Vectors

PHYSICS - I A

The vector, as a mathematical obj ect, is defined as a directed line segment. It should also obey the laws of vector addition.

Examples : – Displacement, velocity, acceleration,

force, momentum, angular momentum , moment of force, Torque, magnetic moment, magnetic induction field, Intensity of electric field, etc ....

To describe a vector quantity we require. a) The specific unit of that quantity.

b) The number of times that unit is contained in that quantity.

c) The orientation of that quantity.

Ex : - A plane is flying from west to east with a

velocity of 50 ms–1. Here ms–1 is the unit, 50 is the number of units of velocity and west to east is the direction.

Note 3.3: 1) A physical quantity having magnitude

and direction but not obeying laws of vector addition is treated as a scalar.

Ex : Electric current is a scalar quantity

Electric current is always associated with direction, but it is not a vector quantity. It does not obey law of vector addition for its addition.

i₁ i₂ (i1 + i2) q 3.1 CLASSIFICATION OF PHYSICAL QUANTI TI ES

All measurable quantities are called physical quantities. Most of the physical Quantities are classified into 'Scalars' and 'Vectors'.

Scal ar: - Physi cal quant i t i es havi ng onl y magnitude are called Scalars.

Exampl es : Length, time, volume, density,

temperature, mass, work, energy, electric charge, electric current, potential ,resistance, capacity, etc...

To describe a scalar quantity we require a) The specific unit of that quantity

b) The number of times that unit is contained in that quantity

Ex: A bag contains 100 kg of sugar. Here kg is the

unit and 100 is the number of units of sugar present in the bag.

Note 3.1: Unit is not a compulsion to represent a scalar Ex : - Specific gravity, Refractive index Note3.2: Mathematical operations of scalar quantities

yield scalar quantities and these quantities are manipulated by ordinary algebraic rules.

Vector : - Physical quantities having both magnitude and direction and that obeys laws of vector addition are called vectors (or)

V E CT O R S

CHAPTER





3

Leonhard Euler (15 April 1707 – 18 September 1783) was a pioneering Swiss mathematician and physicist. He made important discoveries in fields as diverse as infinitesimal calculus and graph theory. He also introduced much of the modern mathematical terminology and notation, particularly for mathematical analysis, such as the notion of a mathematical function. He is also renowned for his work in mechanics, fluid dynamics, optics, and astronomy.

Euler was featured on the sixth series of the Swiss 10-franc banknote and on numerous Swiss, German, and Russian postage stamps. The asteroid 2002 Euler was named in his honor. He is also commemorated by the Lutheran Church on their Calendar of Saints.

L eonhard Euler 1707-1783 h To differentiate between scalar and vector

quantites with examples.

h To find the resultant of vectors, and resolve a

vector into its rectangular components

h To explain the Relative Motion. h To compute the product of vectors.

ELEMENTS OF VECTORS

The resultant of i₁ and i₂ is (i₁ + i₂) by Kirchoff’s current law. The resultant does not depend on angle between currents i₁ and i₂.

Note3.4: Equations in vector form indicate both

mathematical and geometrical relationships among the quantities. Physical laws in vector form are very compact and independent of choice of coordinate system.

3.2 GEOMETRICAL REPRESENTATION OF VECTORS

Any vector A can be represented geometrically as a directed line segment (an arrow), as shown in fig. The magnitude of A is denoted by A , and the direction of A is specified by the sence of the arrow and the angle ‘ ’ that it makes with a fixed reference line.

A q

When using graphical methods, the length of the arrow is proportional to the magnitude of the vector, and the arrow head represents the direction.

Ex:- The velocity vector v is represented by an arrow OP as shown in figure. The initial point of the vector is O, the final point of the vector is P. The length OP is the magnitude of the velocity and its direction is 600 north of east (or) 300 east of north .

0 60 q  N W S E O P 0 3 0  v

600 _{North of East or 30}0 _{Eash of North}

0 4 5 NW N NE W SW SE S E 3.3 TYPES OF VECTORS

a) Polar Vectors : The vector whose direction

does not change even though the co-ordinate system in which it is defined changes is called polar vector.

Ex : Force, momentum, Acceleration.

b) Axial Vectors : The vectors whose direction

changes with the co-ordinate system in which it is defined changes is called axial vector.

Ex:Angular velocity, torque, angular momentum c) Like vectors (or) parallel vectors : Two or

more vectors (representing same physical quantity) are called like vectors if they are parallel to each other, however their magnitudes may be different.

A

B

C



d) Unlike vectors (or) anti parallel vectors : Two

vectors (representing same physical quantity) are called unlike vectors if they act in opposite direction however their magitudes can be different.

A



B

e) Equal vect or s : Two or more vectors

(representing same physical quantity) are called equal if their magnitudes and directions are same.

A

B

C

Ex : Suppose two trains are running on parallel

tracks with same speed and direction. Then their velocity vectors are equal vectors.

Note : If a vector is displaced parallel to itself its magnitude and direction does not change.

A  B  C X Y _{A B C}     

f) Negative Vector : A vector having the same

magnitude and opposite in direction to that of a given vector is called negative vector of the given vector

A

A

PHYSICS - I A

g) Co-initial vectors : The vectors having same

initial point are called co-initial vectors.

C A  B  E  D O

h) Collinear vectors : Two or more vectors are

said to be collinear when they act along the same line however their magnitudes may be different.

Ex : Two vectors A 

and B as shown are collinear vectors.

A



B (or)  A B

i) Coplanar vectors : A number of vectors are

said to be coplanar if they are in the same plane or parallel to the same plane. However their magnitudes may be different. Y O X A B C

j) Unit vector : A vector whose magnitude equals one and used to specify a convenient direction is called a unit vector.

A unit vector has no units and dimensions. Its purpose is to specify the direction of given vector.

In cartesian coordinate system, unit vectors along positive x, y and z axis are symbolised as i , j and k respectively. These three unit vectors are mutually perpendicular and their magnitudes

ˆ ˆ ˆ

i  j  k  .1

If A is a non zero vector, then the unit vector in

the direction of A is given by Aˆ A A 

 

3.4 REPRESENTATI ON OF ANGLE BETWEEN THE TWO VECTORS

The angle between two vectors is represented by the smaller of the two angles between the vectors when they are placed tail to tail by displacing either of the vectors parallel to it self.

Ex: The angle between A 

and B is correctly represented in the following figures.

B  A  q q B  A  B  A  B q (or) Fig.(i) Fig.(ii)

a) If the angle between A 

and B is  , then the angle between A and KB  is also  . Where ‘K’ is a positive constant. q A  B  q A  K B

b) If the angle between A and B is  , then the angle between A and KB

  is (180–  ) . Where K is a positive constant. q A  B  q A  KB   0 180q B 

c) Angle between collinear vectors is always zero or 1800C Q P  P  Q 0 0 q  0 180 q  or Problem : 3.1

Three vectors , ,A B C are shown in the figure. Find  

angle between (i)A and  B (ii) B and  C (iii)A and  C.

300 300 450 x x x A B C

Sol. To find the angle between two vectors we connect the

tails of the two vectors. We can shift the vectors parallel to themselves such that tails of A,B  and C are connected as shown in figure.

ELEMENTS OF VECTORS 300 300 450 A  B  C  x y

Now we observe that angle between A and B is 60, B and C is 15 and between A and C is 75

Problem : 3.2

If , ,A B C represents the three sides of an equilateral  

triangle taken in the same order then find the angle between i)A and  B ii)  B and  C iii) A and  C.

A A B C 1200 1200 1200 B C

Sol. From the diagram the angle between the

vectors A and B is 1200 _{, the angle between B} _{and C}

is 1200_{, the angle between A} _andC _{is 120}0

3.5 ADDITION OF VECTORS (GRAPHICAL METHOD)

1. To add two vectors geometrically represent the vectors by arrow head lines using the same suitable scale, with their proper directions in the choosen co-ordinate system.

2. Join the initial point of the second vector with the final point of the first vector by moving parallel to itself.

3. Now, draw an arrow from the initial point of the first vector to the final point of the second vector. This arrow represents the resultant of the two vectors.

Examples :

1. When two vectors are acting in the same di-rection.

Let the two vectors P and Q be acting in the same direction. RPQ    _R_P_Q (a) (b) (c) P P  Q Q

2. When two vectors are acting at some angle:

First join the initial point of Q with the final point of P and then, to find the resultant of these two, draw a vector R from the initial point of P to the final point of Q . This single vector R

 is the resultant of vectors P and _Q.

R  P _P Q Q R 

represents the resultant of P and Q 

both in magnitude and direction. So, RPQ.

   3.6 LAWS OF VECTOR ADDITION

a) Vector addition obeys commutative law.

Addition of two vectors is independent of the order of the vectors in which they are added. If P and Q are two vectors then PQQP

    as shown in fig.      Q P R      P Q R  P  P  Q  Q O _A B _C

b) Vector addition obeys associative law.

While adding more than two vectors, the resultant is independent of the order in which they are added.

If P, Q  

and R 

are three vectors, then



P Q



 RP



Q R



as shown in fig.  P  Q  R  P Q    Q R        P Q R  P  Q _ R (P Q  )R

c) Vector addition obeys distributive law.

If k, k₁, k₂ are scalars then





k P Q kPkQ and P k

_

₁k₂

_

k P₁k P₂ Note 3.5 : Vector addition is possible only between

PHYSICS - I A

3.7 PARALLELOGRAM LAW OF VECTORS Statement : If two vectors are drawn from a point so as to represent the adj acent sides of a parallelogram both in magnitude and the direc-tion, the diagonal of the parallelogram drawn from the same point represents the resultant of the two vectors both in magnitude and direction.

N A B O C a q  P  Q q R

Let the two vectors P and _Q, inclined at angle  be acting on a particle at the same time. Let they be represented in magnitude and direction by two adjacent sides OA and OB of parallelogram OACB, drawn from a point O.

According to parallelogram law of vectors, their resultant vector R will be represented by the diagonal

OC of the parallelogram.

Magnitude of resultant :

The line OA is extended upto point N and draw a perpendicular from point C on to the extended line as shown in fig.

From the parallelogram OACB. BOA = CAN =  and OA = P, OB = AC = Q, OC = R and

In CNA, AN = AC cos  = Q cos   CN = AC sin  = Q sin   From triangle ONC

OC2 = ON2 + NC2

(OC)2 = (OA+AN)2 + (NC)2

R2 = (OA)2 + 2(OA)(AN) + (AN)2 +(NC)2 R2 = P2 + 2 PQ cos  +



QCos



2



QSin



2

2 2 2

R P 2PQCos Q

R = 2 2

P Q 2PQ cos  ... (3)

The magnitude of the resultant R depends on the magnitudes of the vectors P and Q and also on the angle between them.

Direction of resultant : N A B O C a q  P  Q q R 

If the line of action of the vector P is taken as reference line, the resultant R makes an angle  with it. This angle indicates the direction of R . Then from right angled triangle ONC,

tan CN CN Q sin ON OA AN P Qcos



   

   ...(4)

Note3.6: If  is the angle between the resultant R and the vector Q, then 1 P sin

tan , Q P cos       _{ }    Special Cases:

a) If the magnitude of P > Q then   . i.e., the resultant is closer to the vector of larger magnitude. b) When angle between two vectors increases, the magnitude of their resultant decreases.

c) W hen t wo vect or s ar e i n t he same

direction(parallel).  P Q





     R P Q

then  = 00, cos 00 = 1 and sin 00 = 0 from eq.(3), R = 2 2

P Q 2PQ(1)

=  2

(P Q) = (P + Q)

from eq.(4), tan  = Q 0_

P Q(1) = 0 or  = 0

0_.

Thus for two vectors acting in the same direction, the magnitude of the resultant vector is equal to the sum of the magnitudes of two vectors and acts along the direction of P and Q

 .

ELEMENTS OF VECTORS

(d) When two vector s are acting in opposite directions (antiparallel)._ P  Q      R P Q

then  = 1800, cos 1800 = –1 and sin 1800 = 0. From eq. (3), R = 2 2 

P Q 2PQ( 1)

=  2

(P Q) = (P–Q) or (Q–P) From eq. (4), tan  = _{ }Q 0

P Q( 1) = 0 or  = 00 or 1800.

Thus for two vectors acting in opposite directions, the magnitude of the resultant vector is equal to the difference of the magnitudes of the two vectors and its direction is along the vector of larger magnitude.

(e) When two vectors are perpendicular to each other,  = 900, sin  = 1 and cos  = 0 R  P  Q a From eq(3), R = 2 2 P Q 2PQ(0) = 2 2 P Q

From eq(4), tan  = PQ (1)Q(0) = Q P or  = tan–1   _{ }

 QP

f) If the resultant of two vectors is perpendicular to any one of the vectors, then

i) The angle between the two vectors is greater than 900

ii) The resultant vector is perpendicular to the vector having smaller magnitude.

we know tan Q sin P Q cos      0 Q sin tan 90 P Q cos     P + Q cos  = 0 f

q

P Q R P cos  = P Q  and R = 2 2 Q P , P R P

sin , cos , tan .

Q Q R

     

The angle between P and Q is 0

90     g) max min R P Q R P Q    and max min max min R R P . Q R R   

h) If magnitudes of P and Q are equal and the angle between them is  then their resultant is

2 2 R P Q 2PQ cos  2 2 2 q R P P 2P cos [P=Q] 2 2 R 2P 2P cosq, R 2P 1 c os2



 



2 2 R 2P 2 cos 2     _{ }   , R 2P cos₂   

and the resultant makes an angle ‘  ’ with P is tan Q sin P Q cos      P s i n P P c o s    





P sin tan P 1 cos      2 s i n 2 c o s / 2    2 2 sin cos 2 2 tan 2 cos / 2       tan₂   tan tan 2     , 2    a) If 0 60 ,   then R = 3P and  300 b) If 0 90 ,   then R = 2P and  450 c) If 0 120 ,   then R = P and 0 60  

I) The unit vector parallel to the resultant of two vectors _{P & Q}  is ˆn P Q P Q        .

Note 3.7 : We can add a vector to another vector of

same kind but we can’t add a vector quanity to a scalar quantity.

PHYSICS - I A

Problem : 3.3

The resultant of two forces whose magnitudes are in the ratio 3 : 5 is 28 N. If the angle of their inclination is

600_{, then find the magnitude of each force.}

Sol : Let F₁ and F₂ be the two forces.

Then F₁ = 3x; F₂ = 5x; R = 28 N and 0 60   2 2 1 2 1 2 R F F 2F F cos

 

2

 

2

 

0 28 3x 5x 2 3x (5x) cos60     2 2 2 28 9x 25x 15x 7x      28 x 4 7    , F₁ = 3 × 4 = 12 N, F₂ = 5 × 4 = 20 N. *Problem : 3.4

If vectors A and  B are 3i – 4j + 5k and 2i + 3j – 4k

respectively then find the unit vector parallel to

   A B (Ans :



5



27 i j k ) Hint : nˆ A B_{A B}      *Problem : 3.5

If magnitude of the resultant of two vectors equal of magnitude, is equal to the magnitude of either of the

vectors, what is the angle between them ? (Ans : 1200₎

Hint : 2 cos 2 R P  Problem : 3.6 I f A3i 4j and 724 ,  B i j find a vector having the same magnitude asB and par allel to  A

Sol. The vector parallel to Aand having magnitude of B is

^ A C B B A A        2 2 B 7 24 25 and 





2 2 A 3i 4 j 1 A 3i 4 j A _{3 4} 5        _{ }  





1 C 25 3i 4 j 15i 20 j 5       _{   } Problem : 3. 7

The resultant of two vectorsA and  B is perpendicular

to A and equal to half of the magnitude of  B . Find

the angle between A and  B ?

Sol. Since R is perpendicular to A. figure shows the three

vectors A,B and R . angle between A and

B is  –  A B R p q q A sin  = R B = B 2B = 1 2 = 30°

angle between A and B is 150°.

Problem : 3.8

What is the displacement of the point of a wheel initially in contact with the ground when the wheel rolls forward half a revolution? Take the radius of the wheel as R and the x-axis as the forward direction?

Sol: From figure, during half revolution of the wheel,

the point A covers AC R horizontal distance, and BC = 2R vertical distance. q A x C y B x = R and y = 2R; Displacement,  2 2 s x y 2 2 2 ( R) (2R) R 4       and 1 y 1 2R 1 2

Tan Tan Tan

x R

        

  _{ }  _{ } _{ }

     with

x-axis.

i.e., Displacement has magnitudeR  2 4 and makes an angle 1 2

Tan _ _

 with x-axis.

3.8. TRIANGLE LAW OF VECTORS

" If two vectors are represented in magnitude and direction by the two sides of a triangle taken in one order, their resultant vector is represented in magnitude and direction by the third side of the triangle taken in reverse order" .

Let the two vectors P and Q 

, inclined at an angle  be acting on a particle at the same time. Let

ELEMENTS OF VECTORS

they be represented in magnitude and direction by two sides OA and AB of triangle OAB, taken in the same order shown in fig.

Then, according to triangle law of vector addition, the resultant R is represented by the third side _OB of triangle, taken in opposite order..

p q g q  P A B N O R  Q a RP Q    Magnitude of resultant R: 2 2 R P Q 2PQ cos Direction of resultant R: Q sin Tan . P Q cos     

Statement of triangle law when three forces keep a particle in equilibrium :

When three forces acting at a point can keep a particle in equilibrium the three forces can be represented as the sides of a triangle taken in order both in magnitude and direction.

1 F 3 F 1 F 2 F 2 F 3 F O P Q R

Suppose three fo rece F , F , F  _{1 2 3} are simultaneously acting at point O and the point is in equilibrium. Then the three forces can be represented as three sides of a triangle. The triangle PQR is constructed by drawing parallel lines to the directions in which the forces are applied. The magnitudes of the forces and the corresponding sides of the triangle have equal ratio i.e., F1 _ F2 _ F3

PQ PR QR

.

3.9 POLYGON LAW OF VECTORS

Statement : If a number of vectors are repre-sented by the sides of a polygon both in magni-tude and direction taken in order, their resultant

is represented by the closing side of the polygon taken in reverse order in magnitude and direction .

A B C D E 1  F 2  F 3  F 4  F  R F (or)

If a number of coplanar forces are acting simul-taneously at a point keep the particle in equilib-rium, these forces can be represented as the sides of a polygon taken in order both in magnitude and direction. A B C D E 1  F 2  F 3  F 4  F 5  F 1 2 3 4 ₅ F F F F F 0      

Note 3. 8 : If x is the side of a hexagon ABCDEF as shown in figure A B C D E F 3x 2x 0 60 x 3x x x 900 AB = x, AC = 3x, AD = 2x, AE = 3x, AF= x Problem : 3.9

ABCDEF is a regular hexagon with point O as centre. Find the value of     _{AB AC AD AE AF    .}

A B

C D E

F O

PHYSICS - I A AB AC AD AE   AF      =AB



AB  BC



AD



AD DE



AD DE     EF  







  

3AD 3 2AO 6 AO     

3.10 NULL VECTOR (OR) ZERO VECTOR

A vercotr whose magnitude is equal to zero is called a null vector. Its direction is indeterminate.

Examples of zero vector :

1. The velocity of a particle at rest.

2. The acceleration of a particle moving at uniform velocity.

3. The displacement of a stationary object over any arbitrary interval of time.

4. The position vector of a particle at the origin

Note 3.9 :

A 0 A   

, _{A 0} _{ 0}

3.11 EQUI LIBRIANT

A vect or havi ng same magnitude and opposite in direction to that of the resultant of a number of vectors is called the equilibriant. (or) Negative vector of the resultant of a number of vectors is called the equilibriant (E).

If F , F & F _{1 2} ₃ are the three forces acting on a body, then their resultant force is _F   _{R F1F2 F3}





R 1 2 3 E F , E F F F .              Note 3.10 :

i) Single force cannot keep the particle in equilibrium.

ii) Minimum number of equal forces required to keep the particle in equilibrium is two.

iii) Minimum number of unequal coplanar forces required to keep the particle in equilibrium is

three.

iv) Minimum number of equal or unequal non coplanar forces required to keep the particle in equilibrium is four.

Application 3.1 : Lami's Theorem

" If three coplanar forces acting at a point keeps it in equilibr ium, then each for ce is proportional to the sine of the angle between the other two forces"

If F , F , F are the magnitudes of three forces_{1 2 3} and   , , are the angle between forces F2

 and _{F ,}₃ 3 F  and F1  and F1 

and F₂ respectively, as shown in fig. Then according to lami’s theorem

3

1 2 F

F F

sin sinsin

1  F 3  F a b g 2  F Problem 3.10 :

In the given figure the tesion in the string OB is 30N. Find the weight ‘W’ and the tension in the string OA.

A

B

300

O

W

Sol: Let T₁ and T₂ be the tensions in the strings OA and OB respectively A B 300 O W T1 T₂ 900 900 1500

According to lami’s theorem

1 2

0 0 0

T T W

sin 90 sin150 sin120 (T2 = 30N)

on sloving w30 3 N and T₁= 60N 3.12 SUBTRACTION OF VECTORS

The process of subtracting one vector from another is equivalent to adding vectorially the negative of the vector to be subtracted.

Let P and Q 

be the two vectors as shown in figure. We want to find the difference P Q. Let a

ELEMENTS OF VECTORS

b) the vector subtractio n does not follow associative law ie._P_



_{Q R} _

 

_{ P Q}_ 



_R

 P  Q  R B C O O A  P  Q  R P Q PQR A O C B   A BC A B C B C    O A B

C

Applications 3.2 : If v_i is initial velocity of a particle and v_f is its final velocity. Then change in its velocity is given by  v v_f v_i  _{ } . 2 2 i f i f v v v 2v v cos      

Where '  ' is the angle between initial and final velocities.

Application 3.3 : When a particle is performing

uniform circular motion with a constant speed v, the magnitude of change in velocity when it describes an angle  at the centre is V 2vsin .

2      _{ }  

q

O V V q

Application 3.4 : If velocity of a particle

changes from v_i to v_f in time t then the acceleration

of the particle is given by a vf vi. t      Problem : 3.11

A particle is moving eastwards with a velocity of 5m/s. In 10s the velocity changes to 5 m/s northwards. Find the average acceleration in this time

vector Q be added to the vector P by the laws of vector addition. Their resultant gives the value of



PQ



  .  Q  P  P  Q Q A C O B      S P Q q 180q The magnitude of P Q  is 2 2 0 S  P Q 2PQ cos (180  ) S = 2 2 P Q 2PQ cos  and 0 0 Q sin (180 - ) Q sin tan . P Q cos (180 ) P Q cos          

Note 3.7 : If P  Q then P Q 2P sin . 2      _{ }     S 2P sin 2   

Note 3.8 : When two vectors P and Q have same magnitude and  is the angle between them

Resultant Difference 00 600 900 1200 1800 0 2P P 2P 0 P 2P 2P 3P 3P 2Pcos / 2 2Psin / 2 

3.13 LAWS OF VECTOR SUBTRACTION :

a) the vector subtractio n does not follow commutative law i.e.

PQ Q P, But P Q   Q P  P  Q A B C  Q  P A _B

C

  P Q D    Q P  P  Q

PHYSICS - I A Sol. f i av v v v a t t        _{ }  450 v  _{5 /} f v  m s 5 2 /m s 5 / i v m s   _{5 /} i v m s N S E W av 5j 5i 5 2 1 a 10 10 2        m/s2

north – west direction.

*Problem : 3.12

Two vectors A and  B have precisely equal magni-

tudes. For the magnitude of A B to be larger than 

the magnitude of A B by a factor n, what must be 

the angle between them ? Sol. A B n AB

2A cos n2A sin

2 2

 

 (A = B)

1 1

1 1 1

tan , tan , 2 tan

2 n 2 n n

 

     

   _{ }   _{ }

   

3.14RESOLUTION OF A VECTOR INTO RECTANGULAR COM PONENTS Resolution of a vector is the pr ocess of obtaining the component vectors which when combined, according to laws of vector addition, produce the given vector.

Y Q R y j O

q

i x P X

r



Consider a vector r in the xy plane and it makes an angle  with x axis as shown in figure. i and j are unit vectors along X-axis and Y-axis respectivley. From t he point R, draw RP and RQ perpendicular to X-axis and Y-axis respectively. From the parallelogram law of vector addition, it follows that OROPOQ    ... (1) If OP = x and OQ = y, then OPxi and OQy j From equation (1) rxiy j ... (2) In the above equation xi and y j are called the x-component and the y-component of the vector r, and x and y are called magnitudes of the two component vectors.

from the triganle OPR.

OP x

cos x r cos

OR r

      ... (3)

and sin PR y y r sin

OR r

      ...(4)

 from (3) and (4) tan y x

  ... (5) The equation (3) and (4) gives the magnitudes of the rectangular component vectors in terms of the magnitude of the given vector and its inclination with X-axis.

 from equation (2), r



r cos



ˆi



r sin



ˆj from right angled triangle OPR

2 2

OR OP PR or r x2y2 .... (6) The equation (6) gives the magnitude of the vector in terms of its rectangular components.

Note 3.9 : Method involving resolution of vectors

into components to find the resultant of the vectors is known as analytical method.

Note 3.10 : The components of a vector are

independent of each other and can be handled separately.

Note 3.11 : Theoretically, a given vector can be made

the diagonal of infinite number of parallelograms. Thus there can be infinite number of ways to divide a vector into components.

Note 3.12 : A B C D  o _x y -x -y

ELEMENTS OF VECTORS

a) If the vector A is in first quadrant then it can be written as A A i_xˆA j_y

b) If the vector B is in second quadrant then

xˆ y

B  B iB j c) If the vector C is in third quadrant then

 

 _

xˆ y C C i C j d) If the vector D is in fourth quardant then

 

 _

xˆ y D D i D j

3.15 ADDITION OF TWO VECTORS IN TERMS OF COMPONENTS

Let us consider the addition of two vectors P and Q in terms of their components.

We have PP ixˆP j, Qyˆ Q ixˆQ jyˆ   O Q P R x P Qx y Q y P y R x R

x

y



Let R be the resultant vector with component. R_x and R_y along x and y axis respectively. Then

x xˆ

R R i and R_y R i._yˆ

From the diagram R_x = P_x + Q_x and R_y = P_y+Q_y







_

x x x ˆ y y y ˆ R P Q i, R P Q j       RR ixˆR j; Ryˆ 



PxQx



ˆi



Py Qy



ˆj  



x x



2



y y



2 R  P Q  P Q y y y x x x R P Q Tan , Tan R P Q        y y 1 x x P Q Tan P Q       _{ }    Application 3.5 : sin F q cos F q F q

A block is placed on smooth horizontal surface and pulled by a force ‘F’ making an angle ‘  ’ with horizontal.

The component of force along horizontal = Fcos  . The component of force along vertical = Fsin  .

Application 3.6 : q cos mg q sin mg q q mg m

A block of mass ‘m’ is placed on inclined plane of angle ‘  ’ then the component of weight parallel to the inclined plane is ‘mg sin ’, the component of weight perpendicular to the inclined plane is mgcos .

Application 3.7 : q sin T q cos T q

mg

T

F

O

q  x 2 2 x  

A simple pendulum having a bob of mass ‘m’ is suspended from a rigid support and it is pulled by a horizontal force ‘F’. The string makes an angle _q with the vertical as shown in figure.

The horizontal component of tension = T sin  The vertical component of tension = T cos

PHYSICS - I A

When the bob is in equilibrium

T sin F, ...(1) T cos mg ... (2) 2 2 mg mgl T cos _{l x}    _

From equations (1) and (2)

2 2 F x Tan F mgTan mg mg _{l x}       





2 2 T F  mg

Note- 3.13 : If a vector is rotated through an angle

other than integral multiple of 2 (or 3600_{) it}

changes, but its magnitude does not change.

AB A B

q

Note - 3.14 : If the frame of reference is rotated the

vector does not change (though its components may change). O S Vec tor Problem : 3.13

A vector _{3 i}ˆ _j rotates about its tail through an

angle 300 _{in clock wise direction then the new vector is}

Sol: The magnitude of _{3 i}ˆ_{ j is 3 1 = 2 The angle}

made by the vector with x - axis is

y x A 1 Tan A 3     3i j y o _x 300 0 30  

When the given vector rotates 300 _{in clock wise its} direction changes along x - axis but its magnitude does not change.

 The new vector is 2iˆ

x o

y

2i

Problem : 3.14

A weight mg is suspended from the middle of a rope whose ends are at the same level. The rope is no longer horizontal. What is the minimum tension required to completely straighten the rope is

Sol. From the diagram

2T sin = mg mg 2Tsin T T cosq q q q T cosq T q q mg T 2 sin  

The rope will be staraight when  = 0

0 mg T

2 sin 0

  

The tension required to completely straighten the rope is infinity.

Problem : 3.15

The sum of magnitudes of two forces acting at a point is 16N. If their resultant is normal to the smaller force and has a magnitude of 8N. Then the forces are

q 0 90 2 F 1 F F

Sol. Let F be the resultant of two forces F₁ and F₂ as shown in figure with F₂ > F₁.

F₂ sin = F₁ ... (i) F₂ cos = F = 8 ... (ii)

Squaring and adding Eqs (i) and (ii), we get F₂2 = F₁2 + 64 ... (iii)

Given F₁ + F₂ = 16 ... (iv) Solving Eqs. (iii) and (iv), we get F₁ = 6N and F₂ = 10N

* 3.16 RESOLUTION OF A VECTOR INTOTHREE RECTANGULAR COMPONENTS :

Let us consider a vector A be represented by OP



as shown in fig. With O as origin, construct a rectangular parallelopiped with three edges along the three rectangular axes which meet at O. A becomes the diagonal of the parallelopiped. A , A _{x y} and A_z are three vector intercepts along x, y and z axes re-spectively. These are the three rectangular compo-nents of A .

ELEMENTS OF VECTORS

Applying triangle law of vectors, OPOKKP

  

... (1)

Applying parallelogram law of vectors,

y A  A  x A  z A  S P Q O T K Y Z X OKOTOQ    ... (2) from (1) and (2) OP   OTOQKP But KPOS, OPOTOQOS       z x y AA A A ,     or AA i A j A kx  y  z  _{ } Again, OP2 _{= OK}2 _{+ KP}2 _{= OQ}2_+OT2_+KP2 or A2 _{= A} x 2 _{+ A} z 2_+A y 2 y KP OS A       or A A2xA2yA2z

This gives the magnitude of A in terms of the mag-nitudes of components A , Ax y   and A_z  . * 3.17 DIRECTION COSINES :

The direction cosines , m and n of a vector are the cosines of the angles  , and  which a

given vector makes with x-axis, y-axis and z-axis respectively. A ˆ k a ˆi b ˆj X Z Y O

g

If   , , are angle made byA with x, y, z axis

and cos , cos , cos   are called direction cosines.

y

x A z

A A

cos ,m cos , n cos

A A A

         

Squaring and adding

2 2 2

cos  cos   cos  =

2 2 2 2 2 2 2 y x y z x z 2 2 2 2 2 A A A A A A A 1 A A A A A        and 2 2 2

sin  sin  sin  2 3.18 POSITION VECTOR

Position vector of any point, with respect to an arbitrarily chosen origin, is defined as the vector which connects the origin and the point and is directed towards the point.

Y O X Z P r 



, ,



p x y z

Position vector of a point helps in locating the position of the point. Its magnitude gives the distance between origin and the point. Consider a point ‘P’ having co-ordinates (x,y,z) as shown in figure. If ‘O’ is the origin, then OPis called the position vector (r), r xiy j zk 

Magnitude of r is, r  x2y2z2 The unit vector along r is given by

2 2 2 ˆ ˆ ˆ r xi yj zk ˆr r _{x y z}        

Application 3.8 Displacement vector :

1 r  2 r s x y z2,2,2 X Y B O 1, ,1 1 A x y z

If r₁ is initial position vector of the particle and

2

r 

is final position vector of the particle then the dis-placement of the particle is given by  sr₂r₁.



2 1



ˆ



2 1



ˆ



2 1



ˆ

s x x i y y j z z k 

PHYSICS - I A

The magnitude of the displacement vector is S AB



x₂x₁



2



y₂ y₁



2 



z₂z₁



2

Application 3.9 Condition for collision :

Consider two particles, 1 and 2 move with con-stant velocities _{v and} _{1 v . At initial moment (t=0)}₂ their position vectors are r₁ and r₂.

If particles then collide at the point ‘P’ after time ‘t’ sec.

From the diagram

1 1 2 2 r   r s r s      x y 1 2 P r 2 r 2 2   s v t 1 r 1 1   s v t O 1 1 2 2 r r v t r v t         r₁ r₂ 



v₂v t₁



(or)  r₁r₂  v₂v t₁ ... (1) 1 2 2 1 r r t v v        substitute ‘t’ in equation ...(1)





1 2 1 2 2 1 2 1 r r r r v v v v       _{ }            1 2 2 1 1 2 2 1 r r v v r r v v              Problem : 3.16

Find the resultant of the vectors shown in figure.

Sol. O A B C x y 4cm 3cm 5 cm 370 R OR  AB BC



ˆ ˆ



ˆ ˆ

R  5 cos 37i5sin 37 j 3i4 j

ˆ ˆ ˆ ˆ R 4i3 j 3i 4 j R = 7i + 7j, R7 2cm and 0 45   with horizontal Problem : 3.17

Find the resultant of the vectors   OA OB OC, , as shown in figure. The radius of the circle is R.

Sol : R OA OB OC    

ˆ ˆ ˆ

R RiR cos 45iR sin 45 j Rj

R _ˆ R _ˆ R R i R j 2 2     _  _ _  _      450 450 A B C O





R = 2R+R along _OB. Problem : 3.18

A man walks 40m north, then 30m east and then 40m south. The displacement from the starting point is

Sol. _S_{40 j 30i}ˆ_ ˆ_{40 j}ˆ S N E W ˆ S30i   displacement is 30 m east Problem : 3.19

Vector A is 2 cm long and is 600 _{above the x – axis in}

the first quadrant, vector B is 2cm long and is 600

below the x – axis in the fourth quadrant. Find A B . 

Sol: 600 600 A B Bx B_y Ax Ay O X Y -X -Y R AB 0ˆ ˆ

R 2 cos60 i2 sin 60 j+2cos60 ˆi – 2sin60ˆj

ˆ ˆ R 4 cos 60i2i  2cm, along x – axis

*Problem : 3.20

A vector has x component of -25.0 units and y component of 40.0 units find the magnitude and direction of the vector.

Sol. Consider a vector _A _{A ix}ˆ_{A jy}ˆ

ˆ ˆ A 25i 40 j    





2





2 2 2 x y A  A A  25.0  40.0  = 47.16  47.2 units. and y x A 40.0 tan 1.6 A 25.0      an     (-1.6)= – 58.00with – ve x - axis. (This is in clock wise direction).

This is equivalent to (1220 _{) in anticlockwise direction}

ELEMENTS OF VECTORS y P S x y1 x1 S1 | ps r ps r | s s r

Now to find the velocity of a moving object relative to another moving object, consider a par-ticle P whose position relative to frame S is rps

 and relative to S

|

is r ._ps|



If the position of frame S1 relative to S at any time is r ._{s s}|



Then from the above diagram | |

ps ps s s

r r r

  

Differentiating the above equation with respect to time 1 1 ps d rps d rs s d r dt  dt  dt  _  1 1 1 1 ps ps s s ps ps s s v v v ; v v v       ... (1) Thus, if the velocities of two bodies (here par-ticle and frame S| ) are known with respect to a com-mon frame (s), we can find the velocity of one body relative to the other.

Note 3.15 : I f v_A and v_B are velocities of two bodies A and B relative to earth, then the velocity of A relative to B is v_ABv_Av_B.

a) If two bodies are moving along the same line in the same direction with velocities v_A and v_B relative to earth then the magnitude of velocity of A relative to B is given by v_AB = v_A – v_B

If v_AB is positive the direction of v_AB is that of A and if negative the direction of v_AB is opposite to that of A (or) along the direction of B.

b) If two bodies are moving towards each other or away from each other with velocities v_A and v_B. Then the magnitude of velocity of A relative to B is given by

v_AB = v_A – (– v_B ) = v_A + v_B

and directed towards A (or) away from A

*Problem : 3.21

Find the resultant of the vectors shown in fig by the component method 5 y b x c 3 1 0 37 0 53 6 d a -X -Y O Sol. 0 x ˆ ˆ ˆ

R 1i5 cos 37 i6 cos 53i  x ˆ ˆ ˆ x ˆ R 1i4i3.6iR  6.6i   0 y ˆ ˆ ˆ R 3 j 5sin 37 j 6 sin 53 j   y ˆ ˆ ˆ y ˆ R 3 j 3 j 4.8 j  R 1.2 j  





2

 

2 2 2 x y R R R  6.6  1.2 6.7 x R1.2 R x 6.6 R y 1.2 R  0 10.3 

x

x  y  

y

O

tan y 1

_

_

0 x R 1.2 0.18; tan 0.18 10.3 R 6.6           

with –ve x axis in clock wise direction or 169.70 _with

+ ve x–axis in anti clock wise direction.

*Problem : 3.22

Two vectors a and b have equal magnitudes of 12.7

units. These vectors are making angles 28.20 _and

133.20 _{with the x axis respectively. Their sum is r.}

Find (a) the x and y components of r, (b) the magni-tude of r, (c) the angle r makes with the +x axis.

Ans : 2.5, 15.3, 15.5, 6.120

Hint : same as 3.21

3.19 RELATIVE VELOCITY

When we consider the motion of a particle, we assume a fixed point relative to which the given particle is in motion. For example, if we say that water is flowing or wind is blowing or a person is running with a speed ‘v’, we mean that these all are relative to the earth.(Which we have assumed to be fixed)

PHYSICS - I A

c) If two bodies A and B are moving with ve-locities v_A and v_B in mutually perpendicular direc-tions, then the magnitude of velocity of A relative to

B is given by 2 2

AB A B

v  v v .

The direction of v_AB with v_A is B A

v tan

v  

d) If two bodies A and B are moving with ve-locities v_A and v_B, and  is the angle between the velocities, then the magnitude of velocity of A rela-tive to B is given by

2 2

AB A B A B

v  v v 2v v cos

and the direction of v_AB with v_A is given by









B A B v sin 180 Tan v v cos 180        B A B v sin Tan v v cos     

Differentiating equation (1) with respect to time

 

_PS|  _PS

 

_{S S}| d d d v v v dt dt dt  |   | PS PS S S a a a ... (2)

Application 3.10: If two bodies A and B are

mov-ing with accelerations_a_{A and a}_Bwith respect to ground the acceleration of A relative to B is

     AB A B a a a 2 2 AB A B A B a  a a 2a a cos q 

Where  is the angle between a_A and a_B.

Application 3.11:

Relative Motion on a moving train :

If a boy is running with velocity

BT

V relative to train and train is moving with velocity V_TG relative to ground , then the velocity of the boy relative to ground

BG

V will be given by:

BG BT TG

V V V

So, if the boy is running along the direction of train

V_BG = V_BT + V_TG

If the boy is running on the train in a direction opposite to the motion of train, then

V_BG = V_BT – V_TG

Problem : 3.23

An object A is moving with 5 m/s and B is moving with 20 m/s in the same direction. (Positive x-axis). Find a) velocity of B relative to A. b) Velocity of A relative to B. Sol. a) B ˆ V 20i m/s ; V_A5iˆm/s ; B A ˆ V V 15im/s b) V_B20im / s, Vˆ _A5iˆm/s ; V_ABV_AV_B = – 15 ˆi m/s Problem : 3.24

Two object A and B are moving each with velocities 10 m/s. A is moving towards East and B is moving towards North from the same point as shown. Find

velocity of A relative to B (V )_AB . Sol: 450 4₅ 0 V_AB  E N W S V_A O V_B      AB A B V V V = 10i – 10j 2 2 1 10 10 10 2 AB V _{  } ms along SE Problem : 3.25

Find the speed of two objects if when they move unifomly towards each other, they get 4.0 metres closer each second and when they move uniformly in the same di-rection with original speeds, they get 4.0 metres closer each 10 sec.

Sol. Let their speeds be v₁ and v₂ and let v₁ > v₂ .

In Fir st Case:

Relative velocity, v₁ + v₂ = 4 ... (1)

In Second Case:

Relative velocity, v₁ – v₂ = 4

10= 0.4 ... (2)

Solving eqns. (1) and (2), we get v₁ = 2.2 m s–1, v₂ = 1.8 m s–1

Problem : 3.26

A person walks up a stationary escalator in time t₁. If

he remains stationary on the escalator, then it can take

him up in time t₂. How much time would it take him to

ELEMENTS OF VECTORS

Sol. Let L be the length of escalator. Speed of man w.r.t.

escalator is _me 1 L v t  . Speed of escalator v_e = 2 L t .

Speed of man with respect to ground would be

m me e 1 2 1 1 v v v L t t      _  _  

 The desired time is 1 2

m 1 2 t t L t v t t    . Problem : 3.27

Two ships A and B are 10km apart on a line running south to north. Ship A farther north is streaming west at 20 km/h and ship B is streaming north at 20 km/h. What is their distance of closest approach and how long do they take to reach it ?

Sol. _N E A V B V W S 450 20 2 / BA V  km h 20 / A V  km h 20 /   B V km h 450 B A C AB V Velocity of B relative to A is V_BA V_BV_A

 

2

 

2 BA V  20  20 = 20 2 km/h

i.e., V_BA is 20 2 km/h at an angle of 450 _{from east} towards north.

Condition A is at rest and B is moving with V_BA in the direction shown in figure.

Therefore, the minimum distance between the two is S_min = AC = AB sin 450 ₌ 10 1

2 km    

  = 5 2 km

and the time is

BA BC 5 2 1 t 4 20 2 V     h= 15 min

3.20MOTION OF A BOAT IN THE LINE OF RIVER FLOW

Suppose a boat moves relative to water with velocity VBW



and water is following relative to ground with velocity

WG

V , velocity of boat relative to ground V_BGwill be given by : V_BGV_BWV_WG

a) If the boat is moving along the direction of flow of water (down stream)

A d B

VBW

VWG

V_BG = V_BW + V_WG

The time taken for the boat to move to a dis-tance ‘d’ along the direction of flow of water is

1 BW WG d t V V   ...(1)

b) If the boat is moving opposite to the direc-tion of flow of water (up stream)

A d B

V_BW

VWG

V_BG = V_BW – V_WG

The time taken for the boat to move to a distance ‘d’ opposite to the direction of flow of water is 1 BW WG d t V V   ...(2)

From equations (1) and (2) 1 BW WG

2 BW WG V V t t V V    c) If the engine of boat is switched off , then the boat will be carried by the river current.

V_BG = V_WG (V_BW = 0)

The time taken for travel a distance ‘d’ is

3 WG d t V 

time taken by person to go down streams a dis-tance ‘d’ and comes back

d u Tt t BW WG BW WG d d T V V V V     1 2 2 1Gmm 2 d 

PHYSICS - I A b) Shortest Time : d WG V BW V BG V  A C B q x

To cross the river in shortest time the boat should be rowed along the normal to flow of water (or) by making an angle of 900 with the flow of water The minimum time taken to cross the river is

min BW d t V  .... (5) [from (4)   ]0

The velocity of boat with respect to ground is

2 2

BG BW WG

V  V V ... (6)

In this case the boat reaches the other bank at point ‘C’, due to the flow of water.

The displacement of boat along the direction of flow of water (or) drift is given by

x = V_WG (t_min) ,

_

_

WG BW d x V V    _{ }   .... (7)

The displacement of the boat is

2 2

s d x

Note 3.16 :

The time can be obtained by dividing the distance in a particular direction by velocity in that direction. BW WG BG AB BC AC t V V V   

c) Suppose the boat starts at point ‘A’ on one bank with velocity _V_BW and reaches the other bank at point ‘D’ as shown in the diagram.

d VBW  BE V A D B q C x BW V cosq BW V sinq V_WG

3.21 MOTION OF A BOAT ACROSS A RIVER a) To cross the river over the shotest distance:

d WG V BW V VBG  A B 900 q C

Suppose a boat starts from a point ‘A’ on one bank of the river of width ‘d’. To cross the river over the shortest distance, the boat should move by mak-ing an angle _q with the normal to the flow of water as shown in above figure.

Let _V_WG is the velocity of the water with re-spect to ground, VBW



is the velocity of boat in still water (or) velocity of boat relative to water.

 Velocity of boat with respect to ground is given by _V_{BG V}_BWV_WG

 from the triangle ABC

2 2 BG BW WG V  V V ... (1) and BG BW V cos V   ... (2) WG BW V sin V   ... (3)

To cross the river in shortest path the angle made by the velocity of boat with the flow of water is 90 +  = 90 + sin–1 WG BW V V      

The component of velocity of boat normal to the flow of water is = V_BW cos .

The time taken for the boat to cross the river is

BW d t V cos  ... (4) BG BG BW BW d d t V V V V          2 2 BW BG d t V V   ... (5) [ from (1)]

ELEMENTS OF VECTORS

The component of velocity of boat parallel to the flow of water is VBW



The component of velocity of boat normal to the flow of water is of V Cos_BW

 The time taken for the boat to cross the river is

BW d t V Cos  

The horizontal displacement of the boat (or) drift is



WG BW



x V V sin t



WG BW



BW d x V V sin V Cos    

Note 3.17 : i) If V_WG > V_{BWsin  , the boat reaches} the other end of the river to the right of B.

ii) If V_WG < V_{BWsin  , the boat reaches the} other end of the river to the left of B.

iii) If V_WG = V_{BWsin  , the boat reaches the} exactly opposite point on the other bank. i.e, at B.

Problem : 3.28

A boat is moving with a velocity v_bw = 5 km/hr relative to

water. At time t = 0, the boat passes through a piece of cork floating in water while moving downstream. If it

turns back at time t₁=30 min,

a) when the boat meet the cork again ?

b) The distance travelled by the boat during this time.

Sol. t = 0 B A C vw vw

Liet AB=d is the distance travelled by boat along down stream, in ‘t₁’ sec and it returns back and it meets the cork at point C after ‘t₂’ sec.

 Let AC=x is the distance travelled by the cork dur--ing (t₁ + t₂) sec.



B w



1 d V V t ... (1)



B w



2 dx V V t ... (2) and xV_w



t₁t₂



... (3)

Substitute (1) and (3) in (2) we get t₁ = t₂

 The boat meets the cork again after T = 2t₁ = 60 min

and the distance (AB+BC) travelled by the boat before meets the cork is

D = 2d – x



B W



1 W 1 D2 V V t V 2t B 1 W 1 W 1 D2V t 2V t 2V t B 1 30 D 2V t 2 5 5km 60      Problem : 3.29

A man wishe 00000s to cross a river flowing with velocity u jumps at an angle  with the river flow. Find the velocity of the man with respect to ground if he can swim with speed v. Also find how far from the point directly opposite to the starting point does the boat reach the opposite bank. In what direction does the boat actually move. If the width of the river is d. Sol. q f A B C v u d x C

v

MG MW WG V V V   

and  is the angle between VMG  and VWG  2 2 MG V V u 2Vu cos      drift =



u V cos



d V sin    V sin tan u V cos      Problem : 3.30

A swimmer crosses a flowing stream of width ‘d’ to and

fro in time t₁. The time taken to cover the same

dis-tance up and down the stream is t₂. If t₃ is the time the

swimmer would take to swim a distance 2d in still wa-ter, then

Sol. Let v be the river velocity and u the velocity of

swim-mer in still water. Then

1 2 2 d t 2 u v         ... (i)

PHYSICS - I A 2 _{2 2} d d 2ud t u v u v u v       .... (ii) and t₃ =2d u ... (iii)

from equations (i), (ii) and (iii)

2

1 2 3 1 2 3

t t t t  t t

Problem : 3.31

Two persons P and Q crosses the river starting from point A on one side to exactly opposite point B on the other bank of the river. The person P crosses the river in the shortest path. The person Q crosses the river in shortest time and walks back to point B. Velocity of river is 3 kmph and speed of each boat is 5 kmph w.r.t river. If the two persons reach the point B in the same time, then the speed of walk of Q is

Sol : For per son (P) : For person (Q) :

C B A V_B W V C B A VB x   W V P 2 2 2 2 B W d d d t 4 5 3 V V      Q P Q B d d t , t t t V 5      man d d x , 4  5V But W B d x V V  W B man V d d d , 4  5V V

 

man d d 3d 4 5 5 V      man 1 1 3 , 45 5V _man 1 3 20 5V

  

man 3 20 V 12kmph 5   *Problem : 3.32

A man in a boat attempts to cross a river from point A. If he rows the boat perpendicular to the banks, he reaches point C at a distance s = 120m down stream, from point B, in 10 minutes shown in figure. If the man

rows his boat at an angle  to the straight line AB he

reaches the point B in 12.5 minutes. Find a) the width of the river, (b) velocity of the boat relative to the water, (c)

velocity of the curent and (d) the angle  . Assume the

velocity of the boat relative to the water to be constnat and of the same magnitude in both cases.

Sol : Let the velocity of the boat with respect to water be u,

the velocity of the water current with respect to the banks is v, and the width of the river '_'.

A

B C

u _a

v

When boat is rowed perpendicular to the banks the re-sultant velocity of the boat with respect to the banks is along AC. The boat moves along the river with velocity v. Suppose the boat moves a distnace BC = s along the river in time t₁.

Then BC = s = vt₁ = 120m...(1)

Exactly in this time t₁ boat moves across the river and travels a distnace'_', (width of the river), with velocity u.

Then _ = ut₁ ...(2)

Now we consider the case when the boat moving with velocity u makeing an angle '' with the line AB. The relative velocity of the boat with respect to the bank along the river is v – u sin = 0,

or  v = u sin ...(3)

The relative velocity of the boat with respect to the bank and perpendicular to it is ucosa. If the time taken to reach the point B in the case is t₂

then (ucos)t₂ = _ ... (4) From (1), (2), (3) & (4) t₁ = s/v; t₂ = _/ucos

sin = s/_; cos= t₁/ t₂

We know that sin2_{ + cos}2_{ = 1. From this identity}





2





2

1 2

s /  t / t 1 Solving for '' we get

 = st₂ 2 2

2 1

t t and substituting the values s = 120 m, t₁ = 10 minutes, t₂ = 12.5 minutes, we get  = 200m.

using (1) and (2) we get u = 19.2 mmin; v = 12 m/min and  sin–1_{(v/u) = sin}–1_(12/19.2)