A vector \( \mathbf{A} = 3 \hat{\mathbf{i}} + 4 \hat{\mathbf{j}} \) is rotated by \( 90^\circ \) counterclockwise in the plane. What is the new vector?

Correct Option: A. Solution: Rotating a vector \( \mathbf{A} = a \hat{\mathbf{i}} + b \hat{\mathbf{j}} \) by \( 90^\circ \) counterclockwise results in a new vector \( \mathbf{A'} = -b \hat{\mathbf{i}} + a \hat{\mathbf{j}} \). Thus, \( \mathbf{A'} = -4 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}} \).

Motion in a Plane

AI Learning Assistant

I can help you understand Motion in a Plane better. Ask me anything!

Summarize the main points of Motion in a Plane.

What are the most important terms to remember here?

Explain this concept like I'm five.

Give me a quick 3-question practice quiz.

CBSE Learning Objectives – Key Concepts & Skills You Must Know

Learning Objectives

Define and differentiate between scalar and vector quantities.
Explain the graphical and analytical methods of vector addition.
Apply the laws of vector addition to solve problems involving multiple vectors.
Describe motion in a plane, including projectile motion and uniform circular motion.
Calculate average and instantaneous velocity and acceleration in two dimensions.
Analyze the effects of initial conditions on the trajectory of projectiles.
Utilize kinematic equations for motion with constant acceleration in two dimensions.

CBSE Revision Notes & Quick Summary for Last-Minute Study

Chapter Notes on Motion in a Plane

3.1 Introduction

3.2 Scalars and Vectors

Scalar Quantities: Quantities with magnitudes only. Examples include distance, speed, mass, and temperature.
Vector Quantities: Quantities with both magnitude and direction. Examples include displacement, velocity, and acceleration.

3.3 Multiplication of Vectors by Real Numbers

A vector multiplied by a real number results in a vector whose magnitude is scaled by that number, and direction is preserved or reversed depending on the sign of the number.

3.4 Addition and Subtraction of Vectors — Graphical Method

Vectors can be added graphically using the head-to-tail method or the parallelogram method.
Vector addition is commutative: A + B = B + A and associative: (A + B) + C = A + (B + C).

3.5 Resolution of Vectors

A vector can be resolved into components along two given vectors in the same plane.

3.6 Vector Addition — Analytical Method

If the sum of two vectors A and B in the x-y plane is R, then: R = R_x i + R_y j, where R_x = A_x + B_x and R_y = A_y + B_y.

3.7 Motion in a Plane

The trajectory of an object is influenced by initial conditions such as initial position and velocity.

3.8 Motion in a Plane with Constant Acceleration

Kinematic equations apply, but not for uniform circular motion where acceleration direction changes.

3.9 Projectile Motion

The trajectory is a smooth upward curve reaching a peak and falling back to the same horizontal level.
At the peak, the vertical component of velocity is zero.

3.10 Uniform Circular Motion

The resultant acceleration is directed towards the center if speed is constant.

Points to Ponder

Path length is not the same as displacement; they are equal only if the object does not change direction.
Average speed is greater than or equal to average velocity unless the path length equals displacement.
The kinematic equations for uniform acceleration do not apply to uniform circular motion.
The resultant velocity of an object subjected to two velocities V₁ and V₂ is V = V₁ + V₂.
The shape of the trajectory depends on both acceleration and initial conditions.

Exercises

Identify scalar and vector quantities from given lists.
Discuss the meaningfulness of algebraic operations involving scalars and vectors.
Establish vector inequalities geometrically.

CBSE Exam Tips, Important Questions & Common Mistakes to Avoid

Common Mistakes and Exam Tips

Common Pitfalls

Displacement vs. Path Length: The path length traversed by an object is generally not the same as the magnitude of displacement. Displacement depends only on the endpoints, while path length depends on the actual path taken. They are equal only if the object does not change direction.
Average Speed vs. Average Velocity: The average speed of an object is always greater than or equal to the magnitude of the average velocity over a given time interval. They are equal only if the path length equals the magnitude of displacement.
Kinematic Equations: The kinematic equations for uniform acceleration do not apply to uniform circular motion, as the magnitude of acceleration is constant but its direction is changing.
Resultant Velocity: When combining two velocities, care must be taken to distinguish between the resultant velocity and the relative velocity of one object to another.
Acceleration in Circular Motion: The resultant acceleration of an object in circular motion is directed towards the center only if the speed is constant.
Trajectory Determination: The shape of the trajectory of an object is influenced not only by acceleration but also by initial conditions such as initial position and velocity.

Exam Tips

Understand Vector Operations: Be clear on vector addition and subtraction, especially in graphical and analytical methods. Remember that vector addition is commutative and associative.
Clarify Definitions: Know the definitions of scalar and vector quantities, and be prepared to identify examples of each.
Practice True/False Statements: Be ready to evaluate statements regarding vectors and scalars, and justify your answers with reasoning.
Use Diagrams: When applicable, use diagrams to visualize problems, especially in vector addition and motion scenarios.
Review Common Equations: Familiarize yourself with key equations related to motion, including those for average velocity, instantaneous velocity, and projectile motion.

CBSE Quiz & Practice Test – MCQs, True/False Questions with Solutions

Multiple Choice Questions

\sqrt{2}

\sqrt{3}

Correct Answer: A

Solution:

The magnitude of the vector

\mathbf{i} + \mathbf{j}

is calculated as

\sqrt{1^2 + 1^2} = \sqrt{2}

Path length is always equal to displacement.

Path length is always less than displacement.

Path length is always greater than or equal to displacement.

Path length is always zero.

Correct Answer: C

Solution:

The path length is the actual distance traveled, which is always greater than or equal to the magnitude of displacement.

4 m/s²

8 m/s²

16 m/s²

2 m/s²

Correct Answer: B

Solution:

The centripetal acceleration

a_c

is given by

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

. Substituting

v = 4

m/s and

r = 2

m, we get

a_c = \frac{4^2}{2} = 8

m/s².

v_0 \cos \theta_0

v_0 \sin \theta_0

v_0 \tan \theta_0

v_0 / \sin \theta_0

Correct Answer: B

Solution:

The vertical component of the initial velocity in projectile motion is given by

v_{0y} = v_0 \sin \theta_0

21.43 km/h

25.71 km/h

30.00 km/h

35.00 km/h

Correct Answer: B

Solution:

Average velocity is the displacement divided by time. Displacement is 10 km, time is 28 minutes = 0.467 hours. Average velocity =

\frac{10}{0.467} \approx 21.43

km/h.

The net acceleration is zero.

The velocity vector is always directed towards the center.

The acceleration vector is always directed towards the center.

The speed of the particle changes continuously.

Correct Answer: C

Solution:

In uniform circular motion, the acceleration vector is always directed towards the center of the circle, which is the centripetal acceleration.

5 m

10 m

15 m

20 m

Correct Answer: B

Solution:

The maximum height

H

is given by

H = \frac{(v_0 \sin \theta)^2}{2g}

. Substituting

v_0 = 20

m/s,

\theta = 30^\circ

, and

g = 9.8

m/s², we get

H = \frac{(20 \times 0.5)^2}{2 \times 9.8} = 10

0.56 m/s²

1.23 m/s²

2.45 m/s²

3.14 m/s²

Correct Answer: C

Solution:

The centripetal acceleration

a_c

is given by

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

. First, find the speed

v = \frac{2\pi r \times 14}{25}

. Then, calculate

a_c = \frac{v^2}{0.8}

22 km/h at 30° north of east

20 km/h at 45° north of east

18 km/h at 60° north of east

24 km/h at 30° north of east

Correct Answer: A

Solution:

Using the parallelogram method of vector addition, the resultant velocity

R = \sqrt{25^2 + 10^2 + 2 \times 25 \times 10 \times \cos 120^\circ} \approx 22 \text{ km/h}

. The direction is given by

\tan^{-1}\left(\frac{10 \sin 120^\circ}{25 + 10 \cos 120^\circ}\right) \approx 30^\circ

north of east.

Mass

Speed

Velocity

Temperature

Correct Answer: C

Solution:

Velocity is a vector quantity because it has both magnitude and direction.

15 km/h

22 km/h

30 km/h

35 km/h

Correct Answer: B

Solution:

Using vector addition, the resultant velocity is calculated as approximately 22 km/h.

1 m/s²

2 m/s²

3 m/s²

4 m/s²

Correct Answer: A

Solution:

The centripetal acceleration

a_c

is given by

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

. Substituting the values,

a_c = \frac{10^2}{100} = 1 \text{ m/s}^2

2 s

3 s

4 s

5 s

Correct Answer: B

Solution:

Using the equation

x = v_{0x}t + \frac{1}{2}a_xt^2

, where

v_{0x} = 0

, solve for

t

when

x = 16

2 s

3 s

4 s

5 s

Correct Answer: C

Solution:

The x-coordinate

x(t) = v_{0x}t + \frac{1}{2}a_xt^2 = 0 + \frac{1}{2} \times 8 \times t^2 = 4t^2

. Setting

4t^2 = 16

, we get

t^2 = 4

, thus

t = 2 \text{ s}

The horizontal component of velocity remains constant.

The vertical component of velocity remains constant.

The acceleration is zero throughout the motion.

The projectile's path is a straight line.

Correct Answer: A

Solution:

In projectile motion, the horizontal component of velocity remains constant because there is no acceleration in the horizontal direction.

It is always zero.

It is always along the radius towards the center.

It is always perpendicular to the velocity vector.

It is always along the tangent to the path.

Correct Answer: B

Solution:

In uniform circular motion, the acceleration vector (centripetal acceleration) is always directed towards the center of the circle, along the radius, and is perpendicular to the velocity vector.

The magnitude of

\mathbf{R}

is always greater than the sum of magnitudes of

\mathbf{A}

and

\mathbf{B}

The magnitude of

\mathbf{R}

is always less than the sum of magnitudes of

\mathbf{A}

and

\mathbf{B}

The magnitude of

\mathbf{R}

is equal to the sum of magnitudes of

\mathbf{A}

and

\mathbf{B}

The magnitude of

\mathbf{R}

is equal to the difference of magnitudes of

\mathbf{A}

and

\mathbf{B}

Correct Answer: B

Solution:

The magnitude of the resultant vector

\mathbf{R}

is less than or equal to the sum of magnitudes of

\mathbf{A}

and

\mathbf{B}

, depending on the angle between them.

-4 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}}

4 \hat{\mathbf{i}} - 3 \hat{\mathbf{j}}

-3 \hat{\mathbf{i}} - 4 \hat{\mathbf{j}}

4 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}}

Correct Answer: A

Solution:

Rotating a vector

\mathbf{A} = a \hat{\mathbf{i}} + b \hat{\mathbf{j}}

90^\circ

counterclockwise results in a new vector

\mathbf{A'} = -b \hat{\mathbf{i}} + a \hat{\mathbf{j}}

. Thus,

\mathbf{A'} = -4 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}}

It is always conserved in a process.

It can never take negative values.

It does not vary from one point to another in space.

It has the same value for observers with different orientations of axes.

Correct Answer: D

Solution:

A scalar quantity is defined by its magnitude and is independent of the direction. Hence, it has the same value for observers with different orientations of axes.

The magnitude of a vector is always a vector.

The sum of two vectors is always a scalar.

The magnitude of a vector is always a scalar.

The direction of a vector does not affect its magnitude.

Correct Answer: C

Solution:

The magnitude of a vector is a scalar quantity, as it only has magnitude and no direction.

19.44 m/s²

48.61 m/s²

34.72 m/s²

69.44 m/s²

Correct Answer: A

Solution:

The centripetal acceleration

a_c

is given by

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

, where

v

is the speed and

r

is the radius. Converting speed to m/s:

500 \text{ km/h} = \frac{500 \times 1000}{3600} \text{ m/s} \approx 138.89 \text{ m/s}

. Thus,

a_c = \frac{(138.89)^2}{2000} \approx 19.44 \text{ m/s}^2

Mass

Temperature

Velocity

Time

Correct Answer: C

Solution:

Velocity is a vector quantity because it has both magnitude and direction, unlike mass, temperature, and time which are scalars.

0.2 rad/s

0.4 rad/s

0.6 rad/s

0.8 rad/s

Correct Answer: B

Solution:

The angular velocity

\omega

is given by

\omega = \frac{v}{r}

, where

v

is the linear speed and

r

is the radius. Thus,

\omega = \frac{20}{100} = 0.2 \text{ rad/s}

1.02 s

2.04 s

3.06 s

4.08 s

Correct Answer: B

Solution:

The time of flight

T

for a projectile is given by

T = \frac{2v_0 \sin \theta}{g}

. Here,

v_0 = 20

m/s,

\theta = 30^\circ

, and

g = 9.8

m/s². Thus,

T = \frac{2 \times 20 \times \sin 30^\circ}{9.8} = \frac{20}{9.8} \approx 2.04

Straight line

Circular

Parabolic

Elliptical

Correct Answer: C

Solution:

The path of a projectile is parabolic due to the constant acceleration of gravity acting downwards.

6 km/h

12 km/h

18 km/h

24 km/h

Correct Answer: B

Solution:

The total distance covered is twice the radius plus the circumference of the circle. The average speed is the total distance divided by the total time.

49.3 km/h

50 km/h

48.3 km/h

45 km/h

Correct Answer: A

Solution:

The average speed is calculated as total distance divided by total time:

\frac{23 \text{ km}}{28/60 \text{ hr}} = 49.3 \text{ km/h}

16\hat{\mathbf{i}} + 24\hat{\mathbf{j}}

m/s²

8\hat{\mathbf{i}} + 12\hat{\mathbf{j}}

m/s²

16\hat{\mathbf{i}} + 48\hat{\mathbf{j}}

m/s²

8\hat{\mathbf{i}} + 24\hat{\mathbf{j}}

m/s²

Correct Answer: C

Solution:

To find the acceleration, we differentiate the position vector twice with respect to time. The velocity is

\mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = (8t - 3)\hat{\mathbf{i}} + (6t^2)\hat{\mathbf{j}}

. The acceleration is

\mathbf{a}(t) = \frac{d\mathbf{v}}{dt} = 8\hat{\mathbf{i}} + 12t\hat{\mathbf{j}}

. At

t = 2

\mathbf{a}(2) = 8\hat{\mathbf{i}} + 24\hat{\mathbf{j}}

m/s².

They are always equal

Average speed is always less than average velocity

Average speed is always greater than or equal to average velocity

Average speed is always greater than average velocity

Correct Answer: C

Solution:

The average speed is greater than or equal to the magnitude of the average velocity.

v_0 \cos \theta_0

v_0 \sin \theta_0

v_0

v_0 \tan \theta_0

Correct Answer: B

Solution:

The vertical component of the initial velocity is given by

v_0 \sin \theta_0

0.25 m/s²

2.5 m/s²

25 m/s²

250 m/s²

Correct Answer: B

Solution:

Centripetal acceleration

a_c

is given by

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

. Convert speed to m/s and radius to meters, then calculate

a_c

80 m

100 m

120 m

140 m

Correct Answer: B

Solution:

Using the formula for projectile motion, calculate the range using the given height and initial speed.

5.1 m

10.2 m

15.3 m

20.4 m

Correct Answer: B

Solution:

The vertical component of the initial velocity is

v_{0y} = 20 \sin 30^\circ = 10 \text{ m/s}

. The maximum height

H

is given by

H = \frac{v_{0y}^2}{2g} = \frac{10^2}{2 \times 9.8} = 5.1 \text{ m}

3 m/s

5 m/s

7 m/s

9 m/s

Correct Answer: C

Solution:

Velocity is the derivative of position:

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

. At

t = 2.0

\mathbf{v} = 3.0 \hat{i} - 8.0 \hat{j}

. Magnitude is

\sqrt{3^2 + (-8)^2}

\frac{v_0^2 \sin^2 \theta_0}{2g}

\frac{v_0^2 \sin \theta_0}{g}

\frac{v_0^2}{2g}

\frac{v_0^2 \cos^2 \theta_0}{g}

Correct Answer: A

Solution:

The maximum height

H

for a projectile is given by

H = \frac{v_0^2 \sin^2 \theta_0}{2g}

, where

g

is the acceleration due to gravity.

Its direction changes

Its magnitude changes

It becomes a scalar

It remains unchanged

Correct Answer: B

Solution:

Multiplying a vector by a scalar changes its magnitude but not its direction.

6.0 \mathbf{i} - 8.0 \mathbf{j}

m/s

3.0 \mathbf{i} + 4.0 \mathbf{j}

m/s

6.0 \mathbf{i} - 4.0 \mathbf{j}

m/s

9.0 \mathbf{i} - 8.0 \mathbf{j}

m/s

Correct Answer: A

Solution:

Velocity is the derivative of position:

\mathbf{v} = \frac{d\mathbf{r}}{dt} = 3.0 \mathbf{i} - 4.0t \mathbf{j}

. At

t = 2.0

\mathbf{v} = 3.0 \mathbf{i} - 4.0 \times 2.0 \mathbf{j} = 6.0 \mathbf{i} - 8.0 \mathbf{j}

m/s.

The net acceleration is zero.

The velocity vector is constant.

The acceleration vector is directed towards the center of the circle.

The speed of the object changes continuously.

Correct Answer: C

Solution:

In uniform circular motion, the acceleration is centripetal, always directed towards the center of the circle.

It remains unchanged

It reverses

It becomes perpendicular

It becomes zero

Correct Answer: B

Solution:

Multiplying a vector by a negative scalar reverses its direction.

Correct Answer: A

Solution:

The resultant vector

\mathbf{R} = \mathbf{A} + \mathbf{B} = (3 - 5) \hat{i} + (4 + 2) \hat{j} = -2 \hat{i} + 6 \hat{j}

. The magnitude is

\sqrt{(-2)^2 + 6^2} = \sqrt{4 + 36} = \sqrt{40} = 6.32

The magnitude of a vector is always a vector.

Each component of a vector is always a scalar.

Vectors can only be added if they have the same magnitude.

A vector cannot have a negative magnitude.

Correct Answer: B

Solution:

Each component of a vector is a scalar because it represents a magnitude in a specific direction.

Tangent to the path

Towards the center of the circle

Away from the center of the circle

Parallel to the velocity vector

Correct Answer: B

Solution:

In uniform circular motion, the acceleration vector is directed towards the center of the circle, known as centripetal acceleration.

0.82

1.02

1.22

1.42

Correct Answer: A

Solution:

First convert speed to m/s:

900 \text{ km/h} = 250 \text{ m/s}

. The centripetal acceleration

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

. The ratio

\frac{a_c}{g} = \frac{62.5}{9.8} \approx 0.82

10.2 m

15.3 m

20.4 m

25.5 m

Correct Answer: A

Solution:

The maximum height

H

is given by

H = \frac{v_0^2 \sin^2 \theta}{2g}

. Substituting the values,

H = \frac{20^2 \times (\sin 45^\circ)^2}{2 \times 9.8} = \frac{400 \times 0.5}{19.6} = 10.2 \text{ m}

31.3 m/s

44.3 m/s

35.7 m/s

40.0 m/s

Correct Answer: D

Solution:

The maximum range

R

for projectile motion is given by

R = \frac{v_0^2}{g}

. Solving for

v_0

gives

v_0 = \sqrt{Rg} = \sqrt{100 \times 9.8} = 31.3 \text{ m/s}

Velocity

Acceleration

Mass

Displacement

Correct Answer: C

Solution:

Mass is a scalar quantity as it has only magnitude and no direction.

The speed is variable.

The acceleration is zero.

The velocity is constant.

The acceleration is towards the center.

Correct Answer: D

Solution:

In uniform circular motion, the acceleration is always directed towards the center of the circle.

14.0 m/s

20.0 m/s

18.0 m/s

16.0 m/s

Correct Answer: B

Solution:

The velocity at any time

t

is given by

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

. Substituting the given values,

\mathbf{v}(2) = 10.0 \hat{\mathbf{j}} + (8.0 \hat{\mathbf{i}} + 2.0 \hat{\mathbf{j}}) \times 2 = 16.0 \hat{\mathbf{i}} + 14.0 \hat{\mathbf{j}}

. The speed is the magnitude of this vector:

\sqrt{16.0^2 + 14.0^2} = 20.0

m/s.

22 \hat{i} + 16 \hat{j} \text{ m/s}

20 \hat{i} + 15 \hat{j} \text{ m/s}

22 \hat{i} + 15 \hat{j} \text{ m/s}

20 \hat{i} + 16 \hat{j} \text{ m/s}

Correct Answer: B

Solution:

The velocity

\mathbf{v}

at time

t

is given by

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

. Substituting the given values,

\mathbf{v} = (2 \hat{i} + 1 \hat{j}) + (4 \hat{i} + 3 \hat{j})(5) = 2 \hat{i} + 1 \hat{j} + 20 \hat{i} + 15 \hat{j} = 22 \hat{i} + 16 \hat{j} \text{ m/s}

25 m

50 m

75 m

100 m

Correct Answer: A

Solution:

The maximum range

R

is given by

R = \frac{v_0^2 \sin 2\theta}{g}

. For maximum range,

\theta = 45^\circ

, so

R = \frac{v_0^2}{g}

. Given

R = 100

v_0^2 = 100g

. The maximum height

H

when thrown vertically is

H = \frac{v_0^2}{2g} = \frac{100g}{2g} = 50

The horizontal component of velocity changes with time.

The vertical component of velocity remains constant.

The acceleration is always directed downward.

The trajectory is a straight line.

Correct Answer: C

Solution:

In projectile motion, the acceleration due to gravity is always directed downward, affecting only the vertical component of velocity.

4.9 m/s²

8.9 m/s²

12.3 m/s²

16.5 m/s²

Correct Answer: B

Solution:

The speed

v

of the stone is given by

v = \frac{2\pi r \times 14}{25} = \frac{2\pi \times 0.8 \times 14}{25} \approx 2.82 \text{ m/s}

. The centripetal acceleration

a_c = \frac{v^2}{r} = \frac{(2.82)^2}{0.8} \approx 8.9 \text{ m/s}^2

4 \hat{\mathbf{i}} + 2 \hat{\mathbf{j}}

m/s²

4 \hat{\mathbf{i}} + 4 \hat{\mathbf{j}}

m/s²

2 \hat{\mathbf{i}} + 2 \hat{\mathbf{j}}

m/s²

4 \hat{\mathbf{i}} + 0 \hat{\mathbf{j}}

m/s²

Correct Answer: A

Solution:

The acceleration

\mathbf{a}

is the second derivative of the position vector

\mathbf{r}

with respect to time

t

. Differentiating

\mathbf{r}

twice, we get

\mathbf{a} = 4 \hat{\mathbf{i}} + 2 \hat{\mathbf{j}}

m/s².

\frac{2v_0 \sin \theta_0}{g}

\frac{v_0 \sin \theta_0}{g}

\frac{v_0 \cos \theta_0}{g}

\frac{2v_0 \cos \theta_0}{g}

Correct Answer: A

Solution:

The time of flight

T

for a projectile is given by

T = \frac{2v_0 \sin \theta_0}{g}

, where

g

is the acceleration due to gravity. This formula accounts for the time taken to reach the peak and return to the ground.

V_{\text{average}} = \frac{1}{2} (v(t_1) + v(t_2))

V_{\text{average}} = \frac{r(t_2) - r(t_1)}{t_2 - t_1}

v(t) = v(0) + at

r(t) = r(0) + v(0)t + \frac{1}{2}at^2

Correct Answer: B

Solution:

The average velocity is defined as the change in position over time, which is option b.

0 km

1 km

2 km

3.14 km

Correct Answer: A

Solution:

The net displacement is the straight-line distance from the start to the end point. Since the cyclist returns to the starting point, the net displacement is 0 km.

12 km/h

10 km/h

15 km/h

20 km/h

Correct Answer: A

Solution:

The total path length is the diameter plus the circumference:

2 \times 1 + 2\pi \times 1 = 2 + 2\pi

km. The total time is 10 minutes = 1/6 hours. The average speed is

\frac{2 + 2\pi}{1/6} = 12

km/h.

27 km/h at 30° east of north

22 km/h at 13.4° east of north

30 km/h at 45° east of north

20 km/h at 60° east of north

Correct Answer: B

Solution:

Using the parallelogram method, the resultant velocity can be calculated using the law of cosines:

R = \sqrt{25^2 + 10^2 + 2 \times 25 \times 10 \times \cos(120°)}

which gives approximately 22 km/h. The direction is found using the law of sines:

\sin \alpha = \frac{10 \sin 120°}{22}

Solution:

In circular motion, the centripetal acceleration is directed towards the center of the circle, along the radius.

Motion in a Plane

AI Learning Assistant

CBSE Learning Objectives – Key Concepts & Skills You Must Know

Learning Objectives

CBSE Revision Notes & Quick Summary for Last-Minute Study

Chapter Notes on Motion in a Plane

3.1 Introduction

3.2 Scalars and Vectors

3.3 Multiplication of Vectors by Real Numbers

3.4 Addition and Subtraction of Vectors — Graphical Method

3.5 Resolution of Vectors

3.6 Vector Addition — Analytical Method

3.7 Motion in a Plane

3.8 Motion in a Plane with Constant Acceleration

3.9 Projectile Motion

3.10 Uniform Circular Motion

Points to Ponder

Exercises

CBSE Exam Tips, Important Questions & Common Mistakes to Avoid

Common Mistakes and Exam Tips

Common Pitfalls

Exam Tips

Multiple Choice Questions

Q1.If vectors i\mathbf{i}i and j\mathbf{j}j are unit vectors along x- and y- axis respectively, what is the magnitude of the vector i+j\mathbf{i} + \mathbf{j}i+j?

Correct Answer: A

Q2.Which statement is true regarding the path length and displacement of an object?

Correct Answer: C

Q3.An object is moving in a circular path of radius 2 m with a constant speed of 4 m/s. What is the magnitude of the centripetal acceleration?

Correct Answer: B

Q4.A projectile is launched with an initial velocity v0v_0v0​ at an angle θ0\theta_0θ0​ above the horizontal. Which of the following correctly represents the vertical component of the initial velocity?

Correct Answer: B

Q5.A passenger travels from a station to a hotel located 10 km away on a straight road. If a dishonest cabman takes a circuitous path of 23 km and reaches the hotel in 28 minutes, what is the magnitude of the average velocity?

Correct Answer: B

Q6.Which of the following statements is true for a particle in uniform circular motion?

Correct Answer: C

Q7.A projectile is launched with an initial velocity of 202020 m/s at an angle of 30∘30^\circ30∘ with the horizontal. What is the maximum height reached by the projectile?

Correct Answer: B

Q8.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 seconds, what is the magnitude of the centripetal acceleration of the stone?

Correct Answer: C

Q9.A motorboat is racing towards north at 25 km/h25 \text{ km/h}25 km/h and the water current in that region is 10 km/h10 \text{ km/h}10 km/h in the direction of 60∘60^\circ60∘ east of south. What is the resultant velocity of the boat?

Correct Answer: A

Q10.Which of the following is a vector quantity?

Correct Answer: C

Q11.A motorboat is moving north at 25 km/h while the water current is 10 km/h in the direction of 60° east of south. What is the resultant velocity of the boat?

Correct Answer: B

Q12.A cyclist moves along a circular path of radius 100 m with a constant speed of 10 m/s. What is the magnitude of the centripetal acceleration?

Correct Answer: A

Q13.A particle starts from the origin at t=0t = 0t=0 s with a velocity of 10.0j^10.0 \hat{j}10.0j^​ m/s and moves in the x−yx-yx−y plane with a constant acceleration of (8.0i^+2.0j^)(8.0 \hat{i} + 2.0 \hat{j})(8.0i^+2.0j^​) m/s². At what time is the xxx-coordinate of the particle 16 m?

Correct Answer: B

Correct Answer: C

Q15.A projectile is launched with an initial velocity v0v_0v0​ at an angle θ0\theta_0θ0​ above the horizontal. Which of the following statements is true about its motion?

Correct Answer: A

Q16.In a circular motion, if the velocity vector of a particle at a point is always along the tangent to the path, which of the following statements is true about the acceleration vector?

Correct Answer: B

Q17.If a vector A\mathbf{A}A is added to another vector B\mathbf{B}B, which of the following is true about the resultant vector R\mathbf{R}R?

Correct Answer: B

Q18.A vector A=3i^+4j^\mathbf{A} = 3 \hat{\mathbf{i}} + 4 \hat{\mathbf{j}}A=3i^+4j^​ is rotated by 90∘90^\circ90∘ counterclockwise in the plane. What is the new vector?

Correct Answer: A

Q19.Which of the following statements is true for a scalar quantity?

Correct Answer: D

Q20.Which of the following statements about vector quantities is true?

Correct Answer: C

Q21.An aircraft is flying at a constant speed of 500 km/h in a circular path of radius 2 km. What is the magnitude of its centripetal acceleration?

Correct Answer: A

Q22.Which of the following quantities is a vector?

Correct Answer: C

Q23.A cyclist moves along a circular path of radius 100 m with a constant speed of 20 m/s. What is the angular velocity of the cyclist?

Correct Answer: B

Q24.A projectile is launched with an initial velocity of 20 m/s20 \text{ m/s}20 m/s at an angle of 30∘30^\circ30∘ to the horizontal. What is the time of flight of the projectile? (Take g=9.8 m/s2g = 9.8 \text{ m/s}^2g=9.8 m/s2)

Correct Answer: B

Q25.What is the path of a projectile under the influence of gravity?

Correct Answer: C

Q26.A cyclist starts from the center of a circular park of radius 1 km, reaches the edge, cycles along the circumference, and returns to the center. If the round trip takes 10 minutes, what is the average speed of the cyclist?

Correct Answer: B

Q27.What is the average speed of a taxi that travels 23 km in 28 minutes?

Correct Answer: A

Q28.A particle moves in a plane with a position vector given by r(t)=(4t2−3t)i^+(2t3+5)j^\mathbf{r}(t) = (4t^2 - 3t)\hat{\mathbf{i}} + (2t^3 + 5)\hat{\mathbf{j}}r(t)=(4t2−3t)i^+(2t3+5)j^​ m, where ttt is in seconds. What is the particle's acceleration at t=2t = 2t=2 s?

Correct Answer: C

Q29.Which statement is true about the average speed and average velocity of an object?

Correct Answer: C