The correct option is A 18 m/s2
Given, F=10 N
For a variable mass system where mass is added at a rate μ,
m⋅dvdt=Fext+μu
where v = velocity of car, m0 = mass of car =20 kg and μ=1 kg/s
Here, m=m0+μt after t seconds
and relative velocity of mass added u=−v
(in horizontal direction, sand has zero velocity)
⇒(m0+μt)dvdt=F−μv
Rearranging the terms,
⇒m0dv+μtdv=Fdt−μvdt
⇒m0dv+μ(tdv+vdt)=Fdt
Intergrating v from 0 to v and t from 0 to t:
⇒∫m0dv+∫μ(tdv+vdt)=∫Fdt
⇒m0v+∫μd(vt)=Ft
⇒m0v+μvt=Ft
or v=Ft(m0+μt)=10t20+t
Now, differentiating both sides
dvdt=(m0+μt)F−Ft(μ)(m0+μt)2
⇒ a=m0Fm20(1+μtm0)2=10m0(1+μtm0)2
⇒a=1020(1+1×2020)2
∴a=18 m/s2 after t=20 s