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# Motion in a Plane Class 11 Notes CBSE Physics Chapter 4 (Free PDF Download)

Last updated date: 09th Apr 2024
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## Revision Notes for CBSE Class 11 Physics Chapter 4 (Motion in a Plane) - Free PDF Download

Unlike temperature, work and distance, which can be represented by only magnitude, physical quantities like force, acceleration and displacement are represented by both magnitude and direction. This is what Class 11 Physics Chapter 4 Motion in a Plane is all about. If you have faced any challenge in understanding certain concepts of this chapter, you can refer to Motion in a Plane Class 11 Notes by Vedantu. The Chapter 4 Class 11 Physics Notes can also prove to be of immense help before the exams.

Motion in a Plane is an important chapter in Class 11 Physics. Thus we are providing revision study material that will help the students in revising the Motion in a Plane chapter in a small duration. Motion in a Plane consists of various important questions and answers which are required to be studied by the students, thus we also have incorporated those in our revision study.

Revise chapter 4 with this study material to be well prepared for the CBSE Class 11 exam.

Also, check CBSE Class 11 Physics revision notes for other chapters:

Motion in a Plane Chapter-Related Important Study Materials
It is a curated compilation of relevant online resources that complement and expand upon the content covered in a specific chapter. Explore these links to access additional readings, explanatory videos, practice exercises, and other valuable materials that enhance your understanding of the chapter's subject matter.

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Competitive Exams after 12th Science

## Motion in a Plane Class 11 Notes Physics - Basic Subjective Questions

### Section – A (1 Mark Questions)

1. What is the angle between $\vec{A}\times \vec{B}$ and $\vec{B}\times \vec{A}$ ?

Ans. The given vectors act along two parallel lines in opposite directions i.e. they are anti-parallel, so the angle between them is $180^{\circ}$.

2. What is the minimum number of coplanar vectors of different magnitudes which can give zero resultant?

Ans. Three.

If three vectors can be represented completely by the three sides of a triangle taken in the same order, then their resultant is zero.

3. What is the direction of centripetal force when a particle is following a circular path?

Ans. The direction of the centripetal force is towards the centre of the circle.

4. A gunman always keeps his gun slightly tilted above the line of sight while shooting. Why?

Ans. Because bullets follow parabolic trajectory under constant downward acceleration.

5. Define angular displacement.

Ans. Angle rotated by a position vector of the moving particle in a given time interval with some reference line is called its angular displacement.

### Section – B (2 Marks Questions)

6. A body is projected so that it has maximum range R. What is the maximum height reached during the flight?

Ans. Maximum height is $\dfrac{1}{4}$ of maximum range i.e. $h=\dfrac{R}{4}$ .

7.

(a) Give an example of a body moving with uniform speed but having a variable velocity and an acceleration that remains constant in magnitude but changes in direction.

(b) What is the direction of a centripetal acceleration with reference to the position vector of a particle moving in a circular path?

Ans. (a) A body having uniform circular motion.

(b) The centripetal acceleration vector points toward the centre of the circular path of motion and is an acceleration in radial direction.

8. A body moves from one corner of an equilateral triangle of side 10 cm to the same corner along the sides. Then, find the distance and displacement, respectively.

Ans. Referring the given diagram, we can say that,

Body moved from $A\rightarrow B\rightarrow C\rightarrow A$

Hence, total distance = 10 + 10 + 10 = 30 cm

But, total displacement is shortest distance between starting and ending point, i.e., $\Rightarrow 0$ .

9. If an object changes its position from point P(5,4,6) to point Q(3,7,9). Calculate displacement vector and its magnitude.

Ans. P(5,4,6) and Q(3,7,9)

Hence, displacement vector.

$\vec{r}=\vec{\vec{P}}\vec{\vec{Q}}=(3-5)\hat{i}+(7-4)\hat{j}+(9-6)\hat{k}$

$\Rightarrow \vec{r}=-2\hat{i}+3\hat{j}+3\hat{k}$

Magnitude of displacement $=r=\sqrt{(-2)^{2}+3^{2}+3^{2}}$

$\Rightarrow r=\sqrt{22}$

10. A train of length 200 m travelling at $30ms^{-1}$ overtakes another train of length 300 m travelling at $20ms^{-1}$ Find the time taken by the first train to pass the second.

Ans. Relative velocity of faster train with respect to slower train is 10 m/s. for overtaking the train relative distance to be covered 200+300=500 So, the time taken is 50 sec.

## PDF Summary - Class 11 Physics Motion in a Plane Notes (Chapter 4)

### 1. Scalars and Vectors:

Some quantities can be described by a unique number. For example, mass, time, distance and speed can be described using a single number. These are called scalar quantities.

To express to someone how to get to a location from some other location, one piece of information is not enough. To describe this fully, both distance and displacement are required.

Quantities that require both magnitude and direction to describe a situation fully are known as vectors. For example, displacement and velocity are vectors.

The vectors are denoted by putting an arrow over the symbols representing them.

For example, AB vector can be represented by $\overrightarrow{AB}$.

1.1 Unit Vector:

A unit vector has a magnitude of one and hence, it actually gives just the direction of the vector.

A unit vector can be determined by dividing the original vector by its magnitude

$\Rightarrow \hat{a}=\frac{{\vec{a}}}{\left| {\vec{a}} \right|}$

Unit vectors along different co–ordinate axis are as shown below:

(Image will be updated soon)

1.2 Addition, subtraction and scalar multiplication of vectors:

Consider two vectors as follows:

${{\vec{r}}_{1}}={{a}_{1}}\hat{i}+{{b}_{1}}\hat{j}$

${{\vec{r}}_{2}}={{a}_{2}}\hat{i}+{{b}_{2}}\hat{j}$

Then,

${{\vec{r}}_{1}}+{{\vec{r}}_{2}}=\left( {{a}_{1}}+{{a}_{2}} \right)\hat{i}+\left( {{b}_{1}}+{{b}_{2}} \right)\hat{j}$

${{\vec{r}}_{1}}-{{\vec{r}}_{2}}=\left( {{a}_{1}}-{{a}_{2}} \right)\hat{i}+\left( {{b}_{1}}-{{b}_{2}} \right)\hat{j}$

Multiplication of a vector by scalar quantity:

$c{{\vec{r}}_{1}}=c\left( {{a}_{1}}\hat{i}+{{b}_{1}}\hat{j} \right)=c{{a}_{1}}\hat{i}+c{{b}_{1}}\hat{j}$

Representation of ${{\vec{r}}_{1}}$ on the co–ordinate axis:

(Image will be updated soon)

Magnitude and direction of ${{\vec{r}}_{1}}$:

Magnitude of ${{\vec{r}}_{1}}\left( \left| {{{\vec{r}}}_{1}} \right| \right)=\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}}$

Direction of ${{\vec{r}}_{1}}$ is given by

$\tan \theta =\frac{{{b}_{1}}}{{{a}_{1}}}=\frac{component\text{ }y-axis}{component\text{ }along\text{ }x-axis}$

$\Rightarrow \theta ={{\tan }^{-1}}\left( \frac{{{b}_{1}}}{{{a}_{1}}} \right)$

1.3 Parallel vectors:

Two vectors are said to be parallel vectors if and only if they have the same direction. When any vector is multiplied by a scalar, a vector parallel to the original vector is obtained.

If $b=ka$, then b and a are parallel vectors. Generally, to find if two vectors are parallel or not, we should find their unit vectors.

1.4 Equality of vectors:

Two vectors (representing two values of the same physical quantity) are referred to as equal if their corresponding magnitudes and directions are the same.

For example, $\left( 3i+4j \right)m\,\,and\,\,\left( 3i+4j \right)\frac{m}{s}$ cannot be compared as they represent two different physical quantities.

When two or more vectors are added, the answer is referred to as the resultant. The resultant of two vectors is equivalent to the first vector followed immediately by the second vector.

(Image will be updated soon)

To determine the resultant of vectors a and b, the tail of vector b should be joined to the head of vector a. The resultant a+b is nothing but the direct vector from the tail of vector a to the head of vector b as shown below.

(Image will be updated soon)

This is known as triangle rule of vector addition. Another way to obtain the resultant vector is parallelogram rule of vector addition. Here, we draw vectors $\vec{a}$ and $\vec{b}$, with both the tails coinciding. Taking these two as adjacent sides of a parallelogram, we complete the parallelogram. Now, the diagonal through the common tails gives the sum of two vectors.

(Image will be updated soon)

Finding the magnitude of $\vec{a}+\vec{b}$ and its direction:

${{\left| AD \right|}^{2}}=A{{E}^{2}}+E{{D}^{2}}\text{ }$

Here,

$AE=\left| {\vec{a}} \right|+\left| b\,\cos \,\theta \right|$

$ED=b\,\sin \,\theta$

$\Rightarrow A{{D}^{2}}={{a}^{2}}+{{b}^{2}}{{\cos }^{2}}\theta +2ab\cos \theta +{{b}^{2}}{{\sin }^{2}}\theta$

$\Rightarrow A{{D}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\cos \theta$

$\Rightarrow AD=\sqrt{{{a}^{2}}+{{b}^{2}}+2ab\cos \theta }$

where,

$\theta$is the angle contained between $\vec{a}$ and $\vec{b}$;

Also,

$\tan \alpha =\frac{ED}{AE}=\frac{b\,\sin \theta }{a+b\cos \theta }$

where,

$\alpha$ is the angle which the resultant makes with the positive x-axis.

Subtraction of vectors:

Let $\vec{a}$ and $\vec{b}$ be two vectors. We define $\vec{a}-\vec{b}$ as the sum of vectors $\vec{a}$ and the vector$(-\vec{b})$.

$\Rightarrow \vec{a}-\vec{b}=\vec{a}+(-\vec{b})$

(Image will be updated soon)

Zero vector:

(Image will be updated soon)

In the given triangle, $\overrightarrow{PQ}+\overrightarrow{QR}+\overrightarrow{PR}$ should be equal to zero as the overall journey results in a return to the starting point.

$\Rightarrow \overrightarrow{PQ}+\overrightarrow{QR}+\overrightarrow{PR}=0$

Here are some other resultants:

(Image will be updated soon)

Here, $\vec{a}+\vec{b}+\vec{c}+\vec{d}+\vec{e}=0\Rightarrow \vec{e}=-(\vec{a}+\vec{b}+\vec{c}+\vec{d})$

(Image will be updated soon)

Here, $\vec{a}+\vec{b}+\vec{c}=0$

Resolution of vectors:

(Image will be updated soon)

Consider the given diagram above.

Here,

$\overrightarrow{OA}=\vec{a}$

$\overrightarrow{OA}=\overrightarrow{OB}+\overrightarrow{OC}$

$\left| \overrightarrow{OB} \right|=a\cos \,\theta$

$\left| \overrightarrow{OC} \right|=a\,\sin \theta$

If $\hat{i}$ and $\hat{j}$ denote vectors of unit magnitude along OX and along OY respectively, we get

$\overrightarrow{OB}=a\,\cos \theta \,\hat{i}$

$\overrightarrow{OC}=a\,\sin \theta \,\hat{j}$

$\Rightarrow \vec{a}=\left( a\,\cos \theta \right)\hat{i}+\left( a\,\sin \theta \right)\hat{j}$

(Image will be updated soon)

1.6 Dot product or scalar product of two vectors:

(Image will be updated soon)

Dot product of vectors $\vec{a}$ and $\vec{b}$ is given by

$\vec{a}.\vec{b}=\left| {\vec{a}} \right|\left| {\vec{b}} \right|\,\cos \theta$

If $\theta =0;$

$\Rightarrow \vec{a}.\vec{b}=\left| {\vec{a}} \right|\left| {\vec{b}} \right|\,$

If $\theta ={{90}^{\circ }}$;

$\Rightarrow \vec{a}.\vec{b}=\left| {\vec{a}} \right|\left| {\vec{b}} \right|\,{{90}^{\circ }}=0$

Dot product of unit vectors are given by

$\hat{i}\cdot \hat{i}=\left| {\hat{i}} \right|\left| {\hat{i}} \right|\,\cos {{0}^{\circ }}={{i}^{2}}\times 1=1$

Similarly,

$\hat{j}\cdot \hat{j}=1$;

$\hat{k}\cdot \,\hat{k}=1$

Now,

$\hat{i}\cdot \hat{j}=\left| {\hat{i}} \right|\left| {\hat{j}} \right|\,\cos {{90}^{\circ }}=1\times 1\times 0=0$

Similarly,

$\hat{j}\cdot \hat{k}=0\,$;

$\hat{k}\cdot \hat{i}=0$

Dot products are commutative and distributive:

$\vec{a}\cdot \vec{b}=\vec{b}\cdot \vec{a}$

$\vec{a}\cdot \left( \vec{b}+\vec{c} \right)=\vec{a}\cdot \vec{b}+\vec{a}\cdot \vec{c}$

### 2. MOTION IN 2D (PLANE)

2.1 Position vector and Displacement:

The position vector $\vec{r}$ of a particle P, located in a plane with reference to the origin of on xy–coordinate system is given by $\vec{r}=x\hat{i}+y\hat{j}$, as shown below.

(Image will be updated soon)

Now, if the particle moves along the path as shown to a new position ${{P}_{1}}$ with the position vector ${{\vec{r}}_{1}}$;

${{\vec{r}}_{1}}={{x}_{1}}\hat{i}+{{y}_{1}}\hat{j}$

Change in position of the particle is nothing but its displacement given by,

$\Delta \vec{r}={{\vec{r}}_{1}}-\vec{r}=\left( {{x}_{1}}\hat{i}+{{y}_{1}}\hat{j} \right)-\left( x\hat{i}+y\hat{j} \right)$

$\Rightarrow \Delta \vec{r}==\left( {{x}_{1}}-x \right)\hat{i}+\left( {{y}_{1}}-y \right)\hat{j}$

$\Delta \vec{r}==\Delta x\hat{i}+\Delta y\hat{j}$

From the figure, it can also be seen that

$\vec{r}+\Delta \vec{r}={{\vec{r}}_{1}}\,\,\,\,\,\,\,or\,\,\,\,\,\Delta \vec{r}={{\vec{r}}_{1}}-\,\,\vec{r}$, which is nothing but the triangle law of vector addition.

2.2 Average velocity:

Average velocity is given by,

${{\vec{v}}_{avg}}=\frac{\Delta \vec{r}}{\Delta t}=\frac{\Delta x\hat{i}+\Delta y\hat{j}}{\Delta t}$

$\Rightarrow {{\vec{v}}_{avg}}={{v}_{x}}\hat{i}+{{v}_{y}}\hat{j}$

Note: Direction of the average velocity is the same as that of $\Delta \vec{r}$.

2.3 Instantaneous velocity:

Instantaneous velocity is given by,

$\vec{v}=\underset{\Delta t\to 0}{\mathop{\lim }}\,\frac{\Delta v}{\Delta t}=\frac{d\vec{r}}{dt}$

$\Rightarrow \vec{v}={{v}_{x}}\hat{i}+{{v}_{y}}\hat{j}$

(Image will be updated soon)

Here,

${{v}_{x}}=\frac{dx}{dt}\,\,and\,\,{{v}_{y}}=\frac{dy}{dt}$

$\Rightarrow \left| {\vec{v}} \right|=\sqrt{{{v}_{x}}^{2}+{{v}_{y}}^{2}}$

Also,

$\tan \theta =\frac{{{v}_{y}}}{{{v}_{x}}}$

$\Rightarrow \theta ={{\tan }^{-1}}\left( \frac{{{v}_{y}}}{{{v}_{x}}} \right)$

Note: The direction of instantaneous velocity at any point on the path of an object is the tangent to the path at that point and is in the direction of motion.

2.4 Average acceleration:

$\theta ={{\tan }^{-1}}\left( \frac{{{v}_{y}}}{{{v}_{x}}} \right)$Average acceleration is given by,

${{\vec{a}}_{avg}}=\frac{\Delta \vec{v}}{\Delta t}=\frac{\Delta {{v}_{x}}}{\Delta t}\hat{i}+\frac{\Delta {{v}_{y}}}{\Delta t}\hat{j}$

$\Rightarrow {{\vec{a}}_{avg}}={{a}_{x}}\hat{i}+{{a}_{y}}\hat{j}$

2.5 Instantaneous acceleration:

Instantaneous acceleration is given by,

$\vec{a}=\frac{dv}{dt}=\frac{d{{v}_{x}}}{dt}\hat{i}+\frac{d{{v}_{y}}}{dt}\hat{j}$

$\Rightarrow \vec{a}={{a}_{x}}\hat{i}+{{a}_{y}}\hat{j}$

### 3. PROJECTILE MOTION:

When a particle is projected obliquely close to the surface of the earth, it moves simultaneously in horizontal and vertical directions. Motion of such a particle is referred to as projectile motion.

(Image will be updated soon)

Here, a particle is projected at an angle with an initial velocity ‘u’.

Considering the projectile motion given in the diagram above, let us calculate the following:

(a) time taken to reach A from O

(b) horizontal distance covered (OA)

(c) maximum height reached during the motion

(d) velocity at any time ‘t’ during the motion

 Horizontal axis Vertical axis ${{u}_{x}}=u\,\cos \,\theta$ ${{a}_{x}}=0$ (In the absence of any external force, ${{a}_{x}}$would be assumed to be zero). ${{u}_{y}}=\,u\,\sin \,\theta$ ${{a}_{y}}=-g$ ${{s}_{y}}={{u}_{y}}t+\frac{1}{2}{{a}_{y}}{{t}^{2}}$ $\Rightarrow 0-0=u\,\sin \,\theta t-\frac{1}{2g{{t}^{2}}}$ $\Rightarrow$ ${{s}_{x}}={{u}_{x}}t+\frac{1}{2{{a}_{x}}{{t}^{2}}}$ $\Rightarrow x-0=u\,\cos \theta \,t$ $\Rightarrow x=u\,\cos \,\theta \,\times 2u\sqrt{g}$ $\Rightarrow x=\frac{2{{u}^{2}}\,\cos \theta \sin \theta }{g}$ $\Rightarrow$ $\left( \because 2cos\theta sin\theta =\text{ }sin2\theta \right)$ Horizontal distance covered is known as Range (R). ${{v}_{y}}={{u}_{y}}+{{a}_{y}}t$ $$It depends on time ‘t’. It is not constant.It’s magnitude first decreases; becomes zeroand then increases. {{v}_{x}}={{u}_{x}}+{{a}_{x}}t$$ It is independent of t. It is constant.Time of ascent and time of descent:At the top most point, ${{v}_{y}}=0$ ${{v}_{y}}={{u}_{y}}+{{a}_{y}}t$ $\Rightarrow 0=u\,\sin \theta -gt$ $\Rightarrow {{t}_{1}}=\frac{u\sin \theta }{g}$ $\Rightarrow {{t}_{2}}=T-{{t}_{1}}=\frac{u\,\sin \theta }{g}$ $\Rightarrow$ Maximum height obtained by the particleMethod 1: Using time of ascent;${{s}_{y}}={{u}_{y}}{{t}_{1}}+\frac{1}{2}ay{{t}_{1}}^{2}$ $\Rightarrow$ Method 2: Using third equation of motion${{v}_{y}}^{2}-{{u}_{y}}^{2}=2{{a}_{y}}{{s}_{y}}$ $\Rightarrow 0-{{u}^{2}}\,{{\sin }^{2}}\theta =-2g{{s}_{y}}$ $\Rightarrow$

Maximum Range:

$R=\frac{{{u}^{2}}\sin 2\theta }{g}\,\,and\,\,{{R}_{\max }}=\frac{{{u}^{2}}}{g}$

Range is maximum when $\sin 2\theta$ is maximum;

$\Rightarrow \max \left( \sin 2\theta \right)=1\,\text{ }or\text{ }\theta ={{45}^{\circ }}$

3.1 Analysis of velocity in case of a projectile:

(Image will be updated soon)

From the above equations;

i) ${{v}_{1x}}={{v}_{2x}}={{v}_{3x}}={{v}_{4x}}={{u}_{x}}=u\,\cos \theta$

which suggests that the velocity along x axis remains constant.

[as there is no external force acting along that direction]

ii)

a) The magnitude of velocity along y axis first decreases and then it increases after the top most point.

b) At the top most point, magnitude of velocity is zero.

c) Direction of velocity is in the upward direction while ascending and is in the downward direction while descending.

d) Magnitude of velocity at A is the same as magnitude of velocity at O; but the directions are opposite.

e) Angle which the net velocity makes with the horizontal can be evaluated by,

$\tan \alpha =\frac{{{v}_{y}}}{{{v}_{x}}}=\frac{velocity\text{ }along\text{ }y\text{ }axis}{velocity\text{ }along\text{ }x\text{ }axis}$

f) Net velocity is always along the tangent.

3.2 Equation of trajectory:

Trajectory refers to the path traced by the body. To determine the trajectory, we should find the relation between y and x by eliminating time.

(Image will be updated soon)

 Horizontal Motion Vertical Motion ${{u}_{x}}=u\,\cos \theta$ ${{a}_{x}}=0$ ${{s}_{x}}=u\,\cos \,\theta \,t=x$ $\Rightarrow t=\frac{x}{u\,\cos \,\theta }$ ${{u}_{y}}=u\,\sin \,\theta$ ${{a}_{y}}=-g$ ${{s}_{y}}={{u}_{y}}t+\frac{1}{2}{{a}_{y}}{{t}^{2}}$ $\Rightarrow y=u\,\sin \theta \left( \frac{x}{u\,\cos \,\theta } \right)-\frac{1}{2}g\frac{{{x}^{2}}}{{{u}^{2}}{{\cos }^{2}}\theta }$



(i) This is the equation of a parabola.

(ii) Because the coefficient of ${{x}_{2}}$ is negative, it is an inverted parabola.

(Image will be updated soon)

Path of the projectile is a parabola.

$R=\frac{2{{u}^{2}}\sin \theta \cos \theta }{g}$

$\Rightarrow \frac{2{{u}^{2}}}{g}=\frac{R}{\sin \theta cos\theta }$

Substituting this value in the above equation, we have,

$\Rightarrow$

### 4. RELATIVE MOTION:

Relativity is a very common term. In physics, we use relativity very oftenly.

For example, consider a moving car and yourself (observer) as shown below.

(Image will be updated soon)

Case I: If you are observing a car moving on a straight road, then you say that the velocity of car is 20m/s; which means that velocity of car relative to you is 20m/s; or, velocity of car relative to the ground is 20m/s (as you are standing on the ground.

Case II: If you go inside this car and observe, you would find that the car is at rest while the road is moving backwards. Then, you would say, the velocity of the car relative to the car is 0m/s.

Mathematically, velocity of B relative to A is represented as

${{\vec{v}}_{BA}}={{\vec{v}}_{B}}-{{\vec{v}}_{A}}$

This, being a vector quantity, direction is very important.

$\therefore \,\,{{\vec{v}}_{BA}}\ne {{\vec{v}}_{AB}}$

### 5. RIVERBOAT PROBLEMS:

In riverboat problems, we come across the following three terms:

${{\vec{v}}_{r}}=$ absolute velocity of river.

${{\vec{v}}_{br}}=$ velocity of a boatman with respect to river or velocity of a boatman in still water, and

${{\vec{v}}_{b}}=$ absolute velocity of boatman.

Clearly, it is important to note that ${{\vec{v}}_{br}}$ is the velocity of a boatman with which he steers and ${{\vec{v}}_{b}}$ is the actual velocity of boatman relative to ground. Further,

${{\vec{v}}_{b}}={{\vec{v}}_{br}}+{{\vec{v}}_{r}}$

Now, let us derive a few standard results and their special cases.

A boatman starts from point A on one bank of a river with velocity ${{\vec{v}}_{br}}$ in the direction shown in figure. River is flowing along positive x-direction with velocity ${{\vec{v}}_{r}}$. Width of the river is ‘d’. Then,

${{\vec{v}}_{b}}={{\vec{v}}_{br}}+{{\vec{v}}_{r}}$

Therefore,

${{v}_{bx}}={{v}_{rx}}+{{v}_{brx}}={{v}_{r}}-{{v}_{br}}\sin \theta$

And

${{v}_{by}}={{v}_{by}}+{{v}_{bry}}=0+{{v}_{br}}\cos \theta ={{v}_{br}}\cos \theta$

(Image will be updated soon)

Now, the time taken by the boatman to cross the river is given by,

$t=\frac{d}{{{v}_{by}}}=\frac{d}{{{v}_{br}}\cos \theta }\,$

$\Rightarrow t=\frac{d}{{{v}_{br}}\cos \theta }\,\,\,$…(1)

Further, displacement along the x-axis when he reaches on the other bank (also called as drift) is given by,

$x={{v}_{bx}}t=\left( {{v}_{r}}-{{v}_{br}}\sin \theta \right)\frac{d}{{{v}_{br}}\cos \theta }$

$\Rightarrow x=\left( {{v}_{r}}-{{v}_{br}}\sin \theta \right)\frac{d}{{{v}_{br}}\cos \theta }$…(2)

Condition when the boatman crosses the river in shortest interval of time:

From (1), it can be seen that time (t) will be minimum when $\theta =\text{ }0{}^\circ$ i.e., the boatman should steer his boat perpendicular to the river current.

Condition when the boat wants to reach point B, i.e., at a point just opposite from where he started (shortest distance):

In this case, the drift (x) should be zero.

$\Rightarrow x=0$

$\Rightarrow \left( {{v}_{r}}-{{v}_{br}}\sin \theta \right)\frac{d}{{{v}_{br}}\cos \theta }=0$

$\Rightarrow {{v}_{r}}={{v}_{br}}\sin \theta$

$\Rightarrow \sin \theta =\frac{{{v}_{r}}}{{{v}_{br}}}\,$

$\Rightarrow \theta ={{\sin }^{-1}}\left( \frac{{{v}_{r}}}{{{v}_{br}}} \right)$

Clearly, to reach point B, the boatman should row at an angle $\theta ={{\sin }^{-1}}\left( \frac{{{v}_{r}}}{{{v}_{br}}} \right)$ upstream from AB.

Now, time is given by,

$t=\frac{d}{{{v}_{b}}}=\frac{d}{\sqrt{{{v}_{br}}^{2}-{{v}_{r}}^{2}}}$

As $\sin \theta$ is not greater than 1 and when ${{v}_{r}}\ge {{v}_{br}}$, then the boatman can never reach at point B.

Because, when ${{v}_{r}}=\text{ }{{v}_{br}}$, $\sin \theta ~=1$ or $\sin \theta =90{}^\circ$ and it is just impossible to reach at B if $\theta =90{}^\circ$.

Similarly, when ${{v}_{r}}>{{v}_{br}}$, $sin\theta ~>1$, i.e., no such angle exists. Practically, it can be realized in this manner that it is not possible to reach at B if the river velocity (${{v}_{r}}$) is too high.

6. RELATIVE VELOCITY OF RAIN WITH RESPECT TO A MOVING MAN:

Consider a man walking west with velocity ${{\vec{v}}_{m}}$, represented by OA. Let the rain be falling vertically downwards with velocity ${{\vec{v}}_{r}}$, represented by OB, as shown in the following figure. To find the relative velocity of rain with respect to man (i.e.,${{\vec{v}}_{rm}}$) assume the man to be at rest by imposing a velocity $-{{\vec{v}}_{m}}$ on the man and apply this velocity on rain also. Now, the relative velocity of rain with respect to man would be the resultant velocity of ${{\vec{v}}_{r}}(=\overrightarrow{OB})$ and $-{{\vec{v}}_{m}}\left( =\overrightarrow{OC} \right)$, which would be represented by the diagonal OD of rectangle OBDC.

$\Rightarrow {{v}_{rm}}=\sqrt{{{v}_{r}}^{2}+{{v}_{m}}^{2}+2{{v}_{r}}{{v}_{m}}\cos {{90}^{\circ }}}=\sqrt{{{v}_{r}}^{2}+{{v}_{m}}^{2}}$

(Image will be updated soon)

If $\theta$is the angle which ${{v}_{rm}}$makes with the vertical direction, then

$\tan \theta =\frac{BD}{OB}=\frac{{{v}_{m}}}{{{v}_{r}}}$

$\Rightarrow \theta ={{\tan }^{-1}}\left( \frac{{{v}_{m}}}{{{v}_{r}}} \right)$

Here, angle $\theta$ is from the vertical towards west and is written as $\theta$, west of vertical.

Note: In the above problem, if the man wants to protect himself from the rain, he should hold his umbrella in the direction of relative velocity of rain with respect to man. i.e., the umbrella must be held making an angle $~\left( \theta =ta{{n}^{1}}\left( \frac{{{v}_{m}}}{{{v}_{r}}} \right) \right)$, west of the vertical.

Besides textbooks, you can also download Motion in a Plane Class 11 Notes PDF which comes entirely free of cost. Drafted in a PDF format, you can easily avail the same from Vedantu’s website. Save the file at a location as per your convenience; it can be at your desktop, laptop or tab and study it anytime anywhere.

## Notes of Physics Class 11 Chapter 4: Important Discussions

Motion in a Plane comprises some important topics which are discussed below, students might take a note of these topics:

### Introduction to Plane Motion

The very first section of Class 11 Physics Notes Chapter 4 deals with the meanings of velocity, acceleration and magnitude. Know about the various terms and their definitions in Physics by studying the Motion in a Plane Class 11 Notes.

### Motion in a Plane

Here, students will get to know about different motions in a plane like circular motion, projectile motion, etc. Besides, application of motion in a straight line’s equations are applied in x and y directions to find out the motion in a plane’s equations are also described in Motion in a Plane Class 11 Notes.

### Projectile Motion

By studying from Physics Ch 4 Class 11 Notes, you will get a clear idea about projectile motion (a special type of motion in a plane). Various projectile motion examples are also provided in the Motion in a Plane Class 11 Notes.

### Scalars and Vectors

Along with the definitions of scalar and vector quantities, this section of Physics Class 11 Notes discusses differences and characteristics between the two. Furthermore, explanations of unit vectors, equal vectors, zero vectors, negative of a vector, parallel vectors, displacement vectors and coplanar vectors are also given in Motion in a Plane Class 11 Notes.

### Resolution of Vectors, and Addition and Subtraction of Vectors

In this section of Class 11 Physics Notes Chapter 4, students will get enlightened how a vector can be resolved and what is the process of resolution all about. Specifically in the world of Physics, vectors are resolved as per x, y and z coordinates. Next comes the techniques of addition and subtraction of vectors. Both geometrical and analytical methods are discussed in detail in Motion in a Plane Class 11 Notes, assisting you to have a proper grasp over the same.

### 2D Relative Velocity

This section in Chapter 4 Physics Class 11 Notes starts with the explanation of relative motion velocity. Following which you will get to study 2D relative motion velocity in the Motion in a Plane Class 11 Notes, which is elucidated with a precise and straightforward derivation.

### Uniform Circular Motion

The final section of Motion in a Plane Class 11 Notes focusses on uniform circular motion along with the descriptions of variables used in the same, for instance, angular displacement, angular acceleration, angular velocity and centripetal acceleration. Later in this segment of Class 11th Physics Chapter 4 Notes, more about projectile motion are elaborated comprising 2D projectiles, essential pointers of projectile motion, etc.

You must note that each section is provided with solved numericals based on individual concepts. Practise them thoroughly to obtain full marks. So, without wasting much time, get the Vedantu app on your mobile phone and download the notes of Physics Class 11 Chapter 4 today!

## Motion in a Plane (CBSE) Class 11 Physics Notes - Features of the Notes

Motion in a Plane is curated specially for the CBSE Class 11 Physics students. Let us check what are the features of these notes from the following pointers:

1. The Laws of Motion is a vital chapter and thus important questions are picked from this chapter and the revision notes are prepared.

2. This note consists of all the previous year’s questions and answers.

3. Our master teachers designed these revision notes keeping in mind the CBSE-type questions.

4. After thorough research, the notes are prepared and thus they can be relied on.

5. These notes can be accessed in the form of a PDF incurring zero cost, which makes it free and readable

## Motion in a Plane (CBSE) Class 11 Physics Notes - Importance of Revision Notes

Let us now check how the CBSE Class 11 students will be benefited by revising from these revision notes of Motion in a Plane:

1. The students are not required to make any extra revision material for them, they can anytime refer to this free revision pdf.

2. The students can rely on the notes as they are curated by master teachers.

3. The students will get all the important questions in one capsule, they do not have to search for questions from other study material.

4. The students can focus elsewhere, as revision from this study material can be done lucidly.

5. Revising will help a student to be in touch with the already studied chapters, so this revision material will thus help a student to do the same.

Physics Chapter 4 Class 11 Notes cover all the topics and concepts that are vital for exams along with worked out numericals. Prepared by expert tutors and written in a lucid language, this material ensures a crystal clear comprehension of the chapter. Hence, you can remain assured to receive commendable grades.

## FAQs on Motion in a Plane Class 11 Notes CBSE Physics Chapter 4 (Free PDF Download)

1. What are the types of motion?

Generally, all mechanical devices come with an input motion and later changes to force to produce an output motion, and the types of motion are rotary, linear, oscillating and reciprocating. Firstly, when motion takes place around an axis or a fixed joint like a wheel, it is called as rotary motion. Secondly, when motion occurs in a straight line, it is considered linear - for example, the movement of trains. Thirdly, oscillating motion can be expressed as forward and backward movement of from a fixed axis. Lastly, repeated back and forth or down and up motion is known as reciprocating motion.

2. What are the subtopics of Ch 4 Physics Class 11 Notes?

The various subtopics of class 11 Physics Ch 4 Notes are Introduction about Oscillatory Motion, Scalars and Vectors, Multiplication of Vectors with Real Numbers, Subtraction and Addition of Vectors, Resolution of Vectors, Addition of Vectors with Analytical Method, Motion in a Plane, Motion in a Plane with Constant Acceleration, 2D Relative Velocity and Uniform Circular Motion.

3. What are the laws of motion?

The three fundamental laws of motion are (i) All objects or device move in a straight or linear direction unless an external force acts upon it, (ii) Force exerted on a body is directly proportional to the mass and acceleration of the body and (iii) All actions have an equal and opposite reaction.

4. Explain whether the statement is true or false. A displacement vector is a position vector.

The given statement ‘a displacement vector is a position vector’ is true. A displacement vector provides the position or state of any point similar to the position vector. The position and displacement vector is similar to some extent. The difference between the displacement and position vector is that the displacement vector describes the position of any point concerning the other points instead of origin. While the position vector states the position of any point with respect to the origin. This is how the statement proves to be true.

5. Does the nature of any vector quantity change when it is multiplied by any scalar quantity?

The nature of a vector quantity may or may not be altered when multiplied with a scalar quantity. For instance, when a vector quantity is multiplied by any natural number like 1, 3, 6 its nature doesn’t change. On the other hand, when a scalar quantity is multiplied with a vector quantity, then the nature of vector quantity changes. For instance, when velocity (vector) is multiplied with a scalar quantity like mass, then it gives momentum, which is a quantity whose nature is different from velocity.

6. What do you mean by tensor?

Tensor is a physical quantity that doesn’t have direction. Instead, it has various values in different directions. Hence, it is neither a scalar quantity nor a vector quantity. For example, a moment of inertia of any object does not have any direction but it has various values in different directions. Thus, it is neither a scalar quantity nor a vector quantity. It is an example of a tensor. Stress, density, strain and refractive index are some other examples of tensors.

7. Explain one-dimensional motion

In 1-D motion or one-dimensional motion, only a single coordinate specifies the position of any object. In this type of motion, only one plane out of three planes describes the movement of an object with respect to the starting point or origin. Here are some examples: the motion of a car in a straight line, movement of a train in a straight line, a man cycling on a straight path, an object falling on the ground in a straight line due to the force of gravitation. If you want to understand more about motion along with some interesting examples, then visit the page NCERT Solutions for Class 11 Physics and download the solution PDF free of cost.

8. Explain two-dimensional motion.

In two-dimensional motion or 2-D motion, only two coordinates describe the position of an object. In this type of motion, only two planes out of three coordinates or planes specify the motion of the object with respect to the origin or starting point. Here are some examples: bike riding on a circular path, death well rides, an object thrown at an angle. If you want to understand different types of motions along with some exciting examples, then visit the Vedantu app.