\(\sqrt{x}=t+7\)
\(\Rightarrow x=(t+7)^2\)
\(=t^2+49+14 t \text { (squaring) }\)
\(\frac{d x}{d t}=2 t+14\)
\(v=2 t+14 \Rightarrow v \propto t\)
Acceleration :
\(a=\frac{d v}{d t}\)
\(a=2 \,ms ^{-2} \rightarrow \text { constant }\)
$[g = 10\,m/{s^2}]$