A body travelling with uniform acceleration crosses two points $A$ and $B$ with velocities $20m/sec$ and $30m/sec$ respectively then the speed of the body at mid point of $A$ and $B$ is

Let the acceleration of the car is $=a$  and the distance between $A$ and $B$  is $=d$

$v^2-u^2=2ad$

Given,

$v=30m/s$ and $u=20m/s$

$2ad=30^2-20^2$

$ad=\frac{(900-400)}{2}=250$

when the car is at the mid point of $AB$ then let the speed of the car is 

$v_1$

$v_1^2-20^2=2a(\frac{d}{2})$

$v_1^2=ad+400=250+400=650$

therefore 

$v_1$ is $=25.5m/s$