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⋅d→vdt =−−→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
d→vdt=(m0+μt)→F−→Ft(μ)(m0+μt)2
\(\Rightarrow \overrightarrow{a} = \dfrac{m_0 \overrightarrow{F}{m_0^2\left( 1 + \dfrac{\mu t}{m_0} \right)^2} = \dfrac{10}{m_0 \left(1+\dfrac{\mu t}{m_0} \right)^2} \)
⇒→a=1020(1+1×2020)2
∴→a=18 m/s2 after t=20 s