Motion in a Plane Class 11 NCERT Solutions
NCERT Solutions for Motion in a Plane Class 11 Physics is provided in the form of a free PDF as this chapter is very important from the examination point of view. It would benefit the students to prepare for their assessments. Chapter 5 Physics Class 11 NCERT Solutions have been prepared to keep in mind the latest syllabus prescribed by the CBSE. These solutions are prepared by skilled teachers who have been teaching for years and are very well accustomed to the syllabus and scoring schemes.
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Subject:  
Chapter Name:  Chapter 4  Motion In A Plane 
ContentType:  Text, Videos, Images and PDF Format 
Academic Year:  202425 
Medium:  English and Hindi 
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Class 11 Physics Chapter 5 NCERT Solutions would help the student obtain a better knowledge of the subject. We have added a free PDF to download so that the students can study the notes from anywhere and at any time. The study includes important concepts discussed and tips on how to utilize the NCERT Solutions for Motion in a Plane in the study curriculum of a student.
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Motion in a Plane Chapter at a Glance  Class 11 NCERT Solutions
Scalar quantities are quantities with magnitudes only. Examples are distance, speed, mass and temperature
Vector quantities are quantities with magnitude and direction both. Examples are displacement, velocity and acceleration. They ob key special rules of vector algebra.
A vector $ \vec{A}$ multiplied by a real number λ is also a vector, whose magnitude is λ times the magnitude of the vector $ \vec{A}$ and whose direction is the same or opposite depending upon whether λ is positive or negative.
Two vectors $ \vec{A}$ and $ \vec{B}$ may be added graphically using head to tail method or parallelogram method.
Vector addition is commutative: $\vec{A}+\vec{B}=\vec{B}+\vec{A}$
It also obeys the associative law:
$(\vec{A+B)}+\vec{C}=\vec{A}+(\vec{B+C})$
A null or zero vector is a vector with zero magnitude. Since the magnitude is zero, we don’t have to specify its direction. It has the properties:
$\vec{A}+0=\vec{A}$
$\vec{A}0=0$
40\vec{A}=0$
The subtraction of vector $\vec{B}$ from $\vec{A}$ is defined as the sum of $\vec{A}$ and $\vec{B}$ :
$\vec{A}\vec{B}=\vec{A}+(\vec{B})$
A vector A can be resolved into component along two given vectors a and b lying in the same plane:
$\vec{A}=\lambda \vec{A}+\mu \vec{B}$ where λ and µ are real numbers.
A unit vector associated with a vector $\vec{A}$ has magnitude one and is along the vector $ \vec{A}$ :
$\hat{n}=\dfrac{\vec{A}}{A}$
The unit vectors $ \hat{a},\hat{j}$ and $\hat{k}$ are vectors of unit magnitude and point in the direction of the x, y, and zaxes, respectively in a righthanded coordinate system.
A vector $\vec{A}$ can be expressed as $\vec{A}=A_{x}\hat{i}+A_{y}\hat{j}$ where Ax,Ay are its components along x, and yaxes. If vector $\vec{A}$ makes an angle θ with the xaxis, then $Ax=\vec{A}cos\theta$ , $Ay=\vec{A}sin\theta$ and $A=\vec{A}=\sqrt{A_{x}^{2}+A_{y}^{2}}$ , $tan\theta \dfrac{A_{y}}{A_{x}}$
Vectors can be conveniently added using analytical method. If sum of two vectors $\vec{A}$ and $\vec{B}$ , that lie in xy plane, is R, then: $\vec{R}=R_{x}\hat{i}+R_{y}\hat{j}$ . where, Rx=Ax+Bx and Ry=Ay+By
The position vector of an object in xy plane is given by $r=x\hat{i}+y\hat{j}$ and the displacement from position r to position ${r}'$ is given by
$\Delta \hat{r}=\vec{r'}\vec{r}$
$({x}'x)\hat{i}+({y}'y)\hat{j}$
$\Delta x\hat{i}+\Delta y\hat{j}$
If an object undergoes a displacement $\Delta r$ in time $\Delta t$ , its average velocity is given by $V=\dfrac{\Delta r}{\Delta t}$ . The velocity of an object at time t is the limiting value of the average velocity as $\Delta t$ tends to zero: $V=\lim_{\Delta t\rightarrow 0}\dfrac{\Delta r}{\Delta t}=\dfrac{dr}{dt}$ . It can be written in unit vector notation as: $V=V_{x}\hat{i}+V_{y}\hat{i}+V_{z}\hat{k}$ where $V_{x}=\dfrac{dx}{dt}$ , $V_{y}=\dfrac{dy}{dt}$ , $V_{z}=\frac{dz}{dt}$
When position of an object is plotted on a coordinate system, $\vec{V}$ is always tangent to the curve representing the path of the object.
If the velocity of an object changes from $\vec{V}$ to $\vec{V'}$ in time $\Delta t$ , then its average acceleration is given by: $\vec{a}=\dfrac{\vec{v}\vec{v'}}{\Delta t}=\frac{\Delta \vec{v}}{\Delta t}$
The acceleration $\vec{a}$ at any time t is the limiting value of $\vec{a}$ as $\Delta t\rightarrow 0$ ,
$\vec{a}=\lim_{\Delta t\rightarrow 0}\dfrac{\Delta \vec{v'}}{\Delta t}=\dfrac{d\vec{v}}{dt}$
In component form, we have: $\vec{a}=a_{x}\hat{i}+a_{y}\hat{j}+a_{z}\hat{k}$ Where, $a_{x}=\dfrac{dv_{x}}{dt}$ , $a_{y}=\dfrac{dv_{y}}{dt}$ , $a_{z}=\dfrac{dv_{x}}{dt}$
Relative motion can be defined as the comparison between the motions of a single object to the motion of another object moving with the same velocity. Relative motion can be easily found out with the help of the concept of relative velocity, relative acceleration or relative speed
The relative velocity of an object A with respect to object B is the rate of position of the object A with respect of object B.
If VA and VB be the velocities of objects A and B with respect to the ground, then:
(a) The relative velocity of A with respect to B is
VAB = VA – VB
(b) The relative velocity of B with respect to A is
VBA = VA – VB
SI unit: m/s
Dimensional formula: [LT1]
Relative Acceleration: The relative acceleration (also ar) is the acceleration of an object or observer B in the rest frame of another object or observer A.
Acceleration of B relative to A = aB  aA
SI unit: m/s2
Dimensional formula: [LT2]
Time of crossing: Component (vr + vsr cos 𝜽) will enable the person to drift along the length of river. Hence drift Δx will be
$\Delta t=\dfrac{d}{v_{sr}sin\theta }$ ….(1)
$\Delta x=(v_{r}+v_{sr}cos\theta )\Delta t$ ....(2)
Minimum Time of Crossing
$\therefore \Delta t_{min}=\dfrac{d}{v_{sr}}$ And hence
Drift $\Delta x=v_{r}\left ( \dfrac{d}{v_{sr}} \right )$
Shortest Path: The person should try to swim such that the resultant velocity becomes perpendicular to the river flow.
$\therefore \Delta t=\dfrac{d}{\sqrt{v_{sr}^{2}v_{r}^{2}}}$
Rainman Umbrella Problems
A person standing/running in a particular direction would be needed to be protected by properly directing the axis of the umbrella.
NCERT Solutions for Chapter 4 (Motion of Plane) [Free] PDF to Download  Other Special Features of the Notes
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Motion in a Plane Class 11 NCERT Solutions PDF is what students require for improving their knowledge on in this chapter. They get all the solutions of this chapter in one PDF.
All the concepts and fundamentals are clearly explained in this Motion in a Plane Class 11 solved problems PDF.
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.
Ans:
Scalar: Mass, volume, density, angular frequency, number of moles, speed.
Vector: Acceleration, angular velocity, velocity, displacement.
A scalar quantity is specified by its magnitude. Mass, volume, density, angular frequency, number of moles, speed are some of the scalar physical quantities.
A vector quantity is specified by its magnitude and the direction associated with it.
Acceleration, angular velocity, velocity, displacement belong to this category.
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.
Ans:
Work and current are examples of scalar quantities.
Work done is said to be the dot product of force and displacement. As the dot product of two quantities is always a scalar, work is considered as a scalar physical quantity.
Current is described by its magnitude. Its direction is not considered.
Thus, it is a scalar quantity.
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.
Ans: Impulse
It is given by the product of force and time. As force is a vector quantity, its product with time gives a vector quantity.
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 dimension s,
(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.
Ans:
(a) Not Meaningful.
The addition of two scalar quantities will be meaningful only if they both represent the same physical quantity.
(b) Not Meaningful.
The addition of a vector quantity with a scalar quantity is considered not meaningful.
(c) Meaningful.
A scalar can be multiplied with a vector. Force is multiplied with time to give impulse.
(d) Meaningful.
A scalar, respective of the physical quantity, can be multiplied with another scalar having the same or different dimensions.
(e) Not Meaningful.
The addition of two vector quantities is considered meaningful only if they both represent the same physical quantity.
(f) Meaningful
A component of a vector can be added to the same vector as both of them have the same dimensions.
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 ${\text{m}}$ a plane can never add up to give a null vector.
Ans:
(a) True.
The magnitude of a vector is a number. So, it is a scalar.
(b) False.
Each component of a vector is a vector.
(c) False.
The total path length is scalar, whereas displacement is a vector quantity.
So, the total path length is greater than the magnitude of displacement. It is equal to the magnitude of displacement only when a particle is moving in a straight line.
(d) True.
It is because the total path length is always greater than or equal to the magnitude of displacement of a particle.
(e) True.
Three vectors, which do not lie in a plane, can’t be represented by the sides of a triangle taken in the same order.
6. Establish the following vector inequalities geometrically or otherwise:
(a) ${\text{a}} + {\text{b}} \leqslant {\text{a}} + {\text{b}}$
(b) ${\text{a}} + {\text{b}} \geqslant {\text{a}}  {\text{b}}$
(c) ${\text{a}}  {\text{b}} \leqslant {\text{a}}  {\text{b}}$
(d) ${\text{a}}  {\text{b}} \geqslant {\text{a}}  {\text{b}}$
When does the equality sign above apply?
Ans:
a) Let $\overrightarrow {\text{a}} $ and $\overrightarrow {\text{b}} $ be represented by the adjacent sides of a parallelogram $OMNP$ as shown below:
(Image will be uploaded soon)
$\overrightarrow {{\text{OM}}}  = \overrightarrow a {\text{ }}...\left( {\text{i}} \right)$
$\overrightarrow {{\text{MN}}}  = \overrightarrow {{\text{OP}}}  = \overrightarrow {\text{b}} {\text{ }}...\left( {{\text{ii}}} \right)$
$\overrightarrow {{\text{ON}}}  = \overrightarrow a + \overrightarrow b {\text{ }}...\left( {{\text{iii}}} \right)$
As each side is smaller than the sum of the other two sides in a triangle,
In $\Delta OMN$,
${\text{ON}} < ({\text{OM}} + {\text{MN}})$
\[\overrightarrow {\text{a}} + \overrightarrow {\text{b}}  < \overrightarrow {\text{a}}  + \overrightarrow {\text{b}} {\text{ }}...{\text{(iv)}}\]
If $\vec a$ and $\vec b$ act along a straight line in the same direction, then:
$\overrightarrow {\text{a}} + \overrightarrow {\text{b}}  = \overrightarrow {\text{a}}  + \overrightarrow {\text{b}} {\text{ }}...{\text{(v)}}$
Combine equations ${\text{(iv)}}$ and ${\text{(v)}}$
$\overrightarrow {\text{a}} + \overrightarrow {\text{b}}  \leqslant \overrightarrow {\text{a}}  + \overrightarrow {\text{b}} $
b) Let $\vec a$ and $\vec b$ be represented by the adjacent sides of a parallelogram$OMNP$, as shown below:
(Image will be uploaded soon)
$\overrightarrow {{\text{OM}}}  = \overrightarrow a {\text{ }}...\left( {\text{i}} \right)$
$\overrightarrow {{\text{MN}}}  = \overrightarrow {{\text{OP}}}  = \overrightarrow {\text{b}} {\text{ }}...\left( {{\text{ii}}} \right)$
$\overrightarrow {{\text{ON}}}  = \overrightarrow a + \overrightarrow b {\text{ }}...\left( {{\text{iii}}} \right)$
As each side is smaller than the sum of the other two sides in a triangle,
In $\Delta OMN$,
${\text{ON}} + {\text{MN}} > {\text{OM}}$
${\text{ON}} + {\text{OM}} > {\text{MN}}$
${\text{ON}} > \overrightarrow {{\text{OM}}}  \overrightarrow {{\text{OM}}} \quad (\because {\text{OP}} = {\text{MN}})$
If $\vec a$ and $\vec b$ act along a straight line in the same direction, then:
\[\overrightarrow {\text{a}} + \overrightarrow {\text{b}}  = \overrightarrow {\text{a}}   \overrightarrow {\text{b}} {\text{ }}...{\text{(v)}}\]
Combine equations ${\text{(iv)}}$ and ${\text{(v)}}$
$\overrightarrow {\text{a}} + \overrightarrow {\text{b}}  \geqslant \overrightarrow {\text{a}}   \overrightarrow {\text{b}} $
c) Let $\overrightarrow {\text{a}} $ and $\overrightarrow {\text{b}} $ be represented by the adjacent sides of a parallelogram $PQRS$:
(Image will be uploaded soon)
$\overrightarrow {{\text{OR}}}  = \overrightarrow {{\text{PS}}}  = \overrightarrow {\text{b}} {\text{ }}...{\text{(i)}}$
$\overline {{\text{OP}}}  = \overrightarrow {\text{a}} {\text{ }}...{\text{(ii)}}$
As each side is smaller than the sum of the other two sides in a triangle,
In $\Delta OPS$,
${\text{OS}} < {\text{OP}} + {\text{PS}}$
$\overrightarrow {\text{a}}  \overrightarrow {\text{b}}  < \overrightarrow {\mathbf{a}}  +   \overrightarrow {\text{b}} $
$\overrightarrow {\text{a}}  \overrightarrow {\text{b}}  < \overrightarrow {\mathbf{a}}  + \overrightarrow {\text{b}} {\text{ }}...{\text{(iii)}}$
If the two vectors act in a straight line but in opposite directions, then:
$\overrightarrow {\text{a}}  \overrightarrow {\mathbf{b}}  = \overrightarrow {\text{a}}  + \overrightarrow {\text{b}} $
Combine equations ${\text{(iii)}}$ and ${\text{(iv)}}$
$\overrightarrow {\text{a}}  \overrightarrow {\mathbf{b}}  \leqslant \overrightarrow {\mathbf{a}}  + \overrightarrow {\mathbf{b}} $
d) Let $\vec a$ and $\vec b$ be represented by the adjacent sides of a parallelogram $PQRS$:
(Image will be uploaded soon)
${\text{OS}} + {\text{PS}} > {\text{OP }}...{\text{(i)}}$
${\text{OS}} > {\text{OP  PS }}...{\text{(ii)}}$
$\overrightarrow {\text{a}}  \overrightarrow {\text{b}}  > \overrightarrow {\text{a}}   \overrightarrow {\text{b}} {\text{ }}...{\text{(iii)}}$
The L.H.S is always positive and R.H.S can be positive or negative.
To make both quantities positive, take modulus on both sides.
$\left {\left {\overrightarrow {\text{a}}  \overrightarrow {\text{b}} } \right} \right < \left {\left {\overrightarrow {\text{a}} } \right  \left {\overrightarrow {\text{b}} } \right} \right$
\[\left {\overrightarrow {\text{a}}  \overrightarrow {\text{b}} } \right > \left {\left {\overrightarrow {\text{a}} } \right  \left {\overrightarrow {\text{b}} } \right} \right ...{\text{(iv)}}\]
If the two vectors act in a straight line but in the same direction:
\[\left {\overrightarrow {\text{a}}  \overrightarrow {\text{b}} } \right = \left {\left {\overrightarrow {\text{a}} } \right  \left {\overrightarrow {\text{b}} } \right} \right{\text{ }}...{\text{(v)}}\]
Combine equtaion \[{\text{(iv)}}\] and equation \[{\text{(v)}}\]:
\[\left {\overrightarrow {\text{a}}  \overrightarrow {\text{b}} } \right \geqslant \left {\left {\overrightarrow {\text{a}} } \right  \left {\overrightarrow {\text{b}} } \right} \right\]
7. Given ${\text{a}} + {\text{b}} + {\text{c}} + {\text{d}} = 0$, which of the following statements are correct:
(a) ${\text{a}},{\text{b}},{\text{c}}$, and ${\text{d}}$ must each be a null vector,
(b) The magnitude of $(a + c)$ equals the magnitude of $(b + d)$
(c) The magnitude of a can never be greater than the sum of the magnitudes of ${\text{b}}$, ${\text{c}}$, and ${\text{d}}$,
(d) ${\text{b}} + {\text{c}}$ must lie in the plane of ${\text{a}}$ and ${\text{d}}$ if ${\text{a}}$ and ${\text{d}}$ are not collinear, and in the line of ${\text{a}}$ and ${\text{d}}$.
if they are collinear?
Ans:
(a) Incorrect
To make $\overrightarrow {\text{a}} + \overrightarrow {\text{b}} + \overrightarrow {\text{c}} + \overrightarrow {\text{d}} = 0$, it is not necessary to have all four vectors as null vectors. There are many other combinations which will give the sum zero.
(b) Correct
$\overrightarrow {\text{a}} + \overrightarrow {\text{b}} + \overrightarrow {\text{c}} + \overrightarrow {\text{d}} = 0$
$\overrightarrow {\text{a}} + \overrightarrow {\text{c}} =  (\overrightarrow {\text{b}} + \overrightarrow {\text{d}} )$
Take modulus on both sides:
$\overrightarrow {\text{a}} + \overrightarrow {\text{c}}  =   (\overrightarrow {\text{b}} + \overrightarrow {\text{d}} ) = (\overrightarrow {\text{b}} + \overrightarrow {\text{d}} )$
So, the magnitude of $(\vec a + \vec c)$ is the same as the magnitude of $(\vec b + \vec d)$
(c) Correct
$\overrightarrow {\text{a}} + \overrightarrow {\text{b}} + \overrightarrow {\text{c}} + \overrightarrow {\text{d}} = 0$
$\overrightarrow {\text{a}} =  (\overrightarrow {\text{b}} + \overrightarrow {\text{c}} + \overrightarrow {\text{d}} )$
Take modulus on both sides:
$\vec a = (\vec b + \vec c + \vec d)$
$\vec a \leqslant \vec b + \vec c + \vec d{\text{ }}...{\text{(i)}}$
$(\vec b + \vec c + \vec d)$ is the sum of vectors $\vec b$, $\vec c$ and $\vec d$. The magnitude of $(\vec b + \vec c + \vec d)$ is less than, or equal to the sum of the magnitudes of $\vec b$, $\vec c$ and $\vec d.$ So, the magnitude of $\vec a$ cannot be greater than the sum of the magnitudes of $\vec b$, $\vec c$ and $\overrightarrow {\text{d}} .$ Equation ${\text{(i)}}$ shows that the magnitude of $\overrightarrow {\text{a}} $ is equal to or less than the sum of the magnitudes of $\vec b$, $\vec c$ and $\vec d$
(d) Correct
For, $\overrightarrow {\text{a}} + \overrightarrow {\text{b}} + \overrightarrow {\text{c}} + \overrightarrow {\text{d}} = 0$
$\overrightarrow {\text{a}} + (\overrightarrow {\text{b}} + \overrightarrow {\text{c}} ) + \overrightarrow {\text{d}} = 0$
The resultant sum of the vectors $\vec a,{\text{ }}(\vec b + \vec c)$ and $\vec d$ is zero only if $(\vec b + \vec c)$ lie in the same plane as $\vec a$ and $\vec d$
If $\vec a$ and $\overrightarrow {\text{d}} $ are collinear, then $(\overrightarrow {\text{b}} + \overrightarrow {\text{c}} )$ is in the line of $\overrightarrow {\text{a}} $ and $\overrightarrow {\text{d}} $. This is true in this case and the vector sum of all the vectors will be zero.
8. Three girls skating on a circular ice ground of radius $200\;{\text{m}}$ start from a point ${\text{P}}$ on the edge of the ground and reach a point ${\text{Q}}$ diametrically opposite to ${\text{P}}$ following different paths as shown in Fig. 4.20 . What is the magnitude of the displacement vector for each? For which girl is this equal to the actual length of the path skated?
(Image will be uploaded soon)
Ans:
The magnitudes of displacements are equal to the diameter of the ground.
Radius of the ground $ = 200\;{\text{m}}$
Diameter of the ground $ = 2 \times 200 = 400\;{\text{m}}$
So, the magnitude of the displacement for each girl is $400\;{\text{m}}$which is equal to the actual length of the path skated by girl ${\text{B}}$.
9. A cyclist starts from the centre ${\text{O}}$ of a circular park of radius $1\;{\text{ km}}$, reaches the edge ${\text{P}}$ of the park, then cycles along the circumference, and returns to the centre along ${\text{QO}}$ as shown in the following figure. If the round trip takes $1\;0$ min, what is the
(a) net displacement,
(b) average velocity, and
(c) average speed of the cyclist?
Ans 9:
(a) The cyclist comes to the starting point after cycling for $1\;0$ minutes. So, his net displacement is zero.
(b) Average velocity $ = \dfrac{{{\text{ Net displacement Total time }}}}{{{\text{ Total time }}}}$
As the net displacement of the cyclist is zero, his average velocity is also zero.
(c) ${\text{Average speed }} = \dfrac{{{\text{ Total path length }}}}{{{\text{ Total time }}}}$
${\text{Total path length }} = {\text{OP}} + {\text{PQ}} + {\text{QO}}$
$ = 1 + \dfrac{1}{4}(2\pi \times 1) + 1$
$ = 2 + \dfrac{1}{2}\pi 3.570\;{\text{km}}$
Time taken $ = 10\;{\text{min}}$
$ = \dfrac{{10}}{{60}}$
$ = \dfrac{1}{6}\;{\text{h}}$
$\therefore $ Average speed $ = \dfrac{{3.570}}{1} = 21.42\;{\text{km}}/{\text{h}}$
10. On an open ground, a motorist follows a track that turns to his left by an angle of ${{60}^{0}}$ after every $500\;{\text{m}}$. Starting from a given turn, specify the displacement of the total 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.
Ans:
The path is a regular hexagon with side $500\;{\text{m}}$.
Let the motorist start from ${\text{P}}$.
The motorist takes the third turn at $S$.
$\therefore $ Magnitude of displacement $ = {\text{PS}}$
$ = {\text{PV}} + {\text{VS}}$
$ = 500 + 500$
$ = 1000\;{\text{m}}$
Total path length $ = {\text{PQ}} + {\text{QR}} + {\text{RS}}$
$ = 500 + 500 + 500$
$ = 1500\;{\text{m}}$
The motorist takes the sixth turn at $P$, which is the starting point.
$\therefore $ Magnitude of displacement $ = 0$
Total path length $ = {\text{PQ}} + {\text{QR}} + {\text{RS}} + {\text{ST}} + {\text{TU}} + {\text{UP}}$
$ = 500 + 500 + 500 + 500 + 500 + 500$
$ = 3000\;{\text{m}}$
The motorist takes the eight turn at $R$.
$\therefore {\text{ Magnitude of displacement }} = {\text{PR}}$
$ = \sqrt{{PQ^{2}}+{QR^{2}}+(PQ)(QR)cos 60^{0}}$
$ = \sqrt{{500^{2}}+{500^{2}}+(500)(500)cos 60^{0}} $
$ = \sqrt {250000 + 250000 + \left( {500000 \times \dfrac{1}{2}} \right)} $
$ = 866.03\;{\text{m}}$
$\beta = tan^{1}\left ( \frac{500\times sin60^{0}}{500+500\times cos 60^{0}} \right )$
$\beta = {{30}^{0}}$
Thus, the magnitude of displacement is $866.03\;{\text{m}}$ at an angle of ${{30}^{0}}$ with ${\text{PR}}$.
Total path length $ = $ Circumference of the hexagon $ + {\text{PQ}} + $${\text{QR}}$
$ = 6 \times 500 + 500 + 500$
$ = 4000\;{\text{m}}$
Turn  Magnitude of Displacement  Total Path Length 
Third  $1000$  $1500$ 
Sixth  $0$  $3000$ 
Eighth  $866.03;{\text{ }}{{30}^{0} }\;$  $4000$ 
11. A passenger arriving in a new town wishes to go from the station to a hotel located $10\;{\text{km}}$ away on a straight road from the station. A dishonest cabman takes him along a circuitous path $23\;{\text{km}}$ long and reaches the hotel in $28\;{\text{min}}$. What is
(a) the average speed of the taxi,
(b) the magnitude of average velocity? Are the two equal?
Ans:
(a) Total distance travelled $ = 23\;{\text{km}}$
Total time taken $ = 28\;{\text{min}} = \dfrac{{28}}{{60}}\;{\text{h}}$
Average speed of the taxi $ = \dfrac{{{\text{ Total distance travelled }}}}{{{\text{ Total time taken }}}}$
(b) Distance between the hotel and the station $ = 10\;{\text{km}} = $ Displacement of the car
$\therefore {\text{ Average velocity }} = \dfrac{{10}}{{\dfrac{{28}}{{60}}}}$
$ = 21.43\;{\text{km}}/{\text{h}}$
12. Rain is falling vertically with a speed of $30\;{\text{m}}{{\text{s}}^{  1}}$. A woman rides a bicycle with a speed of $10\;{\text{m}}{{\text{s}}^{  1}}$ in the north to south direction. What is the direction in which she should hold her umbrella?
Ans:
The described situation is:
(Image will be uploaded soon)
${v_{\text{c}}} = $ Velocity of the cyclist
${{\text{v}}_{\text{r}}} = $ Velocity of falling rain
To protect herself from the rain, the woman should hold the umbrella in the direction of the relative velocity of the rain with respect to the woman.
${\text{v}} = {{\text{v}}_{\text{c}}}{\text{ + }}{{\text{v}}_{\text{r}}}$
$ = 30 + (  10)$
$ = 20\;{\text{m}}{{\text{s}}^{  1}}$
$\tan \theta = \dfrac{{{v_c}}}{{{v_r}}}$
$\tan \theta = \dfrac{{10}}{{30}}$
$\theta = {\tan ^{  1}}\left( {\dfrac{1}{3}} \right)$
$\theta = {\tan ^{  1}}(0.333)$
$\theta \approx {{18}^{0} }$
So, the woman should hold the umbrella toward the south, at an angle of ${{18}^{0}}$ with the vertical.
13. A man can swim with a speed of $4.0\;{\text{km}}/{\text{h}}$ in still water. How long does he take to cross a river $1.0\;{\text{km}}$ wide if the river flows steadily at $3.0\;{\text{km}}/{\text{h}}$ and he makes his strokes normal to the river current? How far down the river does he go when he reaches the other bank?
Ans:
Speed of the man ${{\text{v}}_{\text{m}}} = 4\;{\text{km}}{{\text{h}}^{  1}}$
Width of the river $ = 1\;{\text{km}}$
Time taken to cross the river $ = \dfrac{{{\text{ Width of the river }}}}{{{\text{ Speed of the rivera }}}}$
$ = \dfrac{1}{4}h$
$ = \dfrac{1}{4} \times 60$
$ = 15\;{\text{min}}$
Speed of the river, ${{\text{v}}_{\text{r}}} = 3\;{\text{km}}{{\text{h}}^{  1}}$
Distance covered with flow of the river $ = {v_r} \times t$
$ = 3 \times \dfrac{1}{4}$
$ = \dfrac{3}{4}\;{\text{km}}$
$ = \dfrac{3}{4} \times 1000$
$ = 750\;{\text{m}}$
14. In a harbour, wind is blowing at the speed of $72\;{\text{km}}/{\text{h}}$ and the flag on the mast of a boat anchored in the harbour flutters along the NE direction. If the boat starts moving at a speed of $51\;{\text{km}}/{\text{h}}$ to the north, what is the direction of the flag on the mast of the boat?
Ans:
Velocity of the boat, ${{\text{v}}_{\text{b}}} = 51\;{\text{km}}{{\text{h}}^{  1}}$
Velocity of the wind, ${{\text{v}}_{\text{w}}} = 72\;{\text{km}}{{\text{h}}^{  1}}$
As the flag is fluttering in the northeast direction, the wind is blowing towards the northeast direction. When the ship begins sailing toward the north, the flag will start moving along the direction of the relative velocity of the wind with respect to the boat.
The angle between ${{\text{v}}_{\text{w}}}$ and $\left( {  {{\text{v}}_{\text{b}}}} \right) = {{90}^{0}} + {{45}^{0}}$
$\tan \beta = \dfrac{{51\sin (90 + 45)}}{{72 + 51\cos (90 + 45)}}$
$ = \dfrac{{51\sin (90 + 45)}}{{72 + 51(  \cos 45)}}$
$ = \dfrac{{51 \times \dfrac{1}{{\sqrt 2 }}}}{{72  51 \times \dfrac{1}{{\sqrt 2 }}}}$
$ = \dfrac{{51}}{{72\sqrt 2  51}}$
$ = \dfrac{{51}}{{72 \times 1.414  51}}$
$ = \dfrac{{51}}{{50.800}}$
$\therefore \beta = {\tan ^{  1}}(1.0038)$
$\beta = {{4511}^{0}}$
Angle with respect to the east direction is ${{45.11}^{0}}  {{45}^{0}} = $ ${{0.11}^{0}}$
So, the flag will flutter almost due east.
15. The ceiling of a long hall is $25\;{\text{m}}$ high. What is the maximum horizontal distance that a ball thrown with a speed of $40\;{\text{m}}{{\text{s}}^{  1}}$ can go without hitting the ceiling of the hall?
Ans:
Speed of the ball, $40\;{\text{m}}{{\text{s}}^{  1}}$
Maximum height, $h = 25\;{\text{m}}$
In projectile motion, the maximum height reached, by a body projected at an angle $\theta $ is:
$h=\frac{u^{2} sin^{2}\theta}{2g}$
$25=\frac{{{40}^{2}}{sin^{2}\theta}}{{2}\times {9.8}}$
$sin^{2}\theta =0.30625$
$sin \theta = 0.5534$
$\theta =sin^{1}\left ( 0.5534 \right )$
$\theta = {33.60^ \circ }$
The horizontal range is
$R=\frac{{u^{2}} sin2\theta }{g}$
${\text{R = }}\dfrac{{{{\left( {40} \right)}^{\text{2}}} \times {\text{sin2}} \times {\text{33}}{\text{.60}}}}{{9.8}}$
${\text{R = }}\dfrac{{1600 \times {\text{sin67}}{\text{.2}}}}{{9.8}}$
${\text{R = }}\dfrac{{1600 \times 0.922}}{{9.8}}$
${\text{R = }}150.53{\text{m}}$
16. A cricketer can throw a ball to a maximum horizontal distance of $100{\text{m}}$.How much high above the ground can the cricketer throw the same ball?
Ans:
Maximum horizontal distance, ${\text{R = }}100{\text{m}}$
The cricketer will throw the ball to the maximum horizontal distance when the angle of projection is ${{45}^{0}}$, i.e., $\theta = {{45}^{0}}$
The horizontal range for a projection velocity ${\text{v}}$, is:
${\text{R}} = \dfrac{{{{\text{u}}^2}\sin 2\theta }}{{\text{g}}}$
${\text{100}} = \dfrac{{{{\text{u}}^2}\sin {{90}^{0}}}}{{\text{g}}}$
$\dfrac{{{{\text{u}}^2}}}{{\text{g}}} = 100{\text{ }}...{\text{(i)}}$
The ball will reach the maximum height when it is thrown vertically upward. For this type of motion, the final velocity is zero at the maximum height ${\text{H}}$.
Acceleration, ${\text{a}} =  {\text{g}}$
Use the third equation of motion:
${{\text{v}}^2}  {{\text{u}}^2} =  2\;{\text{gH}}$
${\text{H}} = \dfrac{1}{2} \times \dfrac{{{{\text{u}}^2}}}{{\;{\text{g}}}}$
${\text{H}} = \dfrac{1}{2} \times 100$
${\text{H}} = 50\;{\text{m}}$
17. A stone tied to the end of a string $80\;{\text{cm}}$ long is whirled in a horizontal circle with a constant speed. If the stone makes $14$ revolutions in $25\;{\text{s}}$, what is the magnitude and direction of acceleration of the stone?
Ans:
Length of the string, $ = 80\;{\text{cm}} = 0.8\;{\text{m}}$
Number of revolutions $ = 14$
Time taken $ = 25\;{\text{s}}$
Frequency, ${\text{v}} = \dfrac{{{\text{ Number of revolutions }}}}{{{\text{ Time taken }}}} = \dfrac{{14}}{{25}}\;{\text{Hz}}$
Angular frequency, $\omega = 2\pi {\text{v}}$
$ = 2 \times \dfrac{{22}}{7} \times \dfrac{{14}}{{25}}$
$ = \dfrac{{88}}{{25}}{\text{rad}}{{\text{s}}^{  1}}$
Centripetal acceleration, ${{\text{a}}_{\text{e}}} = {\omega ^2}{\text{r}}$
$ = {\left( {\dfrac{{88}}{{25}}} \right)^2} \times 0.8$
$ = 9.91\;{\text{m}}{{\text{s}}^{  2}}$
The direction of centripetal acceleration is always along the string, towards the center, at all points.
18. An aircraft executes a horizontal loop of radius $1.00\;{\text{km}}$ with a steady speed of $900\;{\text{km}}/{\text{h}}.$ Compare its centripetal acceleration with the acceleration due to gravity.
Ans:
Radius of the loop, ${\text{r}} = 1\;{\text{km}} = 1000\;{\text{m}}$
Speed of the aircraft, ${\text{v}} = 900\;{\text{km}}{{\text{h}}^{  1}}$
$ = 900 \times \dfrac{5}{{18}}$
$ = 250\;{\text{m}}{{\text{s}}^{  1}}$
Centripetal acceleration, ${{\text{a}}_{\text{e}}} = \dfrac{{{{\text{v}}^2}}}{{\text{r}}}$
$ = \dfrac{{{{(250)}^2}}}{{1000}}$
$ = 62.5\;{\text{m}}{{\text{s}}^{  2}}$
Acceleration due to gravity, ${\text{g}} = 9.8\;{\text{m}}{{\text{s}}^{  2}}$
$\dfrac{{{a_c}}}{g} = \dfrac{{6.25}}{{1000}}$
${a_c} = 6.38g$
The Centripetal acceleration is $6.38$ times the acceleration due to gravity.
19. Read each statement below carefully and state, with reasons, if it is true or false:
(a) The net acceleration of a particle m 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
Ans:
(a) False
In circular motion, the net acceleration of a particle is not always directed along the radius of the circle toward the centre. It happens only in the case of uniform circular motion.
(b) True
At a point on a circular path, a particle appears to move tangentially to the circular path.
Thus, the velocity vector of the particle is always along the tangent at a point.
(c) True
In uniform circular motion, the acceleration vector points towards the centre of the circle. The average of these vectors over one cycle is a null vector.
20. The position of a particle is given by ${\text{r}} = 3.0{\text{t}}\widehat {\text{i}}  2.0{{\text{t}}^2}\widehat {\text{j}} + 4.0\;{\text{km}}$
Where $t$ is in seconds and the coefficients have the proper units for $r$ to be in metres.
(a) Find the $v$ and $a$ of the particle?
(b) What is the magnitude and direction of velocity of the particle at ${\text{t}} = 2.0\;{\text{s}}?$
Ans:
$\overrightarrow {\text{v}} ({\text{t}}) = (3.0\widehat {\text{i}}  4.0\widehat {\text{j}}):\overrightarrow {\text{a}} =  4.0\widehat {\text{j}}$
The position of the particle is:
$\overrightarrow {\text{r}} = 3.0\widehat {\text{i}}  2.0{{\text{t}}^2}\widehat {\text{j}} + 4.0\widehat {\text{k}}$
Velocity, $\vec v$ of the particle is:
$\overrightarrow {\text{v}} = \dfrac{{{\text{d}}\overrightarrow {\text{r}} }}{{{\text{dt}}}}$
$\overrightarrow {\text{v}} = \dfrac{{\text{d}}}{{{\text{dt}}}}\left( {3.0{\text{t}}\widehat {\text{i}}  2 \cdot {\text{O}}{{\text{t}}^2}\widehat {\text{j}} + 4.0{\text{k}}} \right)$
$\therefore \overrightarrow {\text{v}} = 3.0\widehat {\text{i}}  4.0{\text{t}}$
Acceleration of the particle is:
$\overrightarrow {\text{a}} = \dfrac{{{\text{d}}\overrightarrow {\text{v}} }}{{{\text{dt}}}}$
$\overrightarrow {\text{a}} = \dfrac{{\text{d}}}{{{\text{dt}}}}(3.0\widehat {\text{i}}  4.{\text{O}}\widehat {\text{j}})$
$\therefore \overrightarrow {\text{a}} =  4.{\text{O}}\widehat {\text{j}}$
The velocity vector, $\vec v = 3.0\hat i  4.0t\hat j$
At ${\text{t}} = 2.0\;{\text{s}}$ :
$\overrightarrow {\text{v}} = 3.0\widehat {\text{i}}  8.0\widehat {\text{j}}$
The magnitude of velocity is:
$\overrightarrow {\text{v}}  = \sqrt {{3^2} + {{(  8)}^2}} $
$\overrightarrow {\text{v}}  = \sqrt {73} $
$\overrightarrow {\text{v}}  = 8.54\;{\text{m}}/{\text{s}}$
And $\theta = {\tan ^{  1}}\left( {\dfrac{{{{\text{v}}_{\text{y}}}}}{{{{\text{v}}_{\text{x}}}}}} \right)$
$ = {\tan ^{  1}}\left( {\dfrac{{  8}}{3}} \right)$
$ =  {\tan ^{  1}}(2.667)$
$ =  {{69.45}^{0}}$
The negative sign indicates that the direction of velocity is $8.54\;{\text{m}}{{\text{s}}^{  1}}$, ${{69.45}^{0}}$ below the $x  $axis.
21. A particle starts from the origin at ${\text{t}} = 0\;{\text{s}}$ with a velocity of $10.07$ and moves in the $x  y$ plane with a constant acceleration of $(8.0\widehat {\text{i}} + 2.0\widehat {\text{j}}){\text{m}}{{\text{s}}^{{\text{  2}}}}$
(a) At what time is the $x  $ coordinate of the particle $16{\text{ m}}$? What is the $y  $coordinate of the particle at that time?
(b) What is the speed of the particle at the time?
Ans:
Velocity of the particle, $\overrightarrow {\text{v}} = 10.0\widehat {\text{j}}{\text{ m}}{{\text{s}}^{  1}}$
Acceleration of the particle $\overrightarrow {\text{a}} = (8.0\widehat {\text{i}} + 2.0\widehat {\text{j}})$
But, $\overrightarrow {\text{a}} = \dfrac{{{\text{d}}\overrightarrow {\text{v}} }}{{{\text{dt}}}} = 8.0\widehat {\text{i}} + 2.0\widehat {\text{j}}$
${\text{d}}\overrightarrow {\text{v}} = (8.0\widehat {\text{i}} + 2.0\widehat {\text{j}}){\text{dt}}$
Integrate both sides:
$\overrightarrow {\text{v}} ({\text{t}}) = 8.0\widehat {\text{i}} + 2.0\widehat {\text{j}} + \overrightarrow {\text{u}} $
Where,
$\overrightarrow {\text{u}} = $ Velocity vector of the particle at ${\text{t}} = 0$
$\overrightarrow {\text{v}} = $ Velocity vector of the particle at time $\phi $
But $\overrightarrow {\text{v}} = \dfrac{{{\text{dr}}}}{{{\text{dt}}}}$
${\text{d}}\overrightarrow {\text{r}} = \overrightarrow {\text{v}} {\text{dt}} = (8.0{\text{t}}\widehat {\text{i}} + 2.0{\text{t}}\widehat {\text{j}} + {\text{u}}){\text{dt}}$
Integrate the equations with: at ${\text{t}} = {\text{0}};{\text{ r}} = 0$ and at ${\text{t}} = {\text{t}};{\text{ r}} = {\text{r}}$
$\overrightarrow {\text{r}} = \overrightarrow {\text{u}} {\text{t}} + \dfrac{1}{2}8.0{{\text{t}}^2}\widehat {\text{i}} + \dfrac{1}{2}2.0{{\text{t}}^2}\widehat {\text{j}}$
$ = \overrightarrow {\text{u}} {\text{t + 4 \times 0}}{{\text{t}}^{\text{2}}}\widehat {\text{i}}{\text{ + }}{{\text{t}}^{\text{2}}}\widehat {\text{j}}$
$ = (10.0{\text{j}}){\text{t}} + 4.0{{\text{t}}^2}\widehat {\text{i}} + {{\text{i}}^2}\widehat {\text{j}}$
${\text{x}}\widehat {\text{i}} + \widehat {\text{y}} = 4.0{{\text{t}}^2}\widehat {\text{i}} + \left( {10{\text{t}} + {{\text{t}}^2}} \right)\widehat {\text{j}}$
Equate the coefficients of $\widehat {\text{i}}$ and $\widehat {\text{j}}$:
$x = 4{t^2}$
$t = {\left( {\dfrac{x}{4}} \right)^{\dfrac{1}{2}}}$
And $y = 10t + {t^2}$
When ${\text{x}} = 16\;{\text{m}}$ :
$t = {\left( {\dfrac{{16}}{4}} \right)^{\dfrac{1}{2}}}$
$t = 2s$
$\therefore y = 10 \times 2 + {(2)^2} = 24m$
Velocity of the particle is:
$\overrightarrow {\text{v}} ({\text{t}}) = 8.0{\text{t}}\widehat {\text{i}} + 2.0\widehat {\text{t}} + \overrightarrow {\text{u}} $
At ${\text{t}} = 2\;{\text{s}}$
$\overrightarrow {\text{v}} ({\text{t}}) = 8.0 \times 2\widehat {\text{i}} + 2.0 \times 2\widehat {\text{j}} + 10\widehat {\text{j}}$
$ = 16\widehat {\text{i}} + 14\widehat {\text{j}}$
$\therefore $ Speed of the particle:
$\vec v = \sqrt {{{(16)}^2} + {{(14)}^2}} $
$ = \sqrt {256 + 196} $
$ = \sqrt {452} $
$ = 21.26\;{\text{m}}{{\text{s}}^{  1}}$
22. $\widehat {\text{i}}$ and $\widehat {\text{j}}$ are unit vectors along ${\text{x}}  $ and ${\text{y}}$axis respectively. What is the magnitude and direction of the vectors $\widehat {\text{i}} + \widehat {\text{j}}$ and $\widehat {\text{i}}.\widehat {\text{j}}$? What are the components of a vector ${\text{a}} = 2\widehat {\text{i}} + 3\widehat {\text{j}}$ along the directions of $\widehat {\text{i}} + \widehat {\text{j}}$ and $\widehat {\text{i}}  \widehat {\text{j}}$?
Ans:
Consider a vector $\overrightarrow {\text{P}} $
\[\overrightarrow {\text{P}} = \widehat {\text{i}} + \widehat {\text{j}}\]
\[{{\text{P}}_{\text{x}}}\widehat {\text{i}} + {{\text{P}}_{\text{y}}}\widehat {\text{j}} = \widehat {\text{i}} + \widehat {\text{j}}\]
Compare the components on both sides:
\[{{\text{P}}_{\text{x}}}{\text{ = }}{{\text{P}}_{\text{y}}}{\text{ = 1}}\]
\[\left {\overrightarrow {\text{P}} } \right = \sqrt {{{\text{P}}_{\text{x}}}^2 + {{\text{P}}_{\text{y}}}^2} \]
\[\left {\overrightarrow {\text{P}} } \right = \sqrt {{1^2} + {{\text{1}}^2}} \]
\[\left {\overrightarrow {\text{P}} } \right = \sqrt 2 {\text{ }}...{\text{(i)}}\]
So, the magnitude of the vector $\widehat {\text{i}} + \widehat {\text{j}}$ is $\sqrt 2 $
Let $\theta $ be the angle made by $\overrightarrow {\text{P}} $, with the ${\text{x  }}$axis.
So, \[{\text{tan}}\theta {\text{ = }}\left( {\dfrac{{{{\text{P}}_{\text{x}}}}}{{{{\text{P}}_{\text{y}}}}}} \right)\]
\[\theta {\text{ = ta}}{{\text{n}}^{  1}}\left( {\dfrac{{\text{1}}}{{\text{1}}}} \right)\]
\[\theta {\text{ = 4}}{{\text{5}}^ \circ }{\text{ }}...{\text{(ii)}}\]
So, the vector $\widehat {\text{i}} + \widehat {\text{j}}$ makes an angle of ${{45}^{0}}$ with the ${\text{x  }}$axis
Let $\theta $ be the angle made by $\overrightarrow {\text{Q}} $, with the ${\text{x  }}$axis.
$\overrightarrow {\text{Q}} = \widehat {\text{i}}  \widehat {\text{j}}$
${{\text{Q}}_{\text{x}}}\widehat {\text{i}}  {{\text{Q}}_{\text{y}}}\widehat {\text{j}} = \widehat {\text{i}}  \widehat {\text{j}}$
${{\text{Q}}_{\text{x}}} + {{\text{Q}}_{\text{y}}} = 1$
$\overrightarrow {\text{Q}}  = \sqrt {{\text{Q}}_{\text{x}}^2 + {\text{Q}}_{\text{y}}^2} $
$\overrightarrow {\text{Q}}  = \sqrt 2 $
So, the magnitude of the vector $\widehat {\text{i}}  \widehat {\text{j}}$ is $\sqrt 2 $.
Let $\theta $ be the angle made by the vector $\overrightarrow {\text{Q}} $, with the ${\text{x  }}$axis.
$\therefore \tan \theta = \left( {\dfrac{{{{\text{Q}}_y}}}{{{{\text{Q}}_{\text{x}}}}}} \right)$
$\theta =  {\tan ^{  1}}\left( {  \dfrac{1}{1}} \right)$
$\theta =  {{45}^{0}}$
So, the vector $\widehat {\text{i}}  \widehat {\text{j}}$ makes and angle of $  {{45}^{0}}$ with the axis.
Compare the coefficients of $\widehat {\text{i}}$ and $\widehat {\text{j}}$
$\overrightarrow {\text{A}} = 2\widehat {\text{i}} + 3\widehat {\text{j}}$
${{\text{A}}_{\text{x}}}\widehat {\text{i}}{\text{ + }}{{\text{A}}_{\text{y}}}\widehat {\text{j}}{\text{ = 2}}\widehat {\text{i}}{\text{ + 3}}\widehat {\text{j}}$
Let $\overline {\text{A}} $ make an angle $\theta $ with the ${\text{x  }}$axis
$\therefore \tan \theta = \left( {\dfrac{{{A_x}}}{{{A_y}}}} \right)$
$\theta = {\tan ^{  1}}\left( {\dfrac{3}{2}} \right)$
$ = {\tan ^{  1}}(1.5)$
$ = {{56.31}^{0}}$
Angle between $(2\hat i + 3\hat j)$ and $(\hat i + \hat j)$
$\theta = 56.31  45 = {{11.31}^{0}}$
$\vec A$, along the direction of $\vec P$ making an angle ${\theta ^\prime }$
$\therefore \tan \theta = \left( {\dfrac{{{{\text{A}}_{\text{x}}}}}{{{{\text{A}}_{\text{y}}}}}} \right)$
$\tan \theta = ({\text{A}}\cos \theta ){\text{P}}$
$\tan \theta = ({\text{A}}\cos 11.31)\dfrac{{(\widehat {\text{i}} + \widehat {\text{j}})}}{{\sqrt 2 }}$
$\tan \theta = \sqrt {13} \times \dfrac{{0.9806}}{{\sqrt 2 }}(\widehat {\text{i}} + \widehat {\text{j}})$
$\tan \theta = 2.5(\widehat {\text{i}} + \widehat {\text{j}})$
$\tan \theta = \dfrac{{25}}{{10}} \times \sqrt 2 $
$\tan \theta = \dfrac{5}{{\sqrt 2 }}\quad \ldots ({\text{v}})$
Let $\theta $ be the angle between $(2\hat i + 3\hat j)$ and $(\hat i + \hat j)$
${\theta ^{\prime \prime }} = 45 + 56.31 = {{101.31}^{0}}$
Component of vector ${\text{A}}$, along the direction of ${\text{Q}}$, making an angle $\theta $
$ = (A\cos \theta )\vec Q$
$ = (A\cos \theta )\dfrac{{(\hat i  \hat j)}}{{\sqrt 2 }}$
$ = \sqrt {13} \cos \left( {{{901.31}^{0}}} \right)\dfrac{{{\text{(}}\widehat {\text{i}}{\text{ + \hat j)}}}}{{\sqrt {\text{2}} }}$
$ =  \sqrt {\dfrac{{13}}{2}} \sin {{11.30}^{0}}(\widehat {\text{i}}  \widehat {\text{j}})$
$ =  2.550 \times 0.1961(\widehat {\text{i}}  \widehat {\text{j}})$
$ =  0.5(\widehat {\text{i}}  {\text{j}})$
$ =  \dfrac{5}{{10}} \times \sqrt 2 $
$ =  \dfrac{1}{{\sqrt 2 }}$
23. For any arbitrary motion in space, which of the following relations are true:
(a) ${{\text{v}}_{{\text{average}}}}{\text{ = }}\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right)\left( {{\text{v}}\left( {{{\text{t}}_{\text{1}}}} \right){\text{ + v}}\left( {{{\text{t}}_{\text{2}}}} \right)} \right)$
(b) ${{\text{v}}_{{\text{average}}}}{\text{ = }}\dfrac{{{\text{[r(}}{{\text{t}}_{\text{2}}}{\text{)  r(}}{{\text{t}}_{\text{1}}}{\text{)]}}}}{{{\text{(}}{{\text{t}}_{\text{2}}}{\text{  }}{{\text{t}}_{\text{1}}}{\text{)}}}}$
(c) ${\text{v(t) = v(0) + at}}$
(d) ${\text{r(t) = r(0) + v(0)t + }}\left( {\dfrac{1}{2}} \right){\text{a}}{{\text{t}}^2}$
(e) ${{\text{a}}_{{\text{average}}}}{\text{ = }}\dfrac{{{\text{[v(}}{{\text{t}}_{\text{2}}}{\text{)  v(}}{{\text{t}}_{\text{1}}}{\text{)]}}}}{{{\text{(}}{{\text{t}}_{\text{2}}}{\text{  }}{{\text{t}}_{\text{1}}}{\text{)}}}}$
(The ‘average’ stands for average of the quantity over the time interval ${t_1}$ to ${t_2}$)
Ans:
(a) False. As the motion of the particle is arbitrary, the average velocity of the particle cannot be given by this equation.
(b) True. The arbitrary motion of the particle can be represented by the given equation.
(c) False. The motion of the particle is arbitrary. The acceleration of the particle may also be nonuniform. So, this equation cannot represent the motion of the particle in space.
(d) False. The motion of the particle is arbitrary, acceleration of the particle may also be nonuniform. So, this equation cannot represent the motion of particle in space.
(e) True. The arbitrary motion of the particle can be represented by the given equation.
24. 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
Ans:
(a) False. Energy is not conserved in inelastic collisions.
(b) False. Temperature can take negative values.
(c) False. Total path length is a scalar quantity and has the dimension of length.
(d) False. A scalar quantity like gravitational potential can vary from one point to another in space.
(e) True. The value of a scalar does not change for observers with different orientations of axes.
25. An aircraft is flying at a height of $3400\;{\text{m}}$ above the ground. If the angle subtended at a ground observation point by the aircraft positions $10.0\;{\text{s}}$ apart is ${{30}^{0}}$, what is the speed of the aircraft?
Ans:
The positions of the observer and the aircraft are shown below:
(Image will be uploaded soon)
Height of the aircraft from ground, ${\text{OR}} = 3400\;{\text{m}}$
Angle subtended between the positions, $\angle {\text{POQ}} = {{30}^{0}}$
Time $ = 10\;{\text{s}}$
In $\Delta {\text{PRO}}:$
$\tan {{15}^{0}} = \dfrac{{PR}}{{OR}}$
$PR = OR\tan {{15}^{0}}$
$PR = 3400 \times \tan {{15}^{0} }$
$\Delta $PRO is similar to $\Delta RQO$
$\therefore {\text{PR}} = {\text{RQ}}$
${\text{PQ}} = {\text{PR}} + {\text{RQ}}$
${\text{PQ}} = 2{\text{PR}}$
${\text{PQ}} = 2 \times 3400\tan {{15}^{0}}$
${\text{PQ}} = 6800 \times 0.268$
${\text{PQ}} = 1822.4\;{\text{m}}$
$\therefore $ Speed of the aircraft $ = \dfrac{{1822.4}}{{10}} = 182.24\;{\text{m}}/{\text{s}}$
Additional Exercise
26. A vector has magnitude and direction. Does it have a location in space? Can it vary with tune? Will two equal vectors $a$ and $b$ at different locations in space necessarily have identical physical effects? Give examples in support of your answer.
Ans: No. A vector has no definite locations in space because it remains invariant when displaced in such a way that its magnitude and direction remains the same. A position vector has a definite location in space.
Yes. A vector can vary with time. The displacement vector of a particle moving with a certain velocity varies with time is an example.
No. Two equal vectors present at different locations in space need not produce the same physical effect. Consider the example of two equal forces acting on an object at different points can cause the body to rotate, but their combination cannot produce an equal turning effect.
27. A vector has both magnitude and direction. Does it mean that anything that has magnitude and direction is necessarily a vector? The rotation of a body can be specified by the direction of the axis of rotation, and the angle of rotation about the axis. Does that make any rotation a vector?
Ans:
No. A physical quantity having both magnitude and direction need not necessarily be a vector.
For example, current is a scalar quantity even though it has magnitude and direction. The requirement for a physical quantity to be considered a vector is that it should follow the law of vector addition.
No. The rotation of a body about an axis is not a vector quantity because it does not follow the law of vector addition. However, a rotation by a small angle follows the law of vector addition and is considered a vector.
28. Can you associate vectors with
(a) the length of a wire bent into a loop,
(b) a plane area,
(c) a sphere? Explain.
Ans:
(a) No. A vector cannot be associated with the length of a wire bent into a loop as the length of a loop does not have a definite direction.
(b) Yes. An area vector can be associated with a plane area. The direction of this vector is represented by a normal drawn outward to the area.
(c) No. A vector cannot be associated with the volume of a sphere as it does not have a specific direction. A null vector can be associated with the area of a sphere.
29. A bullet fired at an angle of ${{30}^{0}}$ with the horizontal hits the ground $3.0\;{\text{km}}$ a way. By adjusting its angle of projection, can one hope to hit a target $5.0\;{\text{km}}$ away? Assume the muzzle speed to the fixed, and neglect air resistance.
Ans:
No.
Range, ${\text{R}} = 3\;{\text{km}}$
Angle of projection, $\theta = {{30}^{0}}$
Acceleration due to gravity, ${\text{g}} = 9.8\;{\text{m}}/{{\text{s}}^2}$
Horizontal range for the projection velocity is:
${\text{R}} = \dfrac{{{\text{u}}_0^2\sin 2\theta }}{{\text{g}}}$
$3 = \dfrac{{{\text{u}}_0^2}}{{\;{\text{g}}}}\sin {{60}^{0}}$
$\dfrac{{{\text{u}}_0^2}}{{\;{\text{g}}}} = 2\sqrt 3 {\text{ }}...{\text{(i)}}$
The maximum range $\left( {{R_{\max }}} \right)$ is achieved by the bullet when it is fired ${{45}^{0}}$ with the horizontal.
\[{{\text{R}}_{\max }} = \dfrac{{{\text{u}}_0^2}}{g}{\text{ }}...{\text{(ii)}}\]
Compare ${\text{(i)}}$ and \[{\text{(ii)}}\]:
${{\text{R}}_{\max }} = 3\sqrt 3 $
$ = 2 \times 1.732$
$ = 3.46\;{\text{km}}$
So, the bullet will not hit a target $5\;{\text{km}}$ away.
30. A fighter plane flying horizontally at an altitude of $1.5\;{\text{km}}$ with speed $720\;{\text{km}}{{\text{h}}^{  1}}$ passes directly overhead an antiaircraft gun. At what angle from the vertical should the gun be fired for the shell with muzzle speed $600\;{\text{m}}{{\text{s}}^{  1}}$ to hit the plane? At what minimum altitude should the pilot fly the plane to avoid being hit? (Take ${\text{g}} = 10\;{\text{m}}{{\text{s}}^{  1}}$ ).
Ans:
Height of the fighter plane $ = 1.5\;{\text{km}} = 1500\;{\text{m}}$
Speed of the fighter plane, ${\text{v}} = 720\;{\text{km}}{{\text{h}}^{  1}} = 200\;{\text{m}}{{\text{s}}^{  1}}$
Let $\theta $ be the angle with the vertical so that the shell hits the plane.
Muzzle velocity of the gun, ${\text{u}} = 600\;{\text{m}}/{\text{s}}$
Time taken by the shell to hit the plane $ = t$
Horizontal distance travelled by the shell $ = {{\text{u}}_{\text{x}}}{\text{t}}$
Distance travelled by the plane $ = vt$
The shell hits the plane. So, these two distances should be equal.
${{\text{u}}_{\text{x}}}{\text{t}} = {\text{vt}}$
${\text{u}}\sin \theta = {\text{v}}$
$\sin \theta = \dfrac{{\text{v}}}{{\text{u}}}$
$ = \dfrac{{200}}{{600}}$
$ = \dfrac{1}{3}$
$ = 0.33$
$\theta = {\sin ^{  1}}(0.33)$$ = {{19.5}^{0}}$
To avoid being hit by the shell, the pilot should fly the plane at an altitude ${\text{(H)}}$ higher than maximum height of shell.
$\therefore H = \dfrac{{{u^2}{{\sin }^2}(90  \theta )}}{{2g}}$
$ = \dfrac{{{{(600)}^2}{{\cos }^2}\theta }}{{2g}}$
$ = \dfrac{{360000 \times {{\cos }^2}19.5}}{{2 \times 10}}$
$ = 18000 \times {(0.943)^2}$
$ = 16006.482\;{\text{m}}$$ \approx 16\;{\text{km}}$
31. A cyclist $1{\text{ s}}$ riding with a speed of $27\;{\text{km}}{{\text{h}}^{  1}}$. As he approaches a circular turn on the road of radius $80\;{\text{m}}$, he applies brakes and reduces his speed at the constant rate of $0.50\;{\text{m}}{{\text{s}}^{  1}}$ every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?
Ans:
Speed of the cyclist, ${\text{v}} = 27\;{\text{km}}{{\text{h}}^{  1}} = 7.5\;{\text{m}}{{\text{s}}^{  1}}$
Radius of the circular turn, ${\text{r}} = 80\;{\text{m}}$
Centripetal acceleration is:
${a_c} = \dfrac{{{v^2}}}{r}$
$ = \dfrac{{{{(7.5)}^2}}}{{80}}$
$ = 0.7{\text{m}}{{\text{s}}^{  2}}$
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Assume that the cyclist starts cycling from $P$ and moves toward $Q$. At $Q$, he applies the breaks and decelerates the speed of the bicycle by $0.5\;{\text{m}}{{\text{s}}^{  2}}.$ This acceleration is along the tangent at $Q$ and opposite to the direction of motion of the cyclist.
As the angle between ${a_c}$ and ${a_r}$ is ${{90}^{0}}$, the resultant acceleration is:
${\text{a}} = \sqrt {{\text{a}}_{\text{c}}^2 + {\text{a}}_{\text{r}}^2} $
$ = \sqrt {{{(0.7)}^2} + {{(0.5)}^2}} $
$ = \sqrt {0.74} $
$ = 0.85{\text{m}}{{\text{s}}^{  1}}$
$\tan \theta = \dfrac{{{{\text{a}}_{\text{c}}}}}{{{{\text{a}}_{\text{r}}}}}$
Where $\theta $ is the angle of the resultant with the direction of velocity
$ = \dfrac{{0.7}}{{0.5}}$
$ = 1.4$
$\theta = {\tan ^{  1}}(1.4)$
$\theta = {{54.46}^{0}}$
32. (a) Show that for a projectile the angle between the velocity and the $x  $axis as a function of time is given by
$\theta ({\text{t}}) = {\tan ^{  1}}\left( {\dfrac{{{{\text{v}}_{{\text{oy}}}}  {\text{gt}}}}{{{{\text{v}}_{{\text{ox}}}}}}} \right)$
(b) Show that the projection angle ${\theta _0}$ for a projectile launched from the origin is given by
${\theta _0} = {\tan ^{  1}}\left( {\dfrac{{4{{\text{h}}_{\text{m}}}}}{{\text{R}}}} \right)$
where the symbols have their usual meaning.
Ans:
Let ${{\text{v}}_{{\text{ox}}}}$ and ${{\text{v}}_{{\text{oy}}}}$ be the initial components of the velocity of the projectile along horizontal $({\text{x}})$ and vertical $({\text{y}})$ directions.
Let ${{\text{v}}_{\text{x}}}$ and ${{\text{v}}_{\text{y}}}$ be the horizontal and vertical components of velocity at a point ${\text{P}}$.
Time taken by the projectile to reach point $P = t$
(a) Apply the first equation of motion along the vertical and horizontal directions.
${{\text{v}}_{\text{y}}} = {{\text{v}}_{0{\text{y}}}}{\text{  gt}}$
And ${v_x} = {v_{ax}}$
$\therefore \tan \theta = \dfrac{{{v_y}}}{{{v_x}}}$
$\tan \theta = \dfrac{{{v_{oy}}  gt}}{{{v_{ax}}}}$
$\theta = {\tan ^{  1}}\left( {\dfrac{{{v_{ey}}  gt}}{{{v_{ox}}}}} \right)$
(b) Maximum vertical height, ${{\text{h}}_{\text{m}}}{\text{ = }}\dfrac{{{\text{u}}_{\text{o}}^{\text{2}}{{\sin }^2}2\theta }}{{{\text{2g}}}}{\text{ }}...{\text{(i)}}$
Horizontal range, ${\text{R}} = \dfrac{{{\text{u}}_{\mathbf{o}}^2{{\sin }^2}2\theta }}{{\text{g}}}{\text{ }}...{\text{(ii)}}$
Solve equation ${\text{(i)}}$ and equation ${\text{(ii)}}$:
$\dfrac{{{{\text{h}}_{\text{m}}}}}{{\text{R}}} = \dfrac{{{{\sin }^2}\theta }}{{2{{\sin }^2}\theta }}$
$\dfrac{{{{\text{h}}_{\text{m}}}}}{{\text{R}}} = \dfrac{{\sin \theta \times \sin \theta }}{{2 \times 2\sin \theta \cos \theta }}$
$\dfrac{{{{\text{h}}_{\text{m}}}}}{{\text{R}}} = \dfrac{{1\sin \theta }}{{4\cos \theta }}$
$\dfrac{{{{\text{h}}_{\text{m}}}}}{{\text{R}}} = \dfrac{1}{4}\tan \theta $
$\tan \theta = \left( {\dfrac{{{\text{4}}{{\text{h}}_{\text{m}}}}}{{\text{R}}}} \right)$
$\theta = {\tan ^{  1}}\left( {\dfrac{{{\text{4}}{{\text{h}}_{\text{m}}}}}{{\text{R}}}} \right)$
NCERT Solutions For Class 11 Physics Chapter 4 PDF Download
Class 11 Physics Chapter 4 NCERT Solutions are the best option available for students to gain basic knowledge in science and learn the concepts of motion in a plane. Motion in a Plane Class 11 NCERT Solutions PDF is what students require for improving their knowledge on this chapter. All the concepts and fundamentals are clearly explained in this Motion in a Plane Class 11 solved problems PDF.
NCERT Solutions of Chapter 4 (Motion in a Plane) [CBSE Class 11 Physics]  Important Coverage of the Chapter
Class 11th Physics Chapter 4 NCERT Solutions will cover these topics:
1. Introduction
In the first section, students will be introduced to some concepts that are related to motion in a plane, such as concepts of position, velocity, displacement and acceleration. Explanations relating to these concepts are included in the ch 4 Physics Class 11 NCERT Solutions.
2. Scalars and Vectors
In the second section, students will gain basic knowledge of scalars and vectors. Topics that are included in this section are the difference between scalars and vectors, directions associated with scalars and vectors, definitions, displacement vectors, triangle law of addition, parallelogram law of addition, the magnitude of displacement and some other topics. Solutions to all these topics can be found in Motion in a Plane Class 11 NCERT Solutions.
3. Multiplication of Vectors by Real Numbers
In this section, students will learn how to derive an equation by using magnitudes and different formulas. They will also learn to multiply the vectors by using the real numbers.
4. Addition and Subtraction of Vectors  Graphical Method
In this section, students will learn about the triangle law, parallelogram law of addition and their definitions. They will gain knowledge on of formulas of the headtotail method, triangle method of vector addition, commutative laws, null vectors and subtraction of vectors in this section.
5. Resolution of Vectors
With the help of this section, students will learn about the resolution of vectors and its implementation in a straight line. Getting an equation by using real numbers will be easy after learning the concepts of this section.
6. Vector Addition  Analytical Method
Through this section, a student can learn to add vector numbers by an analytical method along with a graphical method.
7. Motion in a Plane
In this section, a student will learn about velocity, acceleration, finding out averages, and then finally finding out the motion in a plane.
8. Motion in a Plane with Constant Acceleration
Here, students will learn how motion in a plane can be treated as two different simultaneous onedimensional motions, both of which have a constant acceleration in two different perpendicular directions.
9. Relative Velocity in Three Dimension
In this section, a student will learn to find out threedimensional motion in a plane by using relative velocity.
10. Projectile Motion
This section helps in gaining knowledge of the concepts of projectile motion.
11. Uniform Circular Motion
This section gives a basic knowledge of uniform circular motion.
Benefits of Motion in a Plane Class 11 Solutions
NCERT Solutions of Chapter 4  How to Use It and What are the Benefits Rendered?
Vedantu’s CBSE NCERT Solutions Class 11 Physics Chapter 4 is the key to securing good marks in the Class 11 exams. because of the following benefits:
So, how will you use these NCERT Solutions of Chapter 4 in your study plan?
NCERT Solutions are always to be referred for any CBSE subject. These solutions are strictly include all the questions following the answer writing guidelines prescribed by the CBSE. Also, generally questions are asked from NCERT books. Thus, students can study these solutions for completing the chapter in an effective and efficient way.
The NCERT Solutions of Chapter 4 can also be referred to while you solve questions and answers. In order to level up your practice, one can easily use these solutions.
NCERT Solutions can be used to brush up on the lessons learned. If you want to revise Chapter 4, you can do it by referring to these NCERT Solutions for Chapter 4.
Now, let us talk about the benefits rendered from studying the NCERT Solutions Chapter 4:
Motion in a Plane Class 11 NCERT Solutions is prepared by some of the best teachers who are experts in this section.
Class 11 Physics Chapter 4 NCERT Solutions are given with practical problems for better understanding.
NCERT Solutions for Class 11 Physics Chapter 4 covers the syllabus.
Quick Links
Important Questions for CBSE Class 11 Physics, Chapter wise Solutions  Free PDF Download NCERT Solutions for Class 11
Class 11 Revision Notes, Short Key Notes for CBSE (NCERT) Books
Conclusion
Students must refer to NCERT solutions so that they can excel in their Class 11 CBSE Exams. These comprehensive notes consist of many important topics which will be beneficial for the students. Students can also download other study material in PDF format at no cost from our website.
FAQs on NCERT Solutions for Class 11 Physics Chapter 4  Motion In A Plane
1. What is the difference between scalar and vector quantities?
A quantity that has a magnitude but lacks a direction is termed as a scalar quantity. A quantity that has both magnitudes, as well as a direction, is termed as a vector quantity. A scalar quantity is always a dot product of two quantities, whereas the vector is always the cross product of two vector quantities. Examples of scalar quantities are mass, density and current. Some examples of vector quantity are acceleration, velocity and displacement. The difference between these two is very easy to understand.
2. What is linear motion?
Linear motion is also referred to as rectilinear motion. The motion is termed as a onedimensional motion along a straight line, and thus can be explained using only one spatial dimension. There are two types of linear motion: uniform linear motion with a constant velocity and zero acceleration and nonuniform linear motion with variable velocity and nonzero acceleration. It is referred to as the most basic among all the motions. Linear motion, to some extent, can be compared with a general motion. An example of linear motion is an athlete running in on a straight track.
3. What type of questions are present in Chapter 4 ‘Motion in a Plane’ of the NCERT Solutions for Class 11 Physics?
There are a variety of questions present in Chapter 4 ‘Motion in a Plane’ in NCERT Solutions for Class 11 Physics. The questions will range from detailed to short answertype. This will contain pick/choose the one out, true or false, reason stating questions, numerical or problem questions, long answer questions etc. Therefore, you will find everything in one place and don’t have to seek other places to study thoroughly.
4. Define uniform circular motion according to Chapter 4 of the NCERT Solutions for Class 11 Physics.
Circular motion is the path that leads to the circular movement of the body. Likewise, when the motion of the body in the circular path is followed by a constant speed, it is called uniform circular motion. The speed of the body is the same but with a change in velocity. When the object is moving around the circle, it is continuously changing its direction. The object touches the edge of the circle but does not cross it
5. Can I get full marks in Chapter 4 ‘Motion in a plane’ of Class 11 Physics?
Yes, it is possible to get full marks in Class 11 Physics Chapter 4 Motion in a Plane. However, do remember that even a small mistake can lead to the deduction of the marks. Don’t try to apply the shortcuts, especially in the numerical. Try to give the full detailing about the equations used and the process for getting the answer in the numerical.