MOTION IN A PLANE – Guided Solutions

Q1 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.

Concept Brief

In physics, physical quantities are classified as scalars or vectors.

• Scalar quantities have only magnitude and no direction. Example: mass, speed.
• Vector quantities have both magnitude and direction. Example: velocity, acceleration.

Vectors are usually represented graphically using arrows showing both magnitude and direction.

Solution Map

• Step 1: Recall definitions of scalar and vector quantities.
• Step 2: Check whether each quantity requires direction.
• Step 3: If direction is required → Vector.
• Step 4: If only magnitude is required → Scalar.

Illustration: Scalar vs Vector

Solution

The classification of the given quantities is:

Physical Quantity	Type	Reason
Volume	Scalar	Has magnitude only (amount of space occupied)
Mass	Scalar	Measures quantity of matter
Speed	Scalar	Only magnitude of motion
Acceleration	Vector	Rate of change of velocity with direction
Density	Scalar	Mass per unit volume
Number of moles	Scalar	Amount of substance
Velocity	Vector	Speed with direction
Angular frequency	Scalar	Magnitude of rate of angular oscillation
Displacement	Vector	Directed change in position
Angular velocity	Vector	Has direction along axis of rotation

Q2

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Q3 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.

Concept Brief

Physical quantities are classified as scalars or vectors.

• Scalar quantities possess only magnitude (e.g., time, temperature).
• Vector quantities possess both magnitude and direction (e.g., velocity, force).

Impulse is related to the change in momentum and therefore inherits the vector nature of momentum.

Solution Map

• Step 1: Identify whether each quantity requires direction.
• Step 2: Most quantities listed (temperature, energy, power, etc.) depend only on magnitude.
• Step 3: Check the definition of impulse.
• Step 4: Since impulse equals change in momentum (a vector), impulse must also be a vector.

Illustration: Impulse and Change in Momentum

Solution

Let us classify the given quantities:

• Temperature → Scalar
• Pressure → Scalar
• Impulse → Vector
• Time → Scalar
• Power → Scalar
• Total path length → Scalar
• Energy → Scalar
• Gravitational potential → Scalar
• Coefficient of friction → Scalar
• Charge → Scalar

Therefore, the only vector quantity in the list is: Impulse.

Explanation

Impulse is defined as the product of force and the time interval during which the force acts:

\( \vec{J} = \vec{F}\,\Delta t \)

It is also equal to the change in momentum:

\( \vec{J} = \Delta \vec{p} = \vec{p}_2 - \vec{p}_1 \)

Since momentum is a vector quantity, impulse also has both magnitude and direction, making it a vector quantity.

Q4 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.

Concept Brief

In physics, algebraic operations on physical quantities must obey both mathematical rules and physical consistency.

• Scalars combine using ordinary algebra.
• Vectors combine using vector algebra.
• Only quantities representing the same physical type can be added.

Solution Map

• Step 1: Identify whether the quantities are scalars or vectors.
• Step 2: Apply rules of scalar and vector algebra.
• Step 3: Check dimensional and physical compatibility.
• Step 4: Decide whether the operation is meaningful.

Illustration: Vector Addition

Solution

(a) Adding any two scalars

This operation is meaningful only if the scalars represent the same physical quantity and have the same units. For example, adding two masses or two time intervals is valid.

(b) Adding a scalar to a vector of the same dimensions

This operation is not meaningful. Scalars and vectors belong to different mathematical categories and cannot be directly added.

(c) Multiplying any vector by any scalar

This operation is meaningful. If a vector \( \vec{A} \) is multiplied by a scalar \( k \), the result is another vector \( k\vec{A} \) whose magnitude changes by a factor \( k \), while the direction remains the same (or reverses if \( k \) is negative).

(d) Multiplying any two scalars

This operation is meaningful, because the product of two scalars is always another scalar. Example: density × volume = mass.

(e) Adding any two vectors

This operation is meaningful only if the vectors represent the same physical quantity and have the same units (e.g., adding two forces or two displacements). Otherwise the result has no physical meaning.

(f) Adding a component of a vector to the same vector

This operation is not meaningful. A component of a vector represents only a part of the vector along one direction. Adding it directly to the original vector would mix quantities that represent different physical magnitudes.

Q5 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 is either greater than or equal to the magnitude of the average velocity,
(e) Three vectors not lying in a plane can never add up to give a null vector.

Concept Brief

Vectors possess both magnitude and direction, while scalars have only magnitude. When vectors are resolved into components along coordinate axes, each component represents the projection of the vector along that axis.

Also, in motion problems:

• Path length is the total distance travelled.
• Displacement is the shortest directed distance between initial and final positions.

Solution Map

• Step 1: Recall properties of vector magnitude and components.
• Step 2: Compare definitions of distance and displacement.
• Step 3: Compare formulas of average speed and average velocity.
• Step 4: Apply geometric conditions for vectors forming a closed polygon.

Illustration: Path Length vs Displacement

Solution

(a) The magnitude of a vector is always a scalar – True.

The magnitude of a vector represents only its size and has no direction. Therefore it behaves as a scalar quantity.

(b) Each component of a vector is always a scalar – True.

When a vector is resolved into components along coordinate axes, each component represents the projection of the vector along that axis. These components have magnitude only and are therefore scalars.

(c) The total path length is always equal to the magnitude of displacement – False.

Total path length represents the entire distance travelled by the particle, whereas displacement is the shortest distance between the initial and final positions. Hence,

Path length ≥ magnitude of displacement.

They become equal only when motion occurs along a straight line without changing direction.

(d) Average speed ≥ magnitude of average velocity – True.

Average speed is defined as

\( \text{Average speed} = \frac{\text{total path length}}{\text{total time}} \)

Magnitude of average velocity is

\( |\vec{v}_{avg}| = \frac{\text{displacement}}{\text{time}} \)

Since path length ≥ displacement, average speed is always greater than or equal to the magnitude of average velocity.

(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 zero, they must form the sides of a closed triangle taken in sequence. A triangle lies in a single plane. Therefore the vectors must be coplanar. If they are not in the same plane, their vector sum cannot be zero.

Q6 Establish the following vector inequalities geometrically or otherwise:
(a) \( |\vec a+\vec b|\le |\vec a|+|\vec b| \)
(b) \( |\vec a+\vec b|\ge \big||\vec a|-|\vec b|\big| \)
(c) \( |\vec a-\vec b|\le |\vec a|+|\vec b| \)
(d) \( |\vec a-\vec b|\ge \big||\vec a|-|\vec b|\big| \)
When does the equality sign apply?

Concept Brief

These inequalities follow from the triangle law of vector addition. When two vectors are placed head-to-tail, they form two sides of a triangle, while their resultant represents the third side.

From basic geometry of triangles:

• The length of any side of a triangle is less than or equal to the sum of the other two sides.
• The length of any side is greater than or equal to the difference of the other two sides.

Solution Map

• Step 1: Represent vectors \(\vec a\) and \(\vec b\) graphically.
• Step 2: Apply triangle law of vector addition.
• Step 3: Use geometric properties of triangles.
• Step 4: Obtain upper and lower bounds for resultant magnitude.

Illustration: Triangle Law of Vector Addition

Solution

(a) Upper bound inequality

From the triangle law,

\( |\vec a+\vec b|\le |\vec a|+|\vec b| \)

because the length of one side of a triangle cannot exceed the sum of the other two sides.

(b) Lower bound inequality

From triangle geometry,

\( |\vec a+\vec b|\ge \big||\vec a|-|\vec b|\big| \)

since the length of one side of a triangle cannot be smaller than the difference of the other two sides.

(c) Inequality for \( \vec a-\vec b \)

Since

\( \vec a-\vec b=\vec a+(-\vec b) \)

applying the triangle inequality gives

\( |\vec a-\vec b|\le |\vec a|+|\vec b| \)

(d) Lower bound for \( \vec a-\vec b \)

Similarly,

\( |\vec a-\vec b|\ge \big||\vec a|-|\vec b|\big| \)

When does equality occur?

Equality occurs when the vectors are collinear.

• \( |\vec a+\vec b| = |\vec a| + |\vec b| \) when vectors are in the same direction.
• \( |\vec a+\vec b| = \big||\vec a|-|\vec b|\big| \) when vectors are in opposite directions.
• \( |\vec a-\vec b| = |\vec a| + |\vec b| \) when \( \vec a \) and \( \vec b \) are in opposite directions.
• \( |\vec a-\vec b| = \big||\vec a|-|\vec b|\big| \) when \( \vec a \) and \( \vec b \) are in the same direction.

Q7 Given \( \vec a + \vec b + \vec c + \vec d = 0 \), which of the following statements are correct :
(a) \( \vec a,\vec b,\vec c,\vec d \) 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 the magnitudes of \( \vec b,\vec c,\vec d \),
(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 in the line of \( \vec a \) and \( \vec d \) if they are collinear.

Concept Brief

If several vectors add up to zero, they can be represented geometrically as the sides of a closed polygon taken in sequence.

Thus the relation

\( \vec a + \vec b + \vec c + \vec d = 0 \)

means that the vectors \( \vec a,\vec b,\vec c,\vec d \) form a closed quadrilateral.

Solution Map

• Step 1: Interpret the vector equation geometrically.
• Step 2: Rearrange the equation to compare vector sums.
• Step 3: Apply triangle inequality where needed.
• Step 4: Analyze geometric relations between vectors.

Illustration: Closed Vector Polygon

Solution

From the given relation

\( \vec a + \vec b + \vec c + \vec d = 0 \)

the vectors form a closed quadrilateral.

(a) Statement (a)

The vectors need not individually be zero. Non-zero vectors can add to zero when they form a closed polygon.

Therefore, statement (a) is false.

(b) Statement (b)

Rearranging the given equation:

\( \vec a + \vec c = -(\vec b + \vec d) \)

Thus the vectors \( \vec a+\vec c \) and \( \vec b+\vec d \) are equal in magnitude and opposite in direction.

Therefore,

\( |\vec a+\vec c| = |\vec b+\vec d| \)

Hence statement (b) is true.

(c) Statement (c)

Rewriting the equation:

\( \vec a = -(\vec b+\vec c+\vec d) \)

Taking magnitudes and using the triangle inequality,

\( |\vec a| = |\vec b+\vec c+\vec d| \le |\vec b|+|\vec c|+|\vec d| \)

Thus the magnitude of \( \vec a \) cannot exceed the sum of the magnitudes of the other three vectors.

Therefore statement (c) is true.

(d) Statement (d)

From the original equation,

\( \vec a + (\vec b+\vec c) + \vec d = 0 \)

Thus the three vectors \( \vec a,\ \vec d,\ (\vec b+\vec c) \) must form a triangle.

Therefore \( \vec b+\vec c \) must lie in the plane containing \( \vec a \) and \( \vec d \). If \( \vec a \) and \( \vec d \) are collinear, then \( \vec b+\vec c \) must lie along the same line.

Hence statement (d) is true.

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

Q10 On an open ground, a motorist follows a track that turns to his left by an angle of \(60^\circ\) 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.

Concept Brief

Since the motorist turns left by \(60^\circ\) after every equal segment of \(500\,\text{m}\), the successive path segments form the sides of a regular hexagon.

Key idea:

• 6 sides complete one hexagon.
• After 6 sides the motorist returns to the starting point.

Solution Map

• Step 1: Recognize the track as a regular hexagon.
• Step 2: Determine the position after the required number of sides.
• Step 3: Find the displacement from the starting point.
• Step 4: Compare displacement with total path length.

Q11 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?

Concept Brief

Average speed is defined as:

\[ \text{Average speed}=\frac{\text{Total distance travelled}}{\text{Total time}} \]

Magnitude of average velocity is defined as:

\[ |\vec v_{avg}|=\frac{\text{Displacement}}{\text{Total time}} \]

Distance depends on the path taken, whereas displacement depends only on the initial and final positions.

Solution Map

• Step 1: Identify distance travelled and displacement.
• Step 2: Convert time into hours.
• Step 3: Calculate average speed.
• Step 4: Calculate magnitude of average velocity.
• Step 5: Compare the two results.

Illustration: Distance vs Displacement

Solution

Given

Total distance travelled:

\[ d = 23\ \text{km} \]

Displacement:

\[ s = 10\ \text{km} \]

Time taken:

\[ t = 28\ \text{min} \] \[ t = \frac{28}{60}\ \text{h} \]

(a) Average speed

\[ v_{avg} = \frac{\text{distance}}{\text{time}} \] \[ v_{avg} = \frac{23}{28/60} \] \[ v_{avg} = \frac{23 \times 60}{28} \] \[ v_{avg} \approx 49.3\ \text{km h}^{-1} \]

(b) Magnitude of average velocity

\[ |\vec v_{avg}|=\frac{\text{displacement}}{\text{time}} \] \[ |\vec v_{avg}|=\frac{10}{28/60} \] \[ |\vec v_{avg}|=\frac{10 \times 60}{28} \] \[ |\vec v_{avg}| \approx 21.4\ \text{km h}^{-1} \]

Comparison

\[ \text{Average speed} \approx 49.3\ \text{km h}^{-1} \] \[ \text{Magnitude of average velocity} \approx 21.4\ \text{km h}^{-1} \]

The two are not equal because the path taken by the taxi is not a straight line. Hence the distance travelled (23 km) is greater than the displacement (10 km).

Q12 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?

Concept Brief

For a projectile thrown with speed \(u\) at an angle \( \theta \):

• Maximum height

\[ H=\frac{u^{2}\sin^{2}\theta}{2g} \]

• Horizontal range

\[ R=\frac{u^{2}\sin 2\theta}{g} \]

To obtain the maximum horizontal distance without touching the ceiling, the projectile must reach a maximum height exactly equal to the ceiling height.

Solution Map

• Step 1: Set projectile maximum height equal to the ceiling height.
• Step 2: Determine the angle of projection.
• Step 3: Substitute the angle into the range formula.

Projectile Illustration

Solution

Given

\[ u = 40 \text{ m s}^{-1} \] \[ H = 25 \text{ m} \] \[ g \approx 10 \text{ m s}^{-2} \]

Step 1: Maximum height condition

\[ H=\frac{u^{2}\sin^{2}\theta}{2g} \] Substitute the values: \[ 25=\frac{40^{2}\sin^{2}\theta}{2\times10} \] \[ 25=\frac{1600\sin^{2}\theta}{20} \] \[ 25=80\sin^{2}\theta \] \[ \sin^{2}\theta=\frac{5}{16} \] \[ \sin\theta=\frac{\sqrt5}{4} \]

Step 2: Determine \( \cos\theta \)

\[ \cos^{2}\theta=1-\sin^{2}\theta \] \[ \cos^{2}\theta=\frac{11}{16} \] \[ \cos\theta=\frac{\sqrt{11}}{4} \]

Step 3: Range of projectile

\[ R=\frac{u^{2}\sin2\theta}{g} \] \[ \sin2\theta=2\sin\theta\cos\theta \] \[ \sin2\theta=2\left(\frac{\sqrt5}{4}\right)\left(\frac{\sqrt{11}}{4}\right) \] Substituting: \[ R=\frac{40^{2}}{10}\times2\sin\theta\cos\theta \] \[ R=\frac{1600}{10}\times 2\left(\frac{\sqrt5}{4}\right)\left(\frac{\sqrt{11}}{4}\right) \] \[ R=20\sqrt{55} \]

Numerical value

\[ R=20\sqrt{55} \] \[ R\approx148.3\text{ m} \]

Maximum horizontal distance ≈ 148 m

Q13 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?

Concept Brief

For projectile motion:

• Maximum range occurs when the angle of projection \( \theta = 45^\circ \)

\[ R_{\max}=\frac{u^2}{g} \]

• Maximum height when thrown vertically upward:

\[ H=\frac{u^2}{2g} \]

Thus if the initial speed is the same:

\[ H=\frac{R_{\max}}{2} \]

Solution Map

• Step 1: Use the maximum range formula to determine \(u^2\).
• Step 2: Substitute the same speed into the vertical height formula.
• Step 3: Compute the maximum height.

Concept Illustration

Solution

Given

\[ R_{\max}=100\text{ m} \]

Step 1: Determine initial speed

\[ R_{\max}=\frac{u^2}{g} \] \[ 100=\frac{u^2}{g} \] \[ u^2=100g \] Taking \[ g=10\text{ m s}^{-2} \] \[ u^2=100\times10=1000 \]

Step 2: Maximum vertical height

\[ H=\frac{u^2}{2g} \] \[ H=\frac{1000}{2\times10} \] \[ H=\frac{1000}{20} \] \[ H=50\text{ m} \]

Maximum height = 50 m

Q14 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?

Concept Brief

In uniform circular motion, even though the speed remains constant, the velocity continuously changes direction. Hence the particle has an acceleration called centripetal acceleration.

Its magnitude is given by:

a = \frac{v^2}{r}

where \(v\) is the speed of the particle and \(r\) is the radius of the circle. The direction of this acceleration is always toward the centre of the circle.

Solution Map

• Step 1: Determine the time period of one revolution.
• Step 2: Calculate the speed of the stone.
• Step 3: Use the centripetal acceleration formula.
• Step 4: State the direction of acceleration.

Circular Motion Illustration

Solution

Given

\[ r = 80 \text{ cm} = 0.8 \text{ m} \] Number of revolutions = 14 Time taken = 25 s

Step 1: Time period

\[ T = \frac{\text{total time}}{\text{number of revolutions}} \] \[ T = \frac{25}{14} \text{ s} \]

Step 2: Speed of the stone

\[ v = \frac{2\pi r}{T} \] \[ v = \frac{2\pi \times 0.8}{25/14} \] \[ v = 2\pi \times 0.8 \times \frac{14}{25} \] Using \( \pi \approx \frac{22}{7} \): \[ v \approx 2.82 \text{ m s}^{-1} \]

Step 3: Centripetal acceleration

\[ a = \frac{v^2}{r} \] \[ a = \frac{(2.82)^2}{0.8} \] \[ a \approx 9.9 \text{ m s}^{-2} \]

Direction of acceleration

The acceleration is directed towards the centre of the circular path. This is called centripetal acceleration.

Magnitude of acceleration ≈ \(9.9\text{ m s}^{-2}\)
Direction: toward the centre of the circle.

Q15 An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km h⁻¹. Compare its centripetal acceleration with the acceleration due to gravity.

Concept Brief

In uniform circular motion, a body moving with speed \(v\) in a circle of radius \(r\) experiences centripetal acceleration directed toward the centre.

a = \frac{v^2}{r}

To compare with gravity, we evaluate the ratio \(a/g\).

Solution Map

• Step 1: Convert speed to SI units.
• Step 2: Calculate centripetal acceleration.
• Step 3: Compare the value with gravitational acceleration \(g\).

Circular Flight Illustration

Solution

Given

\[ r = 1.00\text{ km} = 1000\text{ m} \] \[ v = 900\text{ km h}^{-1} \]

Step 1: Convert speed to SI units

\[ v = 900 \times \frac{5}{18} \] \[ v = 250\text{ m s}^{-1} \]

Step 2: Centripetal acceleration

\[ a = \frac{v^2}{r} \] \[ a = \frac{250^2}{1000} \] \[ a = \frac{62500}{1000} \] \[ a = 62.5\text{ m s}^{-2} \]

Step 3: Comparison with gravity

\[ g \approx 9.8\text{ m s}^{-2} \] \[ \frac{a}{g}=\frac{62.5}{9.8} \] \[ \frac{a}{g}\approx6.4 \]

Centripetal acceleration ≈ \(62.5\text{ m s}^{-2}\)
This is about \(6.4\,g\), i.e. roughly 6.4 times the acceleration due to gravity.

Q16 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.

Concept Brief

In circular motion, the velocity of a particle is always tangent to the path, while acceleration can have two components:

• Centripetal (radial) acceleration directed towards the centre.
• Tangential acceleration along the tangent when the speed changes.

For uniform circular motion, the speed is constant, so only centripetal acceleration exists.

Circular Motion Illustration

Solution

(a)

Statement: The net acceleration of a particle in circular motion is always along the radius towards the centre.

False.

In general circular motion (non-uniform), two acceleration components exist:

• Centripetal acceleration directed towards the centre.
• Tangential acceleration along the tangent if the speed changes.

Therefore, the resultant acceleration is not always purely radial. Only in uniform circular motion is the acceleration purely centripetal.

(b)

Statement: The velocity vector of a particle at a point is always along the tangent to the path.

True.

Instantaneous velocity gives the direction of motion of the particle. For motion along a curved path, the direction of motion at any point is along the tangent to the trajectory.

(c)

Statement: The acceleration vector in uniform circular motion averaged over one complete cycle is zero.

True.

Although the magnitude of centripetal acceleration remains constant, its direction continuously changes during the motion. After one complete revolution the acceleration vectors in all directions cancel each other.

Hence the vector average of acceleration over a full cycle is

\[ \vec{a}_{avg}=0 \]

(a) False    (b) True    (c) True

Q17 The position of a particle is given by \( \vec r = 3.0t\,\hat{i} + 2.0t^2\,\hat{j} + 4.0\,\hat{k} \) m where \(t\) is in seconds. (a) Find the velocity \( \vec v \) and acceleration \( \vec a \). (b) What is the magnitude and direction of velocity at \( t=2.0\,\text{s} \)?

Concept Brief

In vector kinematics:

• Velocity = time derivative of position
• Acceleration = time derivative of velocity

\[ \vec v=\frac{d\vec r}{dt}, \qquad \vec a=\frac{d\vec v}{dt} \]

Solution Map

• Step 1: Differentiate the position vector to obtain velocity.
• Step 2: Differentiate velocity to obtain acceleration.
• Step 3: Substitute \(t=2\,\text{s}\) to find velocity components.
• Step 4: Compute magnitude and direction.

Velocity Direction Illustration

Solution

(a) Velocity and acceleration

Position vector: \[ \vec r(t)=3.0t\,\hat i + 2.0t^2\,\hat j + 4.0\,\hat k \] Velocity: \[ \vec v(t)=\frac{d\vec r}{dt} \] \[ \vec v(t)=3.0\,\hat i + 4.0t\,\hat j \] Acceleration: \[ \vec a(t)=\frac{d\vec v}{dt} \] \[ \vec a(t)=4.0\,\hat j \]

(b) Velocity at \(t=2\,\text{s}\)

\[ \vec v(2)=3\,\hat i + 4(2)\,\hat j \] \[ \vec v(2)=3\,\hat i + 8\,\hat j \]

Magnitude of velocity

\[ |\vec v|=\sqrt{3^2+8^2} \] \[ |\vec v|=\sqrt{73} \] \[ |\vec v|\approx8.54\ \text{m s}^{-1} \]

Direction

Angle with the positive \(x\)-axis: \[ \tan\theta=\frac{v_y}{v_x} \] \[ \tan\theta=\frac{8}{3} \] \[ \theta=\tan^{-1}\!\left(\frac{8}{3}\right) \] \[ \theta\approx69.4^\circ \]

Velocity vector at \(t=2\,\text{s}\): \( \vec v = 3\hat i + 8\hat j \) Speed ≈ \(8.54\,\text{m s}^{-1}\) Direction ≈ \(69.4^\circ\) above the positive \(x\)-axis.

Q18 The position of a particle is given by \( \vec r = 3.0t\,\hat{i} + 2.0t^2\,\hat{j} + 4.0\,\hat{k} \) m where \(t\) is in seconds. (a) Find the velocity \( \vec v \) and acceleration \( \vec a \). (b) What is the magnitude and direction of velocity at \( t=2.0\,\text{s} \)?

Concept Brief

In vector kinematics:

• Velocity = time derivative of position
• Acceleration = time derivative of velocity

\[ \vec v=\frac{d\vec r}{dt}, \qquad \vec a=\frac{d\vec v}{dt} \]

Solution Map

• Step 1: Differentiate the position vector to obtain velocity.
• Step 2: Differentiate velocity to obtain acceleration.
• Step 3: Substitute \(t=2\,\text{s}\) to find velocity components.
• Step 4: Compute magnitude and direction.

Velocity Direction Illustration

Solution

(a) Velocity and acceleration

Position vector: \[ \vec r(t)=3.0t\,\hat i + 2.0t^2\,\hat j + 4.0\,\hat k \] Velocity: \[ \vec v(t)=\frac{d\vec r}{dt} \] \[ \vec v(t)=3.0\,\hat i + 4.0t\,\hat j \] Acceleration: \[ \vec a(t)=\frac{d\vec v}{dt} \] \[ \vec a(t)=4.0\,\hat j \]

(b) Velocity at \(t=2\,\text{s}\)

\[ \vec v(2)=3\,\hat i + 4(2)\,\hat j \] \[ \vec v(2)=3\,\hat i + 8\,\hat j \]

Magnitude of velocity

\[ |\vec v|=\sqrt{3^2+8^2} \] \[ |\vec v|=\sqrt{73} \] \[ |\vec v|\approx8.54\ \text{m s}^{-1} \]

Direction

Angle with the positive \(x\)-axis: \[ \tan\theta=\frac{v_y}{v_x} \] \[ \tan\theta=\frac{8}{3} \] \[ \theta=\tan^{-1}\!\left(\frac{8}{3}\right) \] \[ \theta\approx69.4^\circ \]

Velocity vector at \(t=2\,\text{s}\): \( \vec v = 3\hat i + 8\hat j \) Speed ≈ \(8.54\,\text{m s}^{-1}\) Direction ≈ \(69.4^\circ\) above the positive \(x\)-axis.

Q19 \( \hat{i} \) and \( \hat{j} \) are unit vectors along the x- and y-axes respectively. Find the magnitude and direction of the vectors \( \hat{i}+\hat{j} \) and \( \hat{i}-\hat{j} \). Also find the components of the vector \( \vec A = 2\hat{i}+3\hat{j} \) along the directions of \( \hat{i}+\hat{j} \) and \( \hat{i}-\hat{j} \).

Concept Brief

A vector with components \( (a,b) \) has magnitude

\[ |\vec v|=\sqrt{a^2+b^2} \]

and direction given by

\[ \tan\theta=\frac{b}{a}. \]

The component of a vector along a direction is obtained using the dot product with the corresponding unit vector.

Geometric Illustration

Solution

1️⃣ Vector \( \hat{i}+\hat{j} \)

Components: \[ (1,1) \] Magnitude: \[ |\hat{i}+\hat{j}|=\sqrt{1^2+1^2}=\sqrt2 \] Direction: \[ \tan\theta=\frac{1}{1}=1 \] \[ \theta=45^\circ \] Thus this vector is directed at \(45^\circ\) in the first quadrant.

2️⃣ Vector \( \hat{i}-\hat{j} \)

Components: \[ (1,-1) \] Magnitude: \[ |\hat{i}-\hat{j}|=\sqrt{1^2+(-1)^2}=\sqrt2 \] Direction: \[ \tan\theta=-1 \] \[ \theta=-45^\circ \] Thus the vector lies \(45^\circ\) below the positive x-axis.

3️⃣ Components of \( \vec A = 2\hat{i}+3\hat{j} \)

Unit vector along \( \hat{i}+\hat{j} \): \[ \hat u_1=\frac{\hat{i}+\hat{j}}{\sqrt2} \] Component of \( \vec A \) along this direction: \[ A_1=\vec A\cdot\hat u_1 \] \[ A_1=(2\hat{i}+3\hat{j})\cdot \frac{\hat{i}+\hat{j}}{\sqrt2} \] \[ A_1=\frac{2+3}{\sqrt2} \] \[ A_1=\frac{5}{\sqrt2} \]

4️⃣ Component along \( \hat{i}-\hat{j} \)

Unit vector: \[ \hat u_2=\frac{\hat{i}-\hat{j}}{\sqrt2} \] Projection: \[ A_2=\vec A\cdot\hat u_2 \] \[ A_2=(2\hat{i}+3\hat{j})\cdot \frac{\hat{i}-\hat{j}}{\sqrt2} \] \[ A_2=\frac{2-3}{\sqrt2} \] \[ A_2=-\frac{1}{\sqrt2} \]

Magnitude of \( \hat{i}+\hat{j} \) = \( \sqrt2 \), direction \(45^\circ\)
Magnitude of \( \hat{i}-\hat{j} \) = \( \sqrt2 \), direction \(-45^\circ\)
Component of \( \vec A \) along \( \hat{i}+\hat{j} \) = \( \frac{5}{\sqrt2} \)
Component along \( \hat{i}-\hat{j} \) = \( -\frac{1}{\sqrt2} \)

Q20 For any arbitrary motion in space, which of the following relations are true?

(a) \( \vec v_{avg} = \tfrac12[\vec v(t_1)+\vec v(t_2)] \)
(b) \( \vec v_{avg}=\dfrac{\vec r(t_2)-\vec r(t_1)}{t_2-t_1} \)
(c) \( \vec v(t)=\vec v(0)+\vec a t \)
(d) \( \vec r(t)=\vec r(0)+\vec v(0)t+\tfrac12\vec a t^2 \)
(e) \( \vec a_{avg}=\dfrac{\vec v(t_2)-\vec v(t_1)}{t_2-t_1} \)

Concept Brief

For motion with arbitrary acceleration:

• Average velocity and average acceleration are defined by their mathematical definitions.
• The familiar kinematic equations \(v=v_0+at\) and \(r=r_0+v_0t+\tfrac12at^2\) are valid only when acceleration is constant.

Velocity–Time Illustration

Solution

(a) \( \vec v_{avg} = \frac12[\vec v(t_1)+\vec v(t_2)] \)
False. This expression is valid only when velocity varies linearly with time (i.e. constant acceleration). For arbitrary motion the velocity–time relation is not linear.

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(b) \[ \vec v_{avg}= \frac{\vec r(t_2)-\vec r(t_1)}{t_2-t_1} \] True. This is the definition of average velocity. It holds for any type of motion.

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(c) \[ \vec v(t)=\vec v(0)+\vec a t \] False in general. This equation assumes constant acceleration. If acceleration varies with time, velocity must be obtained by integration.

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(d) \[ \vec r(t)=\vec r(0)+\vec v(0)t+\frac12\vec a t^2 \] False in general. This kinematic equation is derived assuming constant acceleration. It does not apply when acceleration varies with time.

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(e) \[ \vec a_{avg}= \frac{\vec v(t_2)-\vec v(t_1)}{t_2-t_1} \] True. This is the definition of average acceleration over the interval \([t_1,t_2]\), valid for any motion.

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True relations: (b) and (e)

Q21 Read each statement below carefully and state, with reasons and examples, whether 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

Concept Brief

A scalar quantity is a physical quantity that has magnitude only and no direction. Its value does not depend on the orientation of the coordinate axes.

Examples: mass, time, temperature, energy.

Solution

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

False.

Conservation is a physical law, not a defining property of scalars. Some scalars are conserved while others are not.

Examples:

• Total energy of an isolated system is conserved.
• Temperature is a scalar but it is not conserved in a process.

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(b) A scalar quantity can never take negative values

False.

Scalars can have positive, negative, or zero values depending on the chosen reference.

Examples:

• Electric potential may be negative.
• Temperature in Celsius scale may be negative.

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(c) A scalar quantity must be dimensionless

False.

Many scalars possess physical dimensions.

Examples:

• Mass (kg)
• Time (s)
• Energy (J)

Dimensionless scalars such as refractive index are only a special case.

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

False.

A scalar can vary with position and form a scalar field.

Examples:

• Temperature distribution in a room.
• Electric potential around a charge.

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

True.

The defining property of a scalar is that its value is independent of the orientation of the coordinate system. Rotating the axes does not change the value of scalar quantities.

Examples: mass, time interval, temperature.

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Correct conclusion: (a) False (b) False (c) False (d) False (e) True