having got the permission to integrate...... i ended up in this way

the answer comes.........

but i have to consider the velocity to be constant for some small interval of time dt.

 

i realised my previous mistake.... i took accl^n along the plane something other than gsin30

let v be the velocity over time interval t to t+dt.

we have to consider that the time t is that time by which the sledge goes 120m.

let M = mass of the sand box (sledge)

So (M + 40 - 2t)v - [M + 40 - 2(t + dt) ]v = change in momentum = Force x (time interval)

or 2vdt = Fdt

or 2v = F

F = force on the sledge at time t = mass at time t x gsin30

                                               = (40 - 2t) x 5

So 5(40 - 2t) = 2v

or v = (5/2)(40 - 2t)

or dx/dt = (5/2)(40 - 2t)

or dx = (5/2)(40 - 2t)dt

integrating..

dx = 5/2{ 40dt - 2tdt }

x = 5/2{ 40t - t²}

120 = 5/2. (40t - t²)

Now solving gives t = 1.23s and 38.8s

it took me two and a half hours ......... but still an imperfect answer with an impurity of 1.23s appears with 38.8s