An object starting from rest travels 20m in first 2s and 200m in 40s.What will be the velocity after 7s ffrom the start?

if the body is in uniform accelerated motion then its velocity is constant, in non uniform accelerated, velocity varies.  As nothing is mentioned in question we will assume it uniform accelerated motion in one dimension .

given in question ;; u =0 (initial velocity of particle), s₁ = distance covered in 2 second , s_​2 = distance covered by body in 40 second , v = final velocity of body after 7 second. 

now we have to find velocity of  body after 7 second

using equation of motion
v = u+at
  = 0+ax7
  = 7a ..........     eq 1>

now again using 
$$\begin{gathered} s=ut+\frac{1}{2}a{t}^2 taking total time of journey = 40+2=42 second, and total displacement =20+200=220meter \\ 220 =0 X42+\frac{1}{2}a\times {\left(42\right)}^2 \\ 220 = 0+ \frac{1}{2}a\times 1764 \\ 220= 882a \\ a= \frac{220}{882} \\ a=0.25\frac{m}{s^2} ............. equation 2> \\ now put the value of a in equation 1> we can get \\ v = 7a \\ = 7\times 0.25 m/s \\ 1.75 m/s. \end{gathered}$$
   
   

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