Q7: 3.7 Given a+b+c+d=0, which of the following statements are correct: (a) a, b, c, and d must each be a null vector, (b) The magnitude of (a+c) equals…

Given the relation: a+b+c+d=0 (a) Incorrect. The vectors do not need to be null vectors. For example, if a = hati, b = hatj, c = -hati, and d = -hatj, their sum is zero, but none of them is a null vector. (b) Correct. From the given relation, we can write a+c = -(b+d). Taking the magnitude on both…

Q10: 3.10 On an open ground, a motorist follows a track that turns to his left by an angle of 60^° after every 500 m. Starting from a given turn, specify…

The motorist turns left by 60^° after every 500 m. This means the path is a regular hexagon with a side length of a = 500 m. Let the motorist start at point A. At the third turn: The motorist moves from A to B, B to C, and C to D. The turns happen at B, C, and D. The third turn is at D. Path length…

Q18: 3.18 A particle starts from the origin at t=0 s with a velocity of 10.0 hatj m/s and moves in the x-y plane with a constant acceleration of (8.0…

Given: Initial position, r_0 = 0 (starts from origin) Initial velocity, v_0 = 10.0 hatj m/s (so, v_0x = 0, v_0y = 10.0 m/s) Constant acceleration, a = (8.0 hati + 2.0 hatj) m/s^2 (so, a_x = 8.0 m/s^2, a_y = 2.0 m/s^2 ) Formula: The position of the particle at time t is given by: r(t) = r_0 + v_0 t…

Motion In A Plane Class 11 Physics NCERT Solutions

Q1EXERCISES

3.1 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.

Solution

Scalar Quantities: These quantities have only magnitude and no direction.

Volume: Has magnitude only (e.g., 2 cubic meters).
Mass: Has magnitude only (e.g., 5 kg).
Speed: Has magnitude only (e.g., 10 m/s).
Density: Has magnitude only (e.g., 1000 kg/m³).
Number of moles: Has magnitude only (e.g., 2 moles).
Angular frequency: Has magnitude only (e.g., 10 rad/s).

Vector Quantities: These quantities have both magnitude and direction.

Acceleration: Has both magnitude (e.g., 9.8 m/s²) and direction (e.g., downwards).
Velocity: Has both magnitude (speed) and direction (e.g., 10 m/s due north).
Displacement: Has both magnitude (distance between initial and final points) and direction.
Angular velocity: Has both magnitude (rate of rotation) and direction (given by the right-hand thumb rule along the axis of rotation).

Q2EXERCISES

3.2 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.

Solution

The two scalar quantities in the given list are:

Work: Work is defined as the dot product of force and displacement vectors ( $W = \mathbf{F} \cdot \mathbf{d}$ ). The dot product of two vectors is a scalar quantity. It has magnitude but no direction.
Current: Electric current is a scalar quantity. Although it has a direction of flow, it does not obey the laws of vector addition (e.g., currents at a junction are added algebraically, not vectorially).

Q3EXERCISES

3.3 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.

Solution

The only vector quantity in the given list is Impulse.

Reason: Impulse is defined as the change in momentum of an object, which is a vector quantity. It is also defined as the product of the force (a vector) and the time interval over which the force acts (a scalar). The product of a vector and a scalar is a vector. The direction of impulse is the same as the direction of the force.

\text{Impulse} = \mathbf{F} \times \Delta t = \Delta \mathbf{p}

All other quantities in the list are scalars:

Temperature, Time, Power, Total path length, Energy, Gravitational potential, Coefficient of friction, Charge: These are specified by magnitude only.
Pressure: Pressure is force per unit area. Although force is a vector, pressure is defined as a scalar because it acts equally in all directions at a point in a fluid.

Q4EXERCISES

3.4 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.

Solution

(a) Adding any two scalars: Not meaningful. Two scalars can be added only if they represent the same physical quantity and have the same units. For example, adding mass (2 kg) to temperature (300 K) is meaningless.

(b) Adding a scalar to a vector of the same dimensions: Not meaningful. A scalar cannot be added to a vector because a vector has direction while a scalar does not. Physical quantities of different types cannot be added.

(c) Multiplying any vector by any scalar: Meaningful. When a vector is multiplied by a scalar, the result is a new vector. Its magnitude is the product of the magnitudes of the scalar and the vector, and its direction is the same as (or opposite to) the original vector, depending on whether the scalar is positive or negative. For example, multiplying velocity (a vector) by time (a scalar) gives displacement (a vector).

(d) Multiplying any two scalars: Meaningful. The product of two scalars is a scalar. For example, multiplying mass (a scalar) by specific heat capacity (a scalar) is a meaningful operation in thermodynamics.

(e) Adding any two vectors: Not meaningful. Two vectors can be added only if they represent the same physical quantity. For example, adding a displacement vector to a force vector is meaningless.

(f) Adding a component of a vector to the same vector: Not meaningful. A component of a vector is a scalar value (the magnitude of the component vector). A scalar cannot be added to a vector. One can, however, add the component vector (e.g.,

A_x \hat{\mathbf{i}}

) to another vector.

Q5EXERCISES

3.5 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.

Solution

(a) True. The magnitude of a vector is a pure number with a unit that represents its length or size. It does not have a direction, so it is a scalar.

(b) True. A component of a vector (e.g.,

A_x

or

A_y

) is a scalar quantity. It is the projection of the vector onto an axis and can be positive, negative, or zero. The component vector (e.g.,

A_x \hat{\mathbf{i}}

) is a vector, but the component itself (

A_x

) is a scalar.

(c) False. The total path length is equal to the magnitude of the displacement vector only if the particle moves along a straight line without changing its direction. In all other cases, such as moving along a curved path or changing direction, the path length is greater than the magnitude of the displacement.

(d) True. Average speed is total path length divided by time. The magnitude of average velocity is the magnitude of displacement divided by time. Since the path length is always greater than or equal to the magnitude of displacement, the average speed is always greater than or equal to the magnitude of the average velocity.

(e) True. To get a null vector, the vectors must form a closed polygon when arranged head-to-tail. Three vectors can form a closed triangle, but this requires them to lie in the same plane. If three vectors are not in the same plane (non-coplanar), they cannot form a closed figure, and their resultant can never be a null vector.

Q6EXERCISES

3.6 Establish the following vector inequalities geometrically or otherwise :

(a)

|\mathbf{a}+\mathbf{b}| \leq |\mathbf{a}|+|\mathbf{b}|

(b)

|\mathbf{a}+\mathbf{b}| \geq ||\mathbf{a}|-|\mathbf{b}||

(c)

|\mathbf{a}-\mathbf{b}| \leq |\mathbf{a}|+|\mathbf{b}|

(d)

|\mathbf{a}-\mathbf{b}| \geq ||\mathbf{a}|-|\mathbf{b}||

When does the equality sign above apply?

Solution

Let two vectors be

\mathbf{a}

and

\mathbf{b}

. We can represent them by two sides of a triangle, say

\mathbf{OP}

and

\mathbf{PQ}

. Then the third side,

\mathbf{OQ}

, represents the resultant vector

\mathbf{R} = \mathbf{a} + \mathbf{b}

. The magnitudes of these vectors are the lengths of the sides of the triangle OPQ.

(a)

|\mathbf{a}+\mathbf{b}| \leq |\mathbf{a}|+|\mathbf{b}|

This is the triangle inequality. In any triangle, the length of one side is always less than the sum of the lengths of the other two sides. Therefore,

|\u0009\mathbf{OQ}| < |\mathbf{OP}| + |\mathbf{PQ}|

, which means

|\mathbf{a}+\mathbf{b}| < |\mathbf{a}|+|\mathbf{b}|

. The equality holds when the vectors

\mathbf{a}

and

\mathbf{b}

are collinear and point in the same direction (

\u0009\theta = 0^{\circ}

). In this case, they form a straight line, not a triangle.

(b)

|\mathbf{a}+\mathbf{b}| \geq ||\mathbf{a}|-|\mathbf{b}||

In any triangle, the length of one side is always greater than the magnitude of the difference between the lengths of the other two sides. Therefore,

|\u0009\mathbf{OQ}| > ||\mathbf{OP}| - |\mathbf{PQ}||

, which means

|\mathbf{a}+\mathbf{b}| > ||\mathbf{a}|-|\mathbf{b}||

. The equality holds when the vectors

\mathbf{a}

and

\mathbf{b}

are collinear and point in opposite directions (

\u0009\theta = 180^{\circ}

).

(c)

|\mathbf{a}-\mathbf{b}| \leq |\mathbf{a}|+|\mathbf{b}|

We can write

|\mathbf{a}-\mathbf{b}|

as

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

. Using the triangle inequality from part (a), we have

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

. Since

|-\mathbf{b}| = |\mathbf{b}|

, we get

|\mathbf{a}-\mathbf{b}| \leq |\mathbf{a}|+|\mathbf{b}|

. The equality holds when

\mathbf{a}

and

-\mathbf{b}

are in the same direction, which means

\mathbf{a}

and

\mathbf{b}

are in opposite directions (

\u0009\theta = 180^{\circ}

).

(d)

|\mathbf{a}-\mathbf{b}| \geq ||\mathbf{a}|-|\mathbf{b}||

We can write

|\mathbf{a}-\mathbf{b}|

as

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

. Using the inequality from part (b), we have

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

. Since

|-\mathbf{b}| = |\mathbf{b}|

, we get

|\mathbf{a}-\mathbf{b}| \geq ||\mathbf{a}|-|\mathbf{b}||

. The equality holds when

\mathbf{a}

and

-\mathbf{b}

are in opposite directions, which means

\mathbf{a}

and

\mathbf{b}

are in the same direction (

\u0009\theta = 0^{\circ}

).

Q7EXERCISES

3.7 Given $\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}=\mathbf{0}$ , which of the following statements are correct :

(a)

\mathbf{a}, \mathbf{b}, \mathbf{c}

, and

\mathbf{d}

must each be a null vector,

(b)

The magnitude of

(\mathbf{a}+\mathbf{c})

equals the magnitude of

(\mathbf{b}+\mathbf{d})

,

(c)

The magnitude of a can never be greater than the sum of the magnitudes of

\mathbf{b}, \mathbf{c}

, and

\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?

Solution

Given the relation:

\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}=\mathbf{0}

(a) Incorrect. The vectors do not need to be null vectors. For example, if

\mathbf{a} = \hat{\mathbf{i}}

,

\mathbf{b} = \hat{\mathbf{j}}

,

\mathbf{c} = -\hat{\mathbf{i}}

, and

\mathbf{d} = -\hat{\mathbf{j}}

, their sum is zero, but none of them is a null vector.

(b) Correct. From the given relation, we can write

\mathbf{a}+\mathbf{c} = -(\mathbf{b}+\mathbf{d})

. Taking the magnitude on both sides, we get

|\mathbf{a}+\mathbf{c}| = |-(\mathbf{b}+\mathbf{d})|

. Since the magnitude of a vector and its negative are equal,

|\mathbf{a}+\mathbf{c}| = |\mathbf{b}+\mathbf{d}|

. So, the magnitudes are equal.

(c) Correct. We can write

\mathbf{a} = -(\mathbf{b}+\mathbf{c}+\mathbf{d})

. Taking the magnitude,

|\mathbf{a}| = |-(\mathbf{b}+\mathbf{c}+\mathbf{d})| = |\mathbf{b}+\mathbf{c}+\mathbf{d}|

. Using the generalized triangle inequality for multiple vectors,

|\mathbf{b}+\mathbf{c}+\mathbf{d}| \leq |\mathbf{b}|+|\mathbf{c}|+|\mathbf{d}|

. Therefore,

|\mathbf{a}| \leq |\mathbf{b}|+|\mathbf{c}|+|\mathbf{d}|

. The magnitude of

\mathbf{a}

cannot be greater than the sum of the magnitudes of the other three vectors.

(d) Correct. From the given relation, we can write

\mathbf{b}+\mathbf{c} = -(\mathbf{a}+\mathbf{d})

. The vector sum

(\mathbf{a}+\mathbf{d})

will lie in the plane containing vectors

\mathbf{a}

and

\mathbf{d}

. Since

\mathbf{b}+\mathbf{c}

is equal to the negative of this sum, it will also lie in the same plane. If

\mathbf{a}

and

\mathbf{d}

are collinear, their sum

(\mathbf{a}+\mathbf{d})

will also lie on the same line. Consequently,

\mathbf{b}+\mathbf{c}

must also lie on that line.

Q8EXERCISES

3.8 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. 3.19. What is the magnitude of the displacement vector for each ? For which girl is this equal to the actual length of path skate?

Solution

Given: Radius of the circular ground,

r = 200 \text{ m}

Initial position = P Final position = Q (diametrically opposite to P)

To Find:

The magnitude of the displacement vector for each girl.
For which girl is the path length equal to the displacement magnitude.

Calculation: Displacement is the shortest straight-line distance between the initial and final points, regardless of the path taken. Here, the initial point is P and the final point is Q. Since Q is diametrically opposite to P, the straight line connecting them is the diameter of the circle.

Magnitude of displacement = Length of the diameter

\text{Displacement} = 2 \times r = 2 \times 200 \text{ m} = 400 \text{ m}

Since all three girls have the same initial and final points (P and Q), the magnitude of the displacement vector is the same for all of them.

Path Length vs. Displacement: The path length is the actual distance covered. The magnitude of displacement is the shortest distance, which is 400 m.

Girl A and Girl C travel along curved paths. Their path lengths are greater than the diameter (400 m).
Girl B travels along the straight-line path PQ, which is the diameter.

Therefore, for Girl B, the actual length of the path skated is equal to the magnitude of her displacement.

Final Answer: The magnitude of the displacement vector for each girl is 400 m. For Girl B, the displacement magnitude is equal to the actual length of the path skated.

Q9EXERCISES

3.9 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. 3.20. If the round trip takes 10 min, what is the (a) net displacement, (b) average velocity, and (c) average speed of the cyclist ?

Solution

Given: Radius of the park,

r = 1 \text{ km} = 1000 \text{ m}

Total time taken,

t = 10 \text{ min} = 10 \times 60 \text{ s} = 600 \text{ s}

The cyclist's path is O -> P -> Q -> O.

(a) Net Displacement Displacement is the change in position. The cyclist starts at the centre O and returns to the centre O. Since the initial and final positions are the same, the net displacement is zero.

Net Displacement = 0

(b) Average Velocity Average velocity is defined as the net displacement divided by the total time taken.

\text{Average Velocity} = \frac{\text{Net Displacement}}{\text{Total Time}}

Since the net displacement is zero, the average velocity is also zero.

Average Velocity = 0

(c) Average Speed Average speed is defined as the total path length divided by the total time taken.

Calculation of Total Path Length:

Path from O to P: This is the radius of the park. Length = $r = 1 \text{ km}$ .
Path from P to Q: This is along the circumference, covering one-quarter of the circle. Length = $\frac{1}{4} \times (2\pi r) = \frac{\pi r}{2} = \frac{\pi \times 1}{2} = \frac{\pi}{2} \text{ km}$ .
Path from Q to O: This is also the radius of the park. Length = $r = 1 \text{ km}$ .

Total Path Length =

OP + PQ + QO = 1 + \frac{\pi}{2} + 1 = 2 + \frac{3.14}{2} = 2 + 1.57 = 3.57 \text{ km}

.

Calculation of Average Speed:

\text{Average Speed} = \frac{\text{Total Path Length}}{\text{Total Time}}

Total Path Length =

3.57 \text{ km} = 3570 \text{ m}

Total Time =

10 \text{ min} = \frac{10}{60} \text{ h} = \frac{1}{6} \text{ h}

\text{Average Speed} = \frac{3.57 \text{ km}}{1/6 \text{ h}} = 3.57 \times 6 = 21.42 \text{ km/h}

Alternatively, in m/s:

\text{Average Speed} = \frac{3570 \text{ m}}{600 \text{ s}} = 5.95 \text{ m/s}

Final Answer: (a) Net displacement is 0. (b) Average velocity is 0. (c) Average speed is 21.42 km/h (or 5.95 m/s).

Q10EXERCISES

3.10 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.

Solution

The motorist turns left by

60^{\circ}

after every 500 m. This means the path is a regular hexagon with a side length of

a = 500 \text{ m}

. Let the motorist start at point A.

At the third turn: The motorist moves from A to B, B to C, and C to D. The turns happen at B, C, and D. The third turn is at D. Path length =

AB + BC + CD = 500 + 500 + 500 = 1500 \text{ m}

. The displacement is the vector

\mathbf{AD}

. In a regular hexagon, the distance between vertices separated by two sides (like A and D) is twice the side length. Magnitude of displacement =

2a = 2 \times 500 = 1000 \text{ m}

. Comparison:

\frac{\text{Displacement}}{\text{Path Length}} = \frac{1000}{1500} = \frac{2}{3}

.

At the sixth turn: The motorist completes one full hexagon and returns to the starting point A. The sixth turn is at A. Path length =

6 \times a = 6 \times 500 = 3000 \text{ m}

. Since the final position is the same as the initial position, the displacement is zero. Magnitude of displacement =

0 \text{ m}

. Comparison:

\frac{\text{Displacement}}{\text{Path Length}} = \frac{0}{3000} = 0

.

At the eighth turn: The motorist completes one full hexagon (6 turns) and then takes two more segments. The path is A -> ... -> A -> B -> C. The eighth turn is at C. Path length =

8 \times a = 8 \times 500 = 4000 \text{ m}

. The final position is C. The displacement is the vector

\mathbf{AC}

. The distance AC can be found using the law of cosines on triangle ABC, where angle B is

180^{\circ} - 60^{\circ} = 120^{\circ}

.

|\mathbf{AC}|^2 = |\mathbf{AB}|^2 + |\mathbf{BC}|^2 - 2|\mathbf{AB}||\mathbf{BC}| \cos(120^{\circ})

|\mathbf{AC}|^2 = 500^2 + 500^2 - 2(500)(500)(-\frac{1}{2})

|\mathbf{AC}|^2 = 2 \times 500^2 + 500^2 = 3 \times 500^2

|\mathbf{AC}| = \sqrt{3} \times 500 \approx 1.732 \times 500 = 866 \text{ m}

. Magnitude of displacement =

866 \text{ m}

. Comparison:

\frac{\text{Displacement}}{\text{Path Length}} = \frac{866}{4000} \approx 0.2165

.

Final Answer:

Third turn: Displacement = 1000 m, Path length = 1500 m. Ratio = 2/3.
Sixth turn: Displacement = 0 m, Path length = 3000 m. Ratio = 0.
Eighth turn: Displacement = 866 m, Path length = 4000 m. Ratio $\approx$ 0.217.

Q11EXERCISES

3.11 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 ?

Solution

Given: Magnitude of displacement (straight-line distance),

|\u0009\Delta \mathbf{r}| = 10 \text{ km}

Total path length,

d = 23 \text{ km}

Total time taken,

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

(a) Average speed of the taxi Average speed is the total path length divided by the total time taken.

Formula:

\text{Average Speed} = \frac{\text{Total Path Length}}{\text{Total Time}}

Calculation:

\text{Average Speed} = \frac{23 \text{ km}}{28/60 \text{ h}} = \frac{23 \times 60}{28} = \frac{1380}{28} \approx 49.29 \text{ km/h}

(b) Magnitude of average velocity Average velocity is the net displacement divided by the total time taken. Its magnitude is the magnitude of displacement divided by time.

Formula:

|\text{Average Velocity}| = \frac{|\text{Net Displacement}|}{\text{Total Time}}

Calculation:

|\text{Average Velocity}| = \frac{10 \text{ km}}{28/60 \text{ h}} = \frac{10 \times 60}{28} = \frac{600}{28} \approx 21.43 \text{ km/h}

Are the two equal? No, the average speed (49.29 km/h) and the magnitude of the average velocity (21.43 km/h) are not equal. This is because the cabman took a circuitous path, making the path length (23 km) much larger than the magnitude of the displacement (10 km).

Final Answer: (a) The average speed of the taxi is approximately 49.3 km/h. (b) The magnitude of the average velocity is approximately 21.4 km/h. The two are not equal.

Q12EXERCISES

3.12 The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of $40 \text{ m s}^{-1}$ can go without hitting the ceiling of the hall ?

Solution

Given: Maximum height,

h_m = 25 \text{ m}

Initial speed,

v_o = 40 \text{ m/s}

Acceleration due to gravity,

g = 9.8 \text{ m/s}^2

To Find: Maximum horizontal distance (Range, R).

Formula: The maximum height of a projectile is given by:

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

The horizontal range is given by:

R = \frac{v_o^2 \sin 2\theta_o}{g} = \frac{2v_o^2 \sin \theta_o \cos \theta_o}{g}

Calculation: First, we find the angle of projection

\theta_o

for which the maximum height is 25 m.

25 = \frac{(40 \sin \theta_o)^2}{2 \times 9.8}

25 \times 19.6 = 1600 \sin^2 \theta_o

490 = 1600 \sin^2 \theta_o

\sin^2 \theta_o = \frac{490}{1600} = \frac{49}{160}

\sin \theta_o = \sqrt{\frac{49}{160}} = \frac{7}{\sqrt{160}} \approx 0.5534

Now, we need to find

\cos \theta_o

. Using the identity

\sin^2 \theta_o + \cos^2 \theta_o = 1

:

\cos^2 \theta_o = 1 - \sin^2 \theta_o = 1 - \frac{49}{160} = \frac{111}{160}

\cos \theta_o = \sqrt{\frac{111}{160}} \approx 0.8328

Now, we can calculate the horizontal range R:

R = \frac{2 \times (40)^2 \times (0.5534) \times (0.8328)}{9.8}

R = \frac{2 \times 1600 \times 0.4608}{9.8}

R = \frac{1474.56}{9.8} \approx 150.46 \text{ m}

Final Answer: The maximum horizontal distance the ball can go without hitting the ceiling is approximately 150.5 m.

Q13EXERCISES

3.13 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 ?

Solution

Given: Maximum horizontal distance (range),

R_m = 100 \text{ m}

To Find: Maximum possible height (

h_{max}

) the cricketer can throw the ball.

Formula: The horizontal range of a projectile is given by:

R = \frac{v_o^2 \sin 2\theta_o}{g}

The maximum height is given by:

h = \frac{(v_o \sin \theta_o)^2}{2g}

Calculation: The range R is maximum when

\sin 2\theta_o = 1

, which occurs at

\theta_o = 45^{\circ}

. So, the maximum range

R_m

is:

R_m = \frac{v_o^2}{g}

We are given

R_m = 100 \text{ m}

.

100 = \frac{v_o^2}{g} \quad \Rightarrow \quad v_o^2 = 100g

This gives us the square of the initial speed with which the cricketer can throw the ball.

To throw the ball to the maximum possible height, the cricketer must throw it vertically upwards, which means the angle of projection

\theta_o

is

90^{\circ}

.

The maximum height

h_{max}

is achieved when

\sin \theta_o = \sin 90^{\circ} = 1

.

h_{max} = \frac{(v_o \sin 90^{\circ})^2}{2g} = \frac{v_o^2}{2g}

Now, substitute the value of

v_o^2

we found from the range:

h_{max} = \frac{100g}{2g} = 50 \text{ m}

Final Answer: The cricketer can throw the same ball to a maximum height of 50 m.

Q14EXERCISES

3.14 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?

Solution

Given: Radius of the circle,

r = 80 \text{ cm} = 0.8 \text{ m}

Number of revolutions = 14 Time taken,

t = 25 \text{ s}

To Find: The magnitude and direction of the acceleration of the stone.

Formula: The acceleration in uniform circular motion is the centripetal acceleration,

a_c

. Its magnitude is given by:

a_c = \omega^2 r

where

\omega

is the angular speed.

Calculation: First, we calculate the frequency (

\u0009\nu

) of the revolution.

\nu = \frac{\text{Number of revolutions}}{\text{Time taken}} = \frac{14}{25} \text{ Hz}

Next, we calculate the angular speed (

\u0009\omega

).

\omega = 2\pi\nu = 2\pi \times \frac{14}{25} = \frac{28\pi}{25} \text{ rad/s}

\omega \approx \frac{28 \times 3.14}{25} \approx 3.52 \text{ rad/s}

Now, we can calculate the magnitude of the centripetal acceleration.

a_c = \omega^2 r = \left(\frac{28\pi}{25}\right)^2 \times 0.8

a_c = \frac{784 \pi^2}{625} \times 0.8 \approx \frac{784 \times (3.14)^2}{625} \times 0.8 \approx 12.39 \times 0.8 \approx 9.91 \text{ m/s}^2

Direction of Acceleration: For an object in uniform circular motion, the acceleration is always directed towards the center of the circle. This is called centripetal acceleration.

Final Answer: The magnitude of the acceleration of the stone is approximately 9.91 m/s². Its direction is always towards the center of the circular path.

Q15EXERCISES

3.15 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.

Solution

Given: Radius of the loop,

R = 1.00 \text{ km} = 1000 \text{ m}

Steady speed,

v = 900 \text{ km/h}

Acceleration due to gravity,

g = 9.8 \text{ m/s}^2

To Find: The ratio of the centripetal acceleration (

a_c

) to the acceleration due to gravity (

g

).

Formula: The centripetal acceleration is given by:

a_c = \frac{v^2}{R}

Calculation: First, we need to convert the speed from km/h to m/s.

v = 900 \frac{\text{km}}{\text{h}} = 900 \times \frac{1000 \text{ m}}{3600 \text{ s}} = 900 \times \frac{5}{18} \text{ m/s} = 50 \times 5 = 250 \text{ m/s}

Now, calculate the centripetal acceleration:

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

Finally, we compare this with the acceleration due to gravity by finding the ratio:

\frac{a_c}{g} = \frac{62.5 \text{ m/s}^2}{9.8 \text{ m/s}^2} \approx 6.38

Final Answer: The centripetal acceleration is 62.5 m/s². This is approximately 6.4 times the acceleration due to gravity.

Q16EXERCISES

3.16 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

Solution

(a) False. This statement is only true for uniform circular motion, where the speed is constant. In non-uniform circular motion, the speed changes, so there is a tangential component of acceleration (

a_t

) in addition to the centripetal (radial) acceleration (

a_c

). The net acceleration is the vector sum of these two components and is not directed towards the centre.

(b) True. The instantaneous velocity of a particle at any point in its path gives the direction of its motion at that instant. The direction of motion at any point on a curved path is always along the tangent to the path at that point.

(c) True. In uniform circular motion, the acceleration vector (centripetal acceleration) has a constant magnitude but its direction continuously changes, always pointing towards the center of the circle. Over one complete cycle, for every acceleration vector pointing in a certain direction, there is an equal and opposite acceleration vector at the diametrically opposite point. When averaged over the entire cycle, the vector sum of all these acceleration vectors is a null vector.

Q17EXERCISES

3.17 The position of a particle is given by $\mathbf{r}=3.0 t \hat{\mathbf{i}}-2.0 t^{2} \hat{\mathbf{j}}+4.0 \hat{\mathbf{k}} \text{ m}$ where $t$ is in seconds and the coefficients have the proper units for $\mathbf{r}$ to be in metres.

(a)

Find the

\mathbf{v}

and

\mathbf{a}

of the particle? (b) What is the magnitude and direction of velocity of the particle at

t=2.0 \text{ s}

?

Solution

Given: Position vector:

\mathbf{r}(t) = 3.0t \hat{\mathbf{i}} - 2.0t^2 \hat{\mathbf{j}} + 4.0 \hat{\mathbf{k}}

(a) Find the velocity (v) and acceleration (a) of the particle.

Velocity Vector (v): Velocity is the first derivative of the position vector with respect to time.

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

\mathbf{v}(t) = 3.0 \hat{\mathbf{i}} - 4.0t \hat{\mathbf{j}} + 0 \hat{\mathbf{k}}

\mathbf{v}(t) = (3.0 \hat{\mathbf{i}} - 4.0t \hat{\mathbf{j}}) \text{ m/s}

Acceleration Vector (a): Acceleration is the first derivative of the velocity vector with respect to time.

\mathbf{a}(t) = \frac{d\mathbf{v}}{dt} = \frac{d}{dt}(3.0 \hat{\mathbf{i}} - 4.0t \hat{\mathbf{j}})

\mathbf{a}(t) = 0 \hat{\mathbf{i}} - 4.0 \hat{\mathbf{j}}

\mathbf{a}(t) = -4.0 \hat{\mathbf{j}} \text{ m/s}^2

(b) What is the magnitude and direction of velocity of the particle at t = 2.0 s?

First, find the velocity vector at

t = 2.0 \text{ s}

:

\mathbf{v}(2.0) = 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 of velocity:

|\mathbf{v}| = \sqrt{v_x^2 + v_y^2} = \sqrt{(3.0)^2 + (-8.0)^2}

|\mathbf{v}| = \sqrt{9 + 64} = \sqrt{73} \approx 8.54 \text{ m/s}

Direction of velocity: The direction is given by the angle

\theta

it makes with the positive x-axis.

\tan \theta = \frac{v_y}{v_x} = \frac{-8.0}{3.0} \approx -2.667

\theta = \tan^{-1}(-2.667) \approx -69.44^{\circ}

The angle is approximately

69.4^{\circ}

below the positive x-axis.

Final Answer: (a)

\mathbf{v}(t) = (3.0 \hat{\mathbf{i}} - 4.0t \hat{\mathbf{j}}) \text{ m/s}

and

\mathbf{a}(t) = -4.0 \hat{\mathbf{j}} \text{ m/s}^2

. (b) At

t = 2.0 \text{ s}

, the magnitude of the velocity is 8.54 m/s, and its direction is 69.4° below the positive x-axis.

Q18EXERCISES

3.18 A particle starts from the origin at $t=0 \text{ s}$ with a velocity of $10.0 \hat{\mathbf{j}} \text{ m/s}$ and moves in the $x-y$ plane with a constant acceleration of $(8.0 \hat{\mathbf{i}}+2.0 \hat{\mathbf{j}}) \text{ m s}^{-2}$ . (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 ?

Solution

Given: Initial position,

\mathbf{r}_0 = \mathbf{0}

(starts from origin) Initial velocity,

\mathbf{v}_0 = 10.0 \hat{\mathbf{j}} \text{ 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}}) \text{ m/s}^2

(so,

a_x = 8.0 \text{ m/s}^2, a_y = 2.0 \text{ m/s}^2

)

Formula: The position of the particle at time

t

is given by:

\mathbf{r}(t) = \mathbf{r}_0 + \mathbf{v}_0 t + \frac{1}{2}\mathbf{a}t^2

In component form:

x(t) = x_0 + v_{0x}t + \frac{1}{2}a_x t^2

y(t) = y_0 + v_{0y}t + \frac{1}{2}a_y t^2

(a) Time when x = 16 m and corresponding y-coordinate Substitute the given values into the equation for

x(t)

:

x(t) = 0 + (0)t + \frac{1}{2}(8.0)t^2 = 4.0t^2

We are given

x(t) = 16 \text{ m}

.

16 = 4.0t^2

t^2 = \frac{16}{4.0} = 4

t = 2.0 \text{ s}

(time cannot be negative)

Now, find the y-coordinate at

t = 2.0 \text{ s}

using the equation for

y(t)

:

y(t) = 0 + (10.0)t + \frac{1}{2}(2.0)t^2 = 10.0t + t^2

y(2.0) = 10.0(2.0) + (2.0)^2 = 20 + 4 = 24 \text{ m}

(b) Speed of the particle at this time (t = 2.0 s) First, find the velocity vector

\mathbf{v}(t)

using

\mathbf{v} = \mathbf{v}_0 + \mathbf{a}t

.

\mathbf{v}(t) = (v_{0x} + a_x t)\hat{\mathbf{i}} + (v_{0y} + a_y t)\hat{\mathbf{j}}

\mathbf{v}(t) = (0 + 8.0t)\hat{\mathbf{i}} + (10.0 + 2.0t)\hat{\mathbf{j}}

At

t = 2.0 \text{ s}

:

\mathbf{v}(2.0) = (8.0 \times 2.0)\hat{\mathbf{i}} + (10.0 + 2.0 \times 2.0)\hat{\mathbf{j}}

\mathbf{v}(2.0) = 16.0 \hat{\mathbf{i}} + (10.0 + 4.0)\hat{\mathbf{j}} = 16.0 \hat{\mathbf{i}} + 14.0 \hat{\mathbf{j}} \text{ m/s}

Speed is the magnitude of the velocity vector:

|\mathbf{v}| = \sqrt{v_x^2 + v_y^2} = \sqrt{(16.0)^2 + (14.0)^2}

|\mathbf{v}| = \sqrt{256 + 196} = \sqrt{452} \approx 21.26 \text{ m/s}

Final Answer: (a) The x-coordinate is 16 m at t = 2.0 s. At this time, the y-coordinate is 24 m. (b) The speed of the particle at this time is approximately 21.3 m/s.

Q19EXERCISES

3.19 $\hat{\mathbf{i}}$ and $\hat{\mathbf{j}}$ are unit vectors along $x$ - and $y$ - axis respectively. What is the magnitude and direction of the vectors $\hat{\mathbf{i}}+\hat{\mathbf{j}}$ , and $\hat{\mathbf{i}}-\hat{\mathbf{j}}$ ? What are the components of a vector $\mathbf{A}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}$ along the directions of $\hat{\mathbf{i}}+\hat{\mathbf{j}}$ and $\hat{\mathbf{i}}-\hat{\mathbf{j}}$ ? [You may use graphical method]

Solution

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

For vector $\mathbf{P} = \hat{\mathbf{i}}+\hat{\mathbf{j}}$ :

Magnitude: $|\u0009\mathbf{P}| = \sqrt{1^2 + 1^2} = \sqrt{2}$ .
Direction: The angle $\theta$ with the x-axis is $\tan \theta = \frac{1}{1} = 1$ , so $\theta = 45^{\circ}$ .

For vector $\mathbf{Q} = \hat{\mathbf{i}}-\hat{\mathbf{j}}$ :

Magnitude: $|\u0009\mathbf{Q}| = \sqrt{1^2 + (-1)^2} = \sqrt{2}$ .
Direction: The angle $\theta$ with the x-axis is $\tan \theta = \frac{-1}{1} = -1$ , so $\theta = -45^{\circ}$ or $315^{\circ}$ .

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

\mathbf{P} = \hat{\mathbf{i}}+\hat{\mathbf{j}}

be represented by the unit vector

\hat{\mathbf{p}}

and the direction of

\mathbf{Q} = \hat{\mathbf{i}}-\hat{\mathbf{j}}

be represented by the unit vector

\hat{\mathbf{q}}

.

\hat{\mathbf{p}} = \frac{\mathbf{P}}{|\u0009\mathbf{P}|} = \frac{\hat{\mathbf{i}}+\hat{\mathbf{j}}}{\sqrt{2}}

\hat{\mathbf{q}} = \frac{\mathbf{Q}}{|\u0009\mathbf{Q}|} = \frac{\hat{\mathbf{i}}-\hat{\mathbf{j}}}{\sqrt{2}}

The component of a vector

\mathbf{A}

along a direction given by a unit vector

\hat{\mathbf{n}}

is

\mathbf{A} \cdot \hat{\mathbf{n}}

.

Component of $\mathbf{A}$ along $\hat{\mathbf{i}}+\hat{\mathbf{j}}$ :

A_p = \mathbf{A} \cdot \hat{\mathbf{p}} = (2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}) \cdot \left(\frac{\hat{\mathbf{i}}+\hat{\mathbf{j}}}{\sqrt{2}}\right)

A_p = \frac{1}{\sqrt{2}}(2 \times 1 + 3 \times 1) = \frac{5}{\sqrt{2}}

Component of $\mathbf{A}$ along $\hat{\mathbf{i}}-\hat{\mathbf{j}}$ :

A_q = \mathbf{A} \cdot \hat{\mathbf{q}} = (2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}) \cdot \left(\frac{\hat{\mathbf{i}}-\hat{\mathbf{j}}}{\sqrt{2}}\right)

A_q = \frac{1}{\sqrt{2}}(2 \times 1 + 3 \times (-1)) = \frac{2-3}{\sqrt{2}} = -\frac{1}{\sqrt{2}}

Final Answer:

For $\hat{\mathbf{i}}+\hat{\mathbf{j}}$ : Magnitude is $\sqrt{2}$ , direction is $45^{\circ}$ with the x-axis.
For $\hat{\mathbf{i}}-\hat{\mathbf{j}}$ : Magnitude is $\sqrt{2}$ , direction is $-45^{\circ}$ with the x-axis.
The component of $\mathbf{A}$ along $\hat{\mathbf{i}}+\hat{\mathbf{j}}$ is $\frac{5}{\sqrt{2}}$ .
The component of $\mathbf{A}$ along $\hat{\mathbf{i}}-\hat{\mathbf{j}}$ is $-\frac{1}{\sqrt{2}}$ .

Q20EXERCISES

3.20 For any arbitrary motion in space, which of the following relations are true :

(a)

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

(b)

\mathbf{v}_{\text{average}} = [\mathbf{r}(t_2) - \mathbf{r}(t_1)] / (t_2 - t_1)

(c)

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

(d)

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

(e)

\mathbf{a}_{\text{average}} = [\mathbf{v}(t_2) - \mathbf{v}(t_1)] / (t_2 - t_1)

(The 'average' stands for average of the quantity over the time interval

t_1

to

t_2

)

Solution

For any arbitrary motion, the acceleration is generally not constant.

(a)

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

False. This relation is only true for motion with constant acceleration. For arbitrary motion, it does not hold.

(b)

\mathbf{v}_{\text{average}} = [\mathbf{r}(t_2) - \mathbf{r}(t_1)] / (t_2 - t_1)

True. This is the fundamental definition of average velocity, which is the total displacement divided by the total time interval. This is true for any type of motion, whether acceleration is constant or not.

(c)

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

False. This is one of the kinematic equations of motion which is valid only for constant acceleration

\mathbf{a}

.

(d)

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

False. This is another kinematic equation of motion which is valid only for constant acceleration

\mathbf{a}

.

(e)

\mathbf{a}_{\text{average}} = [\mathbf{v}(t_2) - \mathbf{v}(t_1)] / (t_2 - t_1)

True. This is the fundamental definition of average acceleration, which is the change in velocity divided by the time interval. This is true for any type of motion.

Final Answer: The true relations for any arbitrary motion in space are (b) and (e).

Q21EXERCISES

3.21 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.

Solution

(a) False. Many scalar quantities are not conserved. For example, temperature and energy are scalars, but they are not always conserved in a process. While total energy in an isolated system is conserved, the kinetic energy or potential energy of a part of the system may not be. Mass is a scalar that is conserved in non-relativistic processes, but not all scalars are.

(b) False. A scalar quantity can be negative. For example, temperature can be negative (e.g., -10°C), electric charge can be negative, and work done can be negative (when force opposes displacement).

(c) False. Scalar quantities can have dimensions. For example, mass has the dimension [M], speed has dimensions [LT⁻¹], and volume has dimensions [L³]. Only some scalars, like refractive index or strain, are dimensionless.

(d) False. Scalar quantities can vary from one point to another in space. For example, the temperature in a room is not uniform; it is a scalar field that varies with position. Similarly, the density of the atmosphere varies with altitude.

(e) True. This is a defining characteristic of a scalar. A scalar quantity has only magnitude and is independent of the coordinate system used to describe it. For example, the mass of an object (e.g., 5 kg) is the same regardless of how the x, y, and z axes are oriented.

Q22EXERCISES

3.22 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 $30^{\circ}$ , what is the speed of the aircraft ?

Solution

Given: Height of the aircraft,

h = 3400 \text{ m}

Time interval,

\Delta t = 10.0 \text{ s}

Angle subtended,

\theta = 30^{\circ}

To Find: Speed of the aircraft,

v

.

Assumptions: We assume the aircraft is flying horizontally at a constant speed. Let O be the observation point on the ground. Let A and B be the two positions of the aircraft 10.0 s apart. The triangle OAB is an isosceles triangle with OA = OB. The height from O to the line AB is

h = 3400 \text{ m}

.

Diagram and Calculation: Consider the triangle OAB. Let P be the midpoint of the line segment AB. Then OP is the perpendicular distance from the observer to the aircraft's path, so

OP = h = 3400 \text{ m}

. The angle AOB is

30^{\circ}

. The line OP bisects this angle, so angle AOP = angle BOP =

15^{\circ}

.

The distance traveled by the aircraft in 10 s is the length of the segment AB.

AB = AP + PB = 2 \times AP

In the right-angled triangle OPA:

\tan(\text{angle AOP}) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AP}{OP}

\tan(15^{\circ}) = \frac{AP}{3400}

AP = 3400 \times \tan(15^{\circ})

We know that

\tan(15^{\circ}) \approx 0.2679

.

AP = 3400 \times 0.2679 \approx 910.86 \text{ m}

The total distance traveled is

AB = 2 \times AP = 2 \times 910.86 = 1821.72 \text{ m}

.

Now, we can calculate the speed of the aircraft.

\text{Speed} (v) = \frac{\text{Distance}}{\text{Time}} = \frac{AB}{\Delta t}

v = \frac{1821.72 \text{ m}}{10.0 \text{ s}} \approx 182.17 \text{ m/s}

Final Answer: The speed of the aircraft is approximately 182 m/s.

Motion In A PlaneClass 11 Physics NCERT Solutions

3.1 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.

Solution

3.2 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.

Solution

3.3 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.

Solution

3.4 State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful :

Solution

3.5 Read each statement below carefully and state with reasons, if it is true or false :

Solution

3.6 Establish the following vector inequalities geometrically or otherwise :

Solution

3.7 Given a+b+c+d=0\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}=\mathbf{0}a+b+c+d=0, which of the following statements are correct :

Solution

Solution

Solution

Solution

Solution

3.12 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−140 \text{ m s}^{-1}40 m s−1 can go without hitting the ceiling of the hall ?

Solution

3.13 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 ?

Solution

3.14 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?

Solution

3.15 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.

Solution

3.16 Read each statement below carefully and state, with reasons, if it is true or false :

Solution

3.17 The position of a particle is given by r=3.0ti^−2.0t2j^+4.0k^ m\mathbf{r}=3.0 t \hat{\mathbf{i}}-2.0 t^{2} \hat{\mathbf{j}}+4.0 \hat{\mathbf{k}} \text{ m}r=3.0ti^−2.0t2j^​+4.0k^ m where ttt is in seconds and the coefficients have the proper units for r\mathbf{r}r to be in metres.

Solution

Solution

Solution

3.20 For any arbitrary motion in space, which of the following relations are true :

Solution

3.21 Read each statement below carefully and state, with reasons and examples, if it is true or false : A scalar quantity is one that

Solution

3.22 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∘30^{\circ}30∘, what is the speed of the aircraft ?

Solution

3.7 Given $\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}=\mathbf{0}$ , which of the following statements are correct :

3.12 The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of $40 \text{ m s}^{-1}$ can go without hitting the ceiling of the hall ?

3.17 The position of a particle is given by $\mathbf{r}=3.0 t \hat{\mathbf{i}}-2.0 t^{2} \hat{\mathbf{j}}+4.0 \hat{\mathbf{k}} \text{ m}$ where $t$ is in seconds and the coefficients have the proper units for $\mathbf{r}$ to be in metres.

3.22 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 $30^{\circ}$ , what is the speed of the aircraft ?