{"id":2469,"date":"2023-05-27T14:05:13","date_gmt":"2023-05-27T14:05:13","guid":{"rendered":"https:\/\/vartmaaninstitutesirsa.com\/?page_id=2469"},"modified":"2024-03-12T07:25:49","modified_gmt":"2024-03-12T07:25:49","slug":"chapter-4-motion-in-a-plane","status":"publish","type":"page","link":"https:\/\/vartmaaninstitutesirsa.com\/index.php\/chapter-4-motion-in-a-plane\/","title":{"rendered":"Chapter 4 Motion in a Plane"},"content":{"rendered":"

Chapter 4 Motion in a Plane<\/h1>\n

Motion in a Plane : Scalar and vector quantities; position and displacement vectors, general vectors and their notations; equality of vectors, multiplication of vectors by a real number; addition and subtraction of vectors, Unit vector; resolution of a vector in a plane, rectangular components, Scalar and Vector product of vectors. Motion in a plane, cases of uniform velocity and uniform acceleration- projectile motion, uniform circular motion<\/p>\n

To download free class 11 physics notes of Chapter 4 Motion in a Plane click on the link given below<\/h2>\n

Chapter 4 Motion in a Plane<\/a><\/h2>\n

 <\/p>\n

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2(C) PROJECTILE MOTION:-<\/span><\/strong><\/p>\n

18 PROJECTILE MOTIONS<\/span><\/strong><\/p>\n

A projectile is the name given to a body which once thrown upward with some initials velocity moves only under the effect of gravity alone.<\/span><\/em><\/p>\n

The path followed by the particle is called trajectory.<\/span><\/em><\/strong><\/p>\n

E,g:- (i) A bullet fired from a rifle.<\/span><\/p>\n

(ii)- A stone thrown horizontally from the top of a building.\u00a0<\/span><\/p>\n

QUESTION 16:\u00a0<\/span><\/strong>What do you mean by principal of physical independence of motions?<\/span><\/em><\/strong><\/p>\n

In the absence of air a particle projected followed two motions, motion in horizontal direction and motion in vertical direction.<\/span><\/p>\n

Both the motion in horizontal direction & in vertical direction are independent on each other.<\/span><\/p>\n

This is called principle of physical independence.\u00a0<\/span><\/em><\/p>\n

E.g.:- if two balls dropped from the top of a building one in horizontal direction & one in vertical direction will reach the ground in same time but with different positions.<\/span><\/p>\n

19. ASSUMPTION USED IN PROJECTILE MOTION:-<\/span><\/strong><\/p>\n

(I)<\/span>there is no air resistance on the projectile.<\/span><\/p>\n

(ii)<\/span>The effect due to curvature of earth is negligible.<\/span><\/p>\n

(iii) The effect due to rotation of earth is negligible.<\/span><\/p>\n

(iv) The acceleration due to gravity is constant for throughout motion.<\/span><\/p>\n

20. PROJECTILE GIVEN HORIZONTAL PROJECTION:-<\/span><\/strong><\/p>\n

\"\"Suppose a body is projected horizontally with initial velocity u from a point O at a certain height h from the ground. <\/span><\/p>\n

The body is under the influence of two simultaneously independent motions.<\/span><\/p>\n

(i)- Uniform horizontal motion with velocity u.<\/span><\/p>\n

(ii)- Vertically downwards motion with acceleration g. <\/span><\/p>\n

Under the combined effect of these two motions body moves along OPA.<\/span><\/p>\n

(a)TRAJECTORY OF THE PROJECTILE:-<\/span><\/strong><\/p>\n

Let after the time t the body reaches the point p (x<\/span>,<\/sub><\/span>\u00a0y).<\/span><\/p>\n

The horizontal distance covered by the body in t time is\u00a0<\/span>\"\"or<\/span>\"\"<\/p>\n

The vertical distance travelled by the body in t time is given by\u00a0<\/span><\/p>\n

\"\"<\/p>\n

Or,<\/span>\"\"(initially vertical velocity = 0)<\/span><\/p>\n

Or,<\/span>\"\"<\/p>\n

Or<\/span>\"\"\"\"<\/p>\n

Or<\/span>\"\"<\/p>\n

So<\/span>\"\"<\/p>\n

This is the eq<\/span>n<\/sup><\/span>\u00a0of parabola. Hence the particle will follow parabolic path.<\/span><\/p>\n

(a)TIME OF FLIGHT:-<\/span><\/strong><\/p>\n

The time for which a particle or projectile remains in space is called time of flight. Let t be the time of flight.<\/span><\/em><\/p>\n

For the vertical downward motion we have\u00a0<\/span><\/p>\n

\"\"<\/p>\n

Or\u00a0<\/span>\"\"<\/p>\n

\uf0e8<\/span>\"\"2\u00a0<\/sup><\/span>= S<\/span><\/p>\n

Or<\/span>T<\/span>2\u00a0<\/sup><\/span>=<\/span>\"\"<\/p>\n

Or<\/span>\"\"\"\"( s= height = h)<\/span><\/p>\n

(b)HORIZONTAL RANGE:-<\/span><\/strong><\/p>\n

It is the horizontal distance covered by the projectile during its time of flight<\/span><\/em>. It is equal to OA = R.<\/span><\/p>\n

R= horizontal velocity x time of flight<\/span><\/p>\n

So\u00a0<\/span>\"\",<\/span><\/p>\n

VELOCITY OF THE PROJECTILE AT ANY INSTANT:-<\/span><\/strong><\/p>\n

Let v is the velocity of projectile at any instant of time.<\/span><\/p>\n

It may be divided into two components.<\/span><\/p>\n

    \n
  1. \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Horizontal component \"\"<\/li>\n
  2. \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Vertical component <\/span>\"\"<\/li>\n<\/ol>\n

    So, the resultant velocity at point P is<\/span><\/strong><\/p>\n

    \"\"<\/p>\n

    Or<\/span>\"\"<\/p>\n

    If the velocity v makes an angle \u03b2 with the horizontal than\u00a0<\/span><\/p>\n

    \"\"\"\"\"\"<\/p>\n

    \uf0e8<\/span>\u03b2= tan<\/span><\/strong>-1<\/sup><\/span><\/strong>(<\/span><\/strong>\"\"\u00a0)<\/span><\/strong><\/p>\n

    21. PROJECTILE GIVEN ANGULAR PROJECTION:-Projectile fired at an angle \u0472 with the Horizontal.:-<\/span><\/strong><\/p>\n

    Suppose a body is projected with initial velocity u making an angle \u0472 with the horizontal. <\/span><\/p>\n

    This velocity has two rectangular components<\/span><\/p>\n

      \n
    1. \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ucos\u03b8 along horizontal x axis.<\/li>\n
    2. \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 usin \u03b8 along vertical y axis.<\/li>\n<\/ol>\n

      Under the combined effect of above two components the body will follow the OAB path. As shown in fig.<\/span><\/p>\n

      EQUATION OF TRAJECTORY OF A PROJECTILE:-<\/span><\/strong><\/p>\n

      Supose at time t the body reaches the point P (x, y). <\/span><\/p>\n

      The horizontal distance covered by the body is<\/span><\/p>\n

      \"\"<\/p>\n

      \"\"Or<\/span>\"\"<\/p>\n

      Similarly vertical distance covered by the body in t time is<\/span><\/p>\n

      \"\"<\/p>\n

      Or<\/span>\"\"\u00a0\u00d7\u00a0<\/span>\"\"<\/p>\n

      \"\"<\/p>\n

      Or<\/span>\"\"<\/p>\n

      Which is also a trajectory of parabolic hence particle given angular projection will follow parabolic path.<\/span><\/p>\n

      TIME OF MAXIMUM HEIGHT:-<\/span><\/strong><\/p>\n

      The time taken by the projectile to reach the maximum height is called time of maximum height.<\/span><\/p>\n

      As at the highest point vertical component of the velocity<\/span>\"\"<\/p>\n

      So,<\/span>\"\"<\/p>\n

      \"\"<\/p>\n

      Or<\/span>\"\"\"\"<\/p>\n

      TIME OF FLIGHT:-<\/span><\/strong><\/p>\n

      It is the total time for which a projectile remains in air. Net vertical displacement covered by the projectile during time of flight= 0.<\/span><\/p>\n

      As\u00a0 <\/span>\"\"<\/p>\n

      \"\"<\/p>\n

      \"\"<\/p>\n

      Comparing\u00a0<\/span>\"\"\u00a0&\u00a0<\/span>\"\"<\/p>\n

      We get\u00a0 \u00a0<\/span>\"\"<\/p>\n

      MAXIMUM HEIGHT OF THE PROJECTILE:-<\/span><\/strong><\/p>\n

      The maximum vertical distance covered by the projectile is called maximum height of the projectile.<\/span><\/p>\n

      At the vertical height A, vertical component of velocity = 0<\/span><\/p>\n

      So,\u00a0<\/span>\"\"<\/p>\n

      \"\"<\/p>\n

      Or<\/span>\"\"<\/p>\n

      HORIZONTAL RANGE (R):-<\/span><\/strong><\/p>\n

      It is the Horizontal distance travelled by the projectile during its time of flight. So,\u00a0<\/span><\/p>\n

      \"\"<\/p>\n

      \"\"<\/p>\n

      \"\"<\/p>\n

      Or<\/span>\"\"\"\"<\/p>\n

      * Condition for maximum horizontal range\u00a0<\/span><\/p>\n

      As\u00a0<\/span>\"\"<\/p>\n

      R will be maximum if\u00a0<\/span>\"\"<\/p>\n

      \"\"<\/p>\n

      Thus if a projectile fired at 45<\/span>0<\/sup><\/span>will have maximum horizontal range & the maximum horizontal range can be given as\u00a0<\/span><\/p>\n

      \"\"<\/p>\n

      22. TWO ANGLE OF PROJECTION FOR THE SAME HORIZONTAL RANGE:-<\/span><\/strong><\/p>\n

      \"\"The Horizontal range of a projectile projected at an angle<\/span>\"\"\u00a0with the horizontal with velocity u is given by\u00a0<\/span><\/p>\n

      R =\u00a0<\/span>\"\"<\/p>\n

      Replacing\u00a0<\/span>\"\"<\/p>\n

      \"\"<\/p>\n

      R<\/span>\u2019<\/sup><\/span>\u00a0=\u00a0<\/span>\"\"=\u00a0<\/span>\"\"<\/p>\n

      I,e.<\/span>R = R<\/span>\u2019<\/sup><\/span><\/p>\n

      Hence for a given velocity of the projection, a projectile has same horizontal range for the angle of projection\u00a0<\/span>\"\"\u00a0& (90 \u2013<\/span>\"\").\u00a0<\/span><\/strong>Its value is maximum for<\/span>\"\"=45<\/span>0<\/sup><\/span><\/p>\n

      QUESTION 16:\u00a0<\/span><\/strong>A ball is thrown from a field with a speed of 12.0 m\/s at an angle of 45\u00b0 with the horizontal.\u00a0<\/strong><\/p>\n

      At what distance will it hit the field again? Take g = 10.0 m\/s.<\/strong><\/p>\n

      Solution\u00a0 : The horizontal range<\/p>\n

      \"\"<\/p>\n

      Thus, the ball hits the field at \"\"\u00a0from the point of projection. \u03b8<\/p>\n

      23. VELOCITY OF PROJECTILE AT ANY INSTANT:-<\/span><\/strong><\/p>\n

      Suppose the projectile has velocity v at the time t at point P (x<\/span>1<\/sub><\/span>y).<\/span><\/p>\n

      The<\/span>velocity v has two rectangular components\u00a0<\/span>\"\"&\u00a0<\/span>\"\".<\/sub><\/span><\/strong><\/p>\n

      \"\"<\/p>\n

      The resultant velocity\u00a0<\/span><\/strong><\/p>\n

      \"\"\"\"<\/p>\n

      Or\u00a0<\/span>\"\"\"\"<\/p>\n

      If the velocity v make angle \u03b2 with the vertical, then<\/span><\/p>\n

      \"\"<\/p>\n

      24. VELOCITY OF THE PROJECTILE AT THE END POINT:-<\/span><\/strong><\/p>\n

      At the end of flight<\/span><\/p>\n

      \"\"<\/p>\n

      So the resultant velocity is<\/span><\/p>\n

      \"\"<\/p>\n

      =<\/span>\"\"\u00a0= u<\/span><\/p>\n

      Also\u00a0<\/span>\"\"<\/p>\n

      \"\"<\/p>\n

      Or<\/span>\"\"<\/p>\n

      Here \u2013ve sign shows that the direction is downward.\u00a0<\/span><\/p>\n

      Hence Projectile returns to ground at the same angle at which it is fired.<\/span><\/p>\n

      QUESTION 17: A football is kicked with a velocity of 20 m\/s at an angle of 45\u00b0 with the horizontal. <\/span><\/strong><\/p>\n

      (a) Find the time taken by the ball to strike the ground. <\/span><\/strong><\/p>\n

      (b) Find the maximum height it reaches. <\/span><\/strong><\/p>\n

      (c) How far away from the kick does it hit the ground ? Take g = 10 m\/s<\/span><\/strong>2.\u00a0<\/span><\/strong><\/p>\n

      Solution :<\/span>(a) The initial velocity has the components<\/span><\/p>\n

      \"\"<\/p>\n

      And<\/span>\"\"<\/p>\n

      When the ball reaches the ground, y = 0.\u00a0<\/span><\/p>\n

      Using<\/span>\"\"<\/p>\n

      \"\"<\/p>\n

      Or<\/span>\"\"<\/p>\n

      Thus, it takes 2.8 s for the football to fall on the ground.\u00a0<\/span><\/p>\n

        \n
      1. At the highest point \"\".<\/li>\n<\/ol>\n

        Using the equation\u00a0<\/span>\"\"<\/p>\n

        \"\"<\/p>\n

        or,<\/span>\"\"<\/p>\n

        Thus, the maximum height reached is 10 m.\u00a0<\/span><\/p>\n

        (c) The horizontal distance travelled before falling to the ground<\/span><\/p>\n

        \u00a0<\/span>\"\"<\/p>\n

        \n

        1<\/strong><\/p>\n<\/div>\n<\/div>\n

        Scalar and<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>Vectors<\/span><\/em><\/strong><\/p>\n

        25. PHYSICAL QUANTITY\u00a0<\/span><\/em><\/strong><\/p>\n

        In the study of the two dimensional motion of the physical quantity we need sometime both magnitude and the direction of the quantity so introduced a new term vector.<\/span><\/em><\/p>\n

        A physical quantity may be divided into two types\u00a0<\/span><\/p>\n

        Scalar Quantity:-<\/span><\/strong><\/p>\n

        Physical quantities that have only magnitude and no direction are called scalar quantities or scalars. Eg: mass, time, speed etc.<\/span><\/p>\n

        Vector Quantity:-<\/span><\/strong><\/p>\n

        Physical quantities that have both magnitude and direction are called vector quantities or vectors. Eg: displacement, weight, velocity, force etc.<\/span><\/p>\n

        Difference between scalar and vector quantity<\/span><\/em><\/strong><\/p>\n

        Scalar quantity<\/span><\/em><\/strong><\/p>\n

        1.Scalar quantity have only magnitudes<\/span><\/p>\n

        2.Scalar quantity changes when magnitude of the quantity changes\u00a0<\/span><\/p>\n

        3.Scalar can be added by the ordinary laws of the algebra<\/span><\/p>\n

        Vector quantity<\/span><\/p>\n

        1. Vector quantity have both magnitude and direction<\/span><\/p>\n

        2. Vector quantity changes when either magnitude or direction changes<\/span><\/p>\n

        3. Vectors can be added by only special laws of<\/span>\u00a0\u00a0<\/span>vector<\/span>\u00a0\u00a0<\/span>addition<\/span><\/p>\n

        25. .<\/span>\u00a0\u00a0<\/span>Representation of vectors<\/span><\/strong>\u00a0<\/span><\/p>\n

        A vector is represented by using a straight line with an arrowhead at one end. The length of the line represents the magnitude of the vector and the arrowhead gives its direction.<\/span><\/p>\n

        \"\"<\/p>\n

        \u00a0<\/span><\/p>\n

        26.. Types of vectors\u00a0<\/span><\/strong><\/p>\n

        Equal vectors<\/span><\/strong><\/p>\n

        Vectors are said to be equal if they have the same magnitude and same direction<\/span><\/em>.\u00a0<\/span><\/p>\n

        \"\"<\/span>\u00a0<\/span><\/p>\n

        Negative vector<\/span><\/strong>\u00a0<\/span><\/strong><\/p>\n

        A vector having the same magnitude but direction opposite to that of a given vector is called negative vector of the given vector.<\/span><\/p>\n

        \"\"<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

        .<\/span>Unit vector<\/span><\/strong>\u00a0<\/span><\/strong><\/p>\n

        A vector whose magnitude is unity (one) is called a unit vector. It is represented as<\/span>\"\"<\/p>\n

        \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>As<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

        Unit vector have direction as same as its parent atom.<\/span><\/p>\n

        Orthogonal unit vectors:-<\/span><\/strong>\u00a0<\/span><\/p>\n

        three mutually perpendicular vectors<\/span>\u00a0\u00a0<\/span>\"\"\u00a0are called orthogonal unit vectors.<\/span><\/p>\n

        These vectors may be represented as<\/span>\u00a0\u00a0\u00a0<\/span><\/p>\n

        \u00a0<\/span>\"\"\u00a0,<\/span>\"\"\u00a0,<\/span>\"\"<\/p>\n

        Zero vector or null vector:-<\/span><\/strong><\/p>\n

        It is a vector, which has zero magnitude and an arbitrary direction.<\/span><\/p>\n

        It is represented by<\/span>\"\"\u00a0.<\/span><\/p>\n

        \uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>When a vector is added to zero vector then we get the same vector\u00a0<\/span><\/p>\n

        As\u00a0<\/span>\"\"<\/p>\n

        \uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>When a real number or a vector is multiplied to zero vectors then we get the zero vectors.<\/span>\"\"<\/p>\n

        \uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>If\u00a0<\/span>\"\"\u00a0and \u03bc are two none zero real numbers then the relation\u00a0<\/span>\"\"\u00a0can hold only when both\u00a0<\/span>\"\"\u00a0are zero vectors.<\/span><\/p>\n

        Polar vector:-<\/span><\/strong><\/p>\n

        The vectors which have starting point are called polar vectors like displacement vector, force etc.<\/span><\/p>\n

        Axial vector<\/span><\/strong>:-<\/span><\/p>\n

        \u00a0<\/span>The vectors which represent the rotational effect and act always along the axis of rotation in accordance with the right hand screw rule are called axial vector.<\/span><\/p>\n

        For example angular velocity, torque, angular momentum etc.<\/span><\/p>\n

        Position Vector<\/span><\/strong>:<\/span><\/strong><\/p>\n

        \"C:\\Users\\vartmaan\\AppData\\Local\\Microsoft\\Windows\\TemporaryThe position vector is used to specify the position of a certain body. The position vector of an object is measured from the origin, in general.\u00a0<\/span><\/p>\n

        Suppose an object is placed in the space as shown:\u00a0<\/span><\/p>\n

        Position vector<\/span>\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

        Where,\u00a0<\/span>\"\"= unit vector along x-direction,<\/span><\/p>\n

        \u00a0<\/span>\"\"= unit vector along y-direction<\/span><\/p>\n

        \"\"= unit vector along z-direction<\/span><\/p>\n

        So if an object is at a certain point P (say) at a certain time, its position vector is given as described above.<\/span><\/p>\n

        Displacement Vector:<\/span><\/h3>\n

        \"C:\\Users\\vartmaan\\AppData\\Local\\Microsoft\\Windows\\TemporaryThe change in the position vector of an object is known as the displacement vector. <\/span><\/p>\n

        Suppose an object is at point A at time = 0 and at point B at time = t. <\/span><\/p>\n

        The position vectors of the object at point A and at point B are given as:<\/span><\/p>\n

        Position vector at point\u00a0<\/span>\"\"<\/p>\n

        \"\"=<\/span>\"\"<\/p>\n

        Position vector at point\u00a0<\/span>B is\u00a0<\/span><\/em><\/p>\n

        \u00a0\u00a0<\/span><\/em>\"\"\u00a0<\/span><\/p>\n

        Now, the displacement vector AB of the object will be:<\/span><\/p>\n

        \"\"\u00a0)<\/span><\/p>\n

        Or<\/span>\u00a0\u00a0<\/span>\"\"<\/p>\n

        The displacement of an object can also be defined as the vector distance between the initial point and the final point. <\/span><\/p>\n

        Suppose an object travels from point A to point B in the path shown in the black curve:<\/span><\/p>\n

        27. Addition of vectors<\/span><\/strong>.<\/span><\/p>\n

          \n
        1. \u00a0\u00a0<\/span>\u00a0\u00a0<\/strong>When the vectors are in the same direction<\/em><\/strong>.<\/em><\/li>\n<\/ol>\n

          \"\"Here the magnitude of the resultant vector is equal to the sum of the magnitudes of the two vectors and direction is same as that of the two vectors.<\/span><\/p>\n

          \u00a0<\/span><\/strong><\/p>\n

            \n
          1. \u00a0\u00a0<\/span>\u00a0<\/strong>When the vectors are in opposite directions<\/em><\/strong>.<\/em><\/li>\n<\/ol>\n

            \"\"Here the magnitude of the resultant is the difference in magnitude of the two vectors and the direction is same as that of the bigger vector.\u00a0<\/span><\/p>\n

            \u00a0<\/span><\/p>\n

              \n
            1. \u00a0\u00a0<\/span>\u00a0<\/strong>When the vectors are inclined at an angle<\/em><\/strong>.<\/em><\/li>\n<\/ol>\n

              When the vectors are inclined at an angle, the sum of vectors can be determined using two methods:<\/span><\/p>\n

              \u00a0<\/span>(a)<\/span>\u00a0\u00a0<\/span>Triangle law of vectors:<\/span><\/strong>\u00a0\u00a0<\/span><\/strong><\/p>\n

              \u201c<\/span>If two vectors are represented by the two sides of a triangle, both in magnitude and in direction, and in same order, then the third side of the triangle in the reverse order represents the vector sum of the vectors.<\/span><\/em>\u201d<\/span><\/p>\n

              I.e.<\/span>\u00a0\u00a0<\/span>\"\"<\/p>\n

              Magnitude of\u00a0<\/span><\/em><\/strong>\"\"<\/p>\n

              \"\"As shown in fig. AB and BC are two vectors taken in the same order and AC is the vector taken in the opposite order. <\/span><\/p>\n

              Now to obtain the magnitude of vector R draw CD perpendicular to AD then from the triangle ACD<\/span><\/p>\n

              \u00a0\u00a0<\/span>\"\"<\/p>\n

              Or<\/span>\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

              Or<\/span>\u00a0\u00a0\u00a0<\/span>\"\"\u00a0\u00a0<\/span>(1)<\/span><\/p>\n

              Now from triangle BCD\u00a0<\/span><\/p>\n

              \u00a0\u00a0<\/span>\"\"<\/p>\n

              \"\"\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

              \"\"<\/p>\n

              \"\"<\/p>\n

              Using in equation 1 we get\u00a0<\/span><\/p>\n

              \u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

              Or<\/span>\u00a0\u00a0<\/span>\"\"<\/p>\n

              Or<\/span>\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

              Or<\/span>\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

              Or<\/span>\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

              Or\u00a0<\/span>\u00a0\u00a0\u00a0<\/span><\/strong>\"\"<\/p>\n

              Which is the required expression for the magnitude of the\u00a0<\/span>\"\".<\/span><\/p>\n

              Direction of the\u00a0<\/span><\/em><\/strong>\"\"<\/p>\n

              The direction of the resultant vector can be given by the slope of the resultant or we can say by the<\/span>\"\".<\/span><\/p>\n

              As from the diagram<\/span>\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

              \"\"<\/p>\n

              Or<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

              QUESTION 18: Two vectors of equal magnitude<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>5 unit have an angle 60\u00b0 between them.<\/span><\/strong><\/p>\n

              Find the magnitude of <\/span><\/strong><\/p>\n

              (a) the sum of the vectors and <\/span><\/strong><\/p>\n

              (b) the difference of the vectors.\u00a0<\/span><\/strong><\/p>\n

              \"\"\u00a0Solution (a)\u00a0<\/span>\"\"\u00a0+<\/span>\"\"\u00a0\u00a0<\/span>is the sum of\u00a0<\/span>\"\"\u00a0and\u00a0<\/span>\"\".Both have a magnitude of5unit and the angle between them is 60\u00b0.\u00a0<\/span><\/p>\n

              Thus, the magnitude of the sum is<\/span><\/p>\n

              \"\"<\/p>\n

              Or<\/span>\u00a0\u00a0\u00a0<\/span>\"\"unit.<\/span><\/p>\n

                \n
              1. \u00a0\u00a0<\/span>\"\"– \"\"\u00a0 is the sum of \"\"\u00a0and (-\"\"\u00a0). As shown in the figure, the angle between A and (-\"\"\u00a0 ) is 120\u00b0.<\/li>\n
              2. The magnitudes of both \"\"\u00a0 and (-\"\") is 5 unit.<\/li>\n<\/ol>\n

                So<\/span>\u00a0\u00a0<\/span>\"\"–\u00a0<\/span>\"\"<\/p>\n

                \"\"<\/p>\n

                (b)<\/span>\u00a0\u00a0<\/span>Parallelogram law of vectors.<\/span><\/strong><\/p>\n

                \u201c<\/span>If two vectors are represented both in magnitude and in direction by the adjacent sides of a parallelogram, then the diagonal of the parallelogram starting from the point of intersection of the two vectors represent the vector sum of the two vectors<\/span><\/em>\u201d.<\/span><\/p>\n

                \"\"\u00a0<\/span>\"\"<\/p>\n

                suppose AB and AE are the two vectors representing as the two adjacent sides of a parallelogram. AC is the greater diagonal of parallelogram. <\/span><\/p>\n

                Now draw CD perpendicular to AD.<\/span><\/p>\n

                then from the triangle ACD\u00a0<\/span>\"\"<\/p>\n

                Or<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

                Or<\/span>\u00a0\u00a0\u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0<\/span>(1)<\/span><\/p>\n

                Now from traingle BCD<\/span>\u00a0\u00a0\u00a0<\/span><\/p>\n

                \u00a0<\/span>\"\"<\/p>\n

                \"\"\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                \u00a0<\/span>\"\"<\/p>\n

                \"\"<\/p>\n

                Using in equation 1 we get\u00a0<\/span><\/p>\n

                \u00a0\u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                Or<\/span>\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

                Or<\/span>\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

                Or<\/span>\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

                Or<\/span>\u00a0\u00a0<\/span>\"\"<\/p>\n

                Or\u00a0<\/span>\u00a0\u00a0<\/span><\/strong>\"\"<\/p>\n

                Which is the required expression for the magnitude of<\/span>\u00a0\u00a0<\/span>\"\".<\/span><\/p>\n

                Direction of the\u00a0<\/span><\/em><\/strong>\"\"<\/p>\n

                The direction of the resultant vector can be given by the slope of the resultant or we can say by the\u00a0<\/span>\"\".<\/span><\/p>\n

                As from the diagram\u00a0<\/span><\/p>\n

                \"\"<\/p>\n

                \u00a0<\/span><\/p>\n

                Or<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

                (c )Polygon law of vector addition:-<\/span><\/strong><\/p>\n

                \"https:\/\/www.studyadda.com\/upload\/html_folder\/5_Polygon_Law_of_Vector_Addition\/5_Polygon_Law_of_Vector_Addition_files\/image001.png\"According to the polygon law of vector addition, if a number of the vectors represent the sides of a open polygon taken in same order then their resultant vector is represented by the closed side of the polygon taken in the opposite order.<\/span><\/p>\n

                Suppose a polygon OABCDEO. Sides\u00a0 \"\"\u00a0<\/span>\"\",<\/span>\"\",<\/span>\"\",\u00a0<\/span>\"\"\u00a0are taken in same order and side\u00a0<\/span>\"\"\u00a0is taken in opposite order.<\/span><\/p>\n

                Than according to the polygon law of vector addition\u00a0<\/span><\/p>\n

                \"\"\u00a0<\/span>\"\"+<\/span>\"\"+<\/span>\"\"+\u00a0<\/span>\"\"<\/p>\n

                Proof:- Draw OB then apply triangle law in OAB. We get\u00a0<\/span><\/p>\n

                \u00a0\u00a0<\/span>\"\"\u00a0<\/span>\"\"..(1)<\/span><\/p>\n

                Again from triangle OBC\u00a0<\/span><\/p>\n

                We get\u00a0<\/span>\"\"+<\/span>\"\"\u00a0<\/span><\/p>\n

                Using eq.(1)<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>\"\"\u00a0=<\/span>\"\"\u00a0<\/span>\"\"\u00a0\u00a0\u00a0<\/span>\u2026 (2)<\/span><\/p>\n

                Similarly from OCD<\/span>\u00a0\u00a0\u00a0<\/span>\"\"+<\/span>\"\"\u00a0<\/span><\/p>\n

                Using eq.(2)\u00a0<\/span>\"\"\u00a0=<\/span>\"\"\u00a0<\/span>\"\"\u2026.(3)<\/span><\/p>\n

                Now from triangle ODE we get\u00a0<\/span>\"\"\u00a0<\/span>\"\"<\/p>\n

                Using eq.(3)<\/span>\"\"\u00a0<\/span>\"\"+<\/span>\"\"+<\/span>\"\"+\u00a0<\/span>\"\"<\/p>\n

                28. Scalar Product or Dot product of two vectors<\/span><\/strong>:-<\/span><\/p>\n

                Scalar product is defined as the product of the magnitude of two vectors and the cosine of the angle between them<\/span><\/em>.<\/span><\/p>\n

                If A and B are two vectors inclined at an angle \u03b8, then scalar product is given by<\/span><\/p>\n

                \"\"\u00a0.\u00a0<\/span>\"\"\u00a0=\u00a0<\/span>\"\"\u00a0<\/span>\"\"\u00a0cos\u03b8=A Bcos\u03b8\u00a0<\/span><\/p>\n

                Properties of dot products:-<\/span><\/strong><\/p>\n

                \uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>scalar product of two vectors is always a scalar.<\/span><\/em><\/p>\n

                \uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>It is commutative,\u00a0<\/span><\/em><\/p>\n

                i.e.\u00a0<\/span><\/em>\"\"\u00a0.\u00a0<\/span>\"\"=<\/span>\"\"\u00a0.\u00a0<\/span>\"\"<\/p>\n

                \uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>It is distributive,\u00a0<\/span><\/em><\/p>\n

                i.<\/span><\/em>e.<\/span>\u00a0\u00a0<\/span>\"\"\u00a0.(\u00a0<\/span>\"\"+<\/span>\"\")=<\/span>\"\"\u00a0.\u00a0<\/span>\"\"+<\/span>\"\"\u00a0.\u00a0<\/span>\"\"<\/p>\n

                \uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>The dot product of two identical vector may be given by\u00a0<\/span><\/p>\n

                \"\"\u00a0.\u00a0<\/span>\"\"=<\/span>\"\"\u00a0<\/span>\"\"\u00a0cos0=A<\/span>2<\/sup><\/span>.<\/span><\/p>\n

                \uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>The dot product of<\/span>\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

                \uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>The dot product of<\/span>\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

                Dot product in term of Cartesian coordinate system<\/span><\/strong><\/p>\n

                suppose\u00a0<\/span>\"\"\u00a0&\u00a0<\/span>\"\" are the two vectors, such that<\/span><\/p>\n

                \u00a0<\/span>\"\"\u00a0\u00a0<\/span>and<\/span>\u00a0\u00a0<\/span>\"\"\u00a0<\/span><\/p>\n

                then there dot product may be given as\u00a0<\/span><\/p>\n

                \"\"\u00a0.\u00a0<\/span>\"\"=(<\/span>\"\").(<\/span>\"\")<\/span><\/p>\n

                \u00a0<\/span>\"\"\u00a0.\u00a0<\/span>\"\"=<\/span>\"\"<\/p>\n

                \"\"\u00a0.\u00a0<\/span>\"\"=<\/span>\"\"\u00a0\u00a0\u00a0<\/span>(\u00a0<\/span>\"\")<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                \"\"\u00a0.\u00a0<\/span><\/strong>\"\"=<\/span><\/strong>\"\"(<\/span>\"\")<\/span><\/p>\n

                QUESTION 19: The work done by a force\u00a0<\/span><\/strong>\"\"\u00a0\u00a0<\/span><\/strong>during a displacement r is given by\u00a0<\/span><\/strong>\"\". <\/span><\/strong><\/p>\n

                Suppose a force of<\/span><\/strong>\u00a0<\/span><\/strong>12 N acts on a particle in vertically upward direction and the particle is displaced through<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>2.0 m<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>in vertically downward direction.\u00a0 <\/span><\/strong>Find the work done by the force during this displacement.\u00a0<\/span><\/strong><\/p>\n

                Solution :<\/span>\u00a0\u00a0<\/span>The angle between displacement r and the force<\/span>\u00a0\u00a0<\/span>F<\/span>\u00a0\u00a0<\/span>is 180\u00b0.\u00a0<\/span><\/p>\n

                \u00a0<\/span>Thus, the work done is<\/span><\/p>\n

                \u00a0<\/span>\"\"<\/p>\n

                \"\"<\/p>\n

                \"\"<\/p>\n

                QUESTION 20: The vector\u00a0<\/span><\/strong>\"\"\u00a0\u00a0\u00a0<\/span><\/strong>has a magnitude of<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>5 unit,\u00a0<\/span><\/strong>\"\"\u00a0\u00a0<\/span><\/strong>has a magnitude of<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>6 unit and the cross product of\u00a0<\/span><\/strong>\"\"\u00a0and has a magnitude of<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>15 unit. <\/span><\/strong><\/p>\n

                Find the angle between\u00a0<\/span><\/strong>\"\"\u00a0and\u00a0<\/span><\/strong>\"\".\u00a0<\/span><\/strong><\/p>\n

                Solution :<\/span>\u00a0\u00a0<\/span>If the angle between<\/span>\u00a0\u00a0<\/span>\"\"\u00a0and\u00a0<\/span>\"\"is 9, the cross product will have a magnitude\u00a0<\/span><\/p>\n

                \"\"\u00a0\u00d7\u00a0<\/span>\"\"<\/p>\n

                \"\"<\/p>\n

                \"\"<\/p>\n

                \"\"<\/p>\n

                Question21: suppose\u00a0<\/span><\/strong>\"\"\u00a0&\u00a0<\/span><\/strong>\"\"\u00a0are the two vectors, such that\u00a0<\/span><\/strong>\"\"\u00a0\u00a0<\/span><\/strong>and<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>\"\"\u00a0<\/span><\/strong>then obtain there dot product.<\/span><\/strong><\/p>\n

                Question22: suppose\u00a0<\/span><\/strong>\"\"\u00a0&\u00a0<\/span><\/strong>\"\"\u00a0are the two mutually perpendicular vectors, <\/span><\/strong>such that\u00a0<\/span><\/strong>\"\"\u00a0\u00a0<\/span><\/strong>and<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>\"\"\u00a0then obtain value of n.<\/span><\/strong><\/p>\n

                29.\u00a0<\/span>Vector Product of two vectors or Cross Product of two vectors:-<\/span><\/strong><\/p>\n

                Vector product is defined as the product of the magnitude of the two vectors and the sine of the angle between them.<\/span><\/em><\/p>\n

                \u00a0<\/span><\/em>If A and B are two vectors inclined at an angle \u03b8, then vector product is given by<\/span><\/p>\n

                \"\"\u00a0\u00d7\u00a0<\/span>\"\"\u00a0=\u00a0<\/span>\"\"\u00a0<\/span>\"\"sin \u03b8<\/span>\"\"\u00a0= ABsin\u03b8<\/span>\"\"<\/p>\n

                The direction of vector product is given by unit vector\u00a0<\/span>\"\"\u00a0\u00a0<\/span>perpendicular to the plane containing A and B.<\/span><\/p>\n

                \u00a0<\/span>Properties of cross products:-<\/span><\/em><\/strong><\/p>\n

                \uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>cross<\/span><\/em><\/strong>\u00a0product of two vectors is always a vector.<\/span><\/em><\/p>\n

                \uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>It is not commutative,\u00a0<\/span><\/em><\/p>\n

                i.e.\u00a0<\/span><\/em>\"\"\u00a0\u00d7\u00a0<\/span>\"\"=<\/span>\"\"\u00a0\u00d7\u00a0<\/span>\"\"<\/p>\n

                \uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>It is distributive,<\/span><\/em><\/p>\n

                \u00a0<\/span><\/em>i.<\/span><\/em>e.<\/span>\u00a0\u00a0<\/span>\"\"\u00a0\u00d7 (\u00a0<\/span>\"\"+<\/span>\"\")=<\/span>\"\"\u00a0\u00d7\u00a0<\/span>\"\"+<\/span>\"\"\u00a0\u00d7\u00a0<\/span>\"\"<\/p>\n

                \uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>The\u00a0<\/span>cross<\/span><\/em><\/strong>\u00a0product of two identical vector may be given by<\/span><\/p>\n

                \u00a0<\/span>\"\"\u00a0\u00d7\u00a0<\/span>\"\"=<\/span>\"\"\u00a0<\/span>\"\"\u00a0cos90=0.<\/span><\/p>\n

                \uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>The\u00a0<\/span>cross<\/span><\/em><\/strong>\u00a0product of<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                \u00a0\u00a0<\/span>\"\"<\/p>\n

                \uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>The\u00a0<\/span>cross<\/span><\/em><\/strong>\u00a0product of<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                \u00a0<\/span>\"\"<\/p>\n

                cross product in term of Cartesian coordinate system<\/span><\/em><\/strong><\/p>\n

                suppose\u00a0<\/span>\"\"\u00a0&\u00a0<\/span>\"\"\u00a0are the two vectors,\u00a0<\/span><\/p>\n

                such that\u00a0<\/span>\"\"\u00a0\u00a0<\/span>and<\/span>\u00a0\u00a0<\/span>\"\"\u00a0<\/span><\/p>\n

                then there\u00a0<\/span>cross<\/span><\/em><\/strong>\u00a0product may be given as\u00a0<\/span><\/p>\n

                \"\"\u00a0\u00d7\u00a0<\/span>\"\"=(<\/span>\"\") \u00d7 (<\/span>\"\")<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                \u00a0<\/span>\"\"\u00a0\u00d7\u00a0<\/span>\"\"=<\/span>\"\"<\/p>\n

                \"\"\u00a0\u00d7\u00a0<\/span>\"\"=\u00a0<\/span>\"\"\u00a0<\/span>\"\"\u00a0\u00a0<\/span>\"\"\u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0<\/span>(<\/span>\"\")\u00a0<\/span><\/p>\n

                As we know<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>(<\/span>\"\")<\/span>\u00a0\u00a0<\/span><\/p>\n

                So<\/span>\u00a0\u00a0\u00a0<\/span>\"\"\u00a0\u00d7\u00a0<\/span>\"\"=<\/span>\"\"\u00a0<\/span>\"\"\u00a0<\/span>\"\"\u00a0<\/span>\"\"\u00a0\u00a0<\/span><\/p>\n

                Or<\/span><\/strong>\u00a0\u00a0\u00a0<\/span><\/strong>\"\"\u00a0\u00d7\u00a0<\/span><\/strong>\"\"=<\/span><\/strong>\"\"\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/strong><\/p>\n

                30. Rectangular components of a vector.<\/span><\/strong><\/p>\n

                \"\"The process of breaking one vector into two or more vectors is called resolution and the vectors so obtained are called components of the vector.<\/span><\/em><\/p>\n

                Consider a vector<\/span>\u00a0\u00a0<\/span>\"\"\u00a0inclined to x axis at an angle \u03b8.\u00a0<\/span><\/p>\n

                If<\/span>\u00a0\u00a0\u00a0<\/span>\"\"\u00a0\u00a0<\/span><\/strong>and\u00a0<\/span>\"\"\u00a0\u00a0<\/span><\/strong>are the component of Vector<\/span>\u00a0\u00a0<\/span>\"\"\u00a0along x axis and y axis then by parallelogram law\u00a0<\/span><\/p>\n

                \"\"\u00a0\u00a0<\/span>\u2026\u2026\u2026(1)<\/span>\u00a0\u00a0\u00a0<\/span>here\u00a0<\/span>\"\"<\/p>\n

                So<\/span><\/em>\u00a0\u00a0\u00a0\u00a0<\/span><\/em>\"\"\u2026..(2)<\/span><\/em><\/strong><\/p>\n

                Also\u00a0<\/span><\/em>\"\"\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/em><\/p>\n

                \u00a0<\/span><\/em>\"\"\u2026.(4)<\/span><\/em><\/p>\n

                Then<\/span><\/em>\u00a0\u00a0<\/span><\/em>\"\"\u00a0\u00a0<\/span><\/em><\/p>\n

                \u00a0\u00a0<\/span><\/em>\"\"<\/p>\n

                Or<\/span><\/em>\u00a0\u00a0\u00a0\u00a0<\/span><\/em>\"\"<\/p>\n

                Squaring and adding eqn (3) & (4) we get<\/span><\/p>\n

                \"\"=<\/span>\"\"<\/p>\n

                So<\/span><\/em>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/em>\"\"\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/em><\/p>\n

                And<\/span><\/em>\u00a0\u00a0\u00a0<\/span><\/em>\"\"<\/p>\n

                \n

                1<\/strong><\/p>\n

                \u00a0<\/em><\/strong><\/p>\n<\/div>\n<\/div>\n

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                Chapter 4 Motion in a Plane Motion in a Plane : Scalar and vector quantities; position and displacement vectors, general vectors and their notations; equality of vectors, multiplication of vectors by a real number; addition and subtraction of vectors, Unit vector; resolution of a vector in a plane, rectangular components, Scalar and Vector product of […]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"_uag_custom_page_level_css":"","site-sidebar-layout":"default","site-content-layout":"default","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"set","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"var(--ast-global-color-4)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"class_list":["post-2469","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_hostinger_reach_plugin_has_subscription_block":false,"_hostinger_reach_plugin_is_elementor":false,"uagb_featured_image_src":{"full":false,"thumbnail":false,"medium":false,"medium_large":false,"large":false,"1536x1536":false,"2048x2048":false},"uagb_author_info":{"display_name":"sandeep.soni484@gmail.com","author_link":"https:\/\/vartmaaninstitutesirsa.com\/index.php\/author\/sandeep-soni484gmail-com\/"},"uagb_comment_info":0,"uagb_excerpt":"Chapter 4 Motion in a Plane Motion in a Plane : Scalar and vector quantities; 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