if the velocity v of a paticle moving along a straight line decreases linearly with its displacement from 20m/s to a value apporoaching zero at s=30m,determine the acceleration of the particle when s=15m.

consider attatched graph...

the relation b/w v & x comes to be 

v = -20/30 * x + 20

now differentiate with respect to time(t)

we get dv/dt = (-20/30)*(dx/dt)

as dv/dt = a & dx/dt= v

put the respective values in the eqn

=> a = (-20/30)v

=> a = (-20/30)*(-20/30 * x + 20)

now to find a at x= 15

put x= 15 in above eqn. we get : a = -20/3

 Given : Decreasing velocity is = -20 m/s

             Its value approaches to zero i.e. final velocity = 0 m/s

             Distance travelled = 30 m

           Acceleration a = v² - u^{2 / 2s}

                                =  0-400 / 2*30

                                 = -400/60

                                   =-40/6

                                     = -6.6 m/s²

   At s=15m a = 6.6m/s² only as velocity while decelerating changes with constant rate by which velocity remains constant.

 HOPE MY ANSWER IS RIGHT & YOU WOULD GIVE ME A POINT