CBSE Tutorials

Physics 9th: Motion | Derivation of Equations of Motion by Graphical Method

Derivation of Equations of Motion by Graphical Method

TO DERIVE v = u + at BY GRAPHICAL METHOD

This is a graph of uniform acceleration with ‘u’ as initial velocity and ‘v’ as final velocity.

    Initial velocity = u = OP

    Final velocity = v = RN

             = RQ + QN

             v = u + QN    …..(i)

    Acceleration, a = slope of line PN

                 …..(ii)

    Putting the value of QN from equation (ii) into equation (i), we get v = u + at        

TO DERIVE BY GRAPHICAL METHOD

    In the above speed-time graph, the distance travelled is given by

            Distance travelled = Area of figure OPNR

                     = Area of DPNQ + Area of rectangle OPQR

    (1)     Area of triangle PNQ =

                     =

                     =     [As v = u + at and v – u = at]

             Area of DPNQ =

    (2)    Area of rectangle OPQR = OP ´ PQ

                      = u ´ t

                      = ut    

    Distance travelled = Area of DPNQ + Area of rectangle OPQR

TO DERIVE BY GRAPHICAL METHOD

    In the above speed time graph distance travelled (S) = Area of trapezium OPNR

     (sum of parallel sides) ´ height

     (OP + RN) ´ OR

     (u + v) ´ t

     (v + u) t            …(i)

    But v = u + at

    at = v – u

                    … (ii)

    Putting this value of ‘t’ from equation (ii) into equation (i) we get that ___

     (v + u)

            [As (a + b) (a – b) = ]

    2as =