The correct option is A Fm0(m0+μt)2
Answer is A.
Initial velocity of the flat car is zero. Let v be its velocity at time t and m its mass at that instant. Then
At t=0,v=0 and m=mo att=t,v=v and m=mo
Here, vr=v(backwards).
dmdt=μ
Therefore, Ft=vrdmdt=μv(backwards)
Net force on the flat car at time t is fnet=F−Ft
or mdvdt=F−μv...(i)
or (m0+μt)dvdt=F−μvor∫v0dvF−μv=∫t0dtm0+μt
Therefore, −1μ[ln(F−μv)]v0=1μ[ln(m0+μt)]t0
⇒ln(FF−μv)=ln(m0+μtm0)
So, FF−μv=m0+μtm0orv=Ftm0+μt
from Eq.(i), dvdt = acceleration of flat car at time t
or =F−μvma=⎛⎜
⎜
⎜
⎜⎝F−Fμtm0+μtm0+μt⎞⎟
⎟
⎟
⎟⎠ora=Fm0(m0+μt)2.
Hence, the time dependence of the acceleration of the flat car in the process of loading is Fm0(m0+μt)2.