]$. Dimensionless quantities (like refractive index, strain) are a special subset of scalars.

(d) A scalar quantity does not vary from one point to another in space.

False. Scalar quantities can vary from point to point in space. For example, temperature, pressure, and gravitational potential energy all vary with position. These are called scalar fields.

(e) A scalar quantity has the same value for observers with different orientations of axes.

True. This is the defining property of a scalar. A scalar quantity is invariant under rotation of the coordinate system. For example, the mass of an object, its temperature, or the distance between two points remains the same regardless of how the coordinate axes are oriented. This distinguishes scalars from vectors (whose components change with axis orientation, though the magnitude remains the same).

Given:
- Height of aircraft above ground, $h = 3400$ m
- Time interval, $\Delta t = 10.0$ s
- Angle subtended at the observation point, $\angle = 30°$

Setting up the geometry:

Let $O$ be the observation point on the ground directly below the aircraft's path. Let $A$ and $B$ be the two positions of the aircraft, 10 s apart.

The aircraft flies horizontally at height $h = 3400$ m. The observation point $P$ is on the ground.

Assume the observation point is directly below the midpoint of $AB$ (for simplicity, or the aircraft passes directly overhead).

Actually, the standard interpretation: The aircraft flies horizontally. Let the observation point be $O$ on the ground. The aircraft is at height $h$. The two positions $A$ and $B$ subtend an angle of $30°$ at $O$.

Let the foot of the perpendicular from the aircraft's path to the ground be directly above $O$ (i.e., the aircraft passes directly overhead). Then by symmetry, each position makes an angle of $15°$ with the vertical.

The horizontal distance from $O$ to each position:
$$x = h \tan 15°$$

Total horizontal distance $AB$:
$$AB = 2h\tan 15°$$

Using $\tan 15° = \tan(45° - 30°) = \frac{\tan 45° - \tan 30°}{1 + \tan 45°\tan 30°} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = \frac{\sqrt{3}-1}{\sqrt{3}+1} = \frac{(\sqrt{3}-1)^2}{2} = 2 - \sqrt{3} \approx 0.2679$

$$AB = 2 \times 3400 \times 0.2679 = 6800 \times 0.2679 \approx 1821.7 \text{ m}$$

Speed of aircraft:
$$v = \frac{AB}{\Delta t} = \frac{1821.7}{10.0} \approx 182.2 \text{ m/s}$$

Converting to km/h: $182.2 \times 3.6 \approx 655.9$ km/h

The speed of the aircraft is approximately $\boxed{182}$ m/s (or about 656 km/h).