A body of mass 200 g is moving with a velocity of 5ms^-1 along the positive $x$-direction. At time $t=0$ when the body is at  $x=0,$ a constant force of 0.4 N directed along the negative $x$-direction is applied to the body for 10s.

Given u=+5 ms^-1 along positive x-direction

F=-0.4 N along negative x-direction

 M=200 g =0.2 kg

The acceleration $a=\frac{F}{m}=\frac{-0.4}{0.2}=-2m{s}^{-2}$.  

The negative sign shows that the motion is retarded.

The position of the body at time t is given by $x=x_0+ut+\frac{1}{2}a{t}^2$

At t=0, the body is at x=0.  Therefore, $x_0=0$.  

Hence  $x=ut+\frac{1}{2}a{t}^2$

Since the force acts during the time interval from t=0 to t=10s, the motion is acceleration only between t=0 and t=10 s.  The position of the body t=2.5 s is given by

  $x=5\times 2.5\times \frac{1}{2}\times \left(-2\right)\times {\left(2.5\right)}^2=1.25m$

The velocity of the body at t = 2.5 s is

$v=u+at=5+\left(-2\right)\times \left(2.5\right)=5-5=0$

 During the first ten seconds (i.e. from t = 0 to t = 10 s) the motion is accelerated.  During this time a = -2 ms^-2.  

Putting u = 5 ms^-1, a = - 2 ms^-2 and t = 10 s in equation $x=ut+\frac{1}{2}a{t}^2$.  

We have $x_1=5\times 10+\frac{1}{2}\times \left(-2\right)\times \left(10\right)2=-50m\left(i\right)$

The velocity of the body at t = 10 s is

$v=u+at=5+\left(-2\right)\times 10=-15m{s}^{-1}$            

During the remaining 20 seconds, i.e. from t=10 s to t=30 s, the acceleration a=0, because the force ceases to act after t=10 s. Then velocity of the body remains constant at $-15m{s}^{-1}$ during the last 20 seconds.  The distance covered by the body during the last 20 seconds is $x_2=-15\times 20=-300m$            

 $\therefore$ Position of the body at t=30 s is

$x=x_1+x_2=-50-300$ m

The magnitude of the velocity (i.e. speed) of the body at t = 30 s is 15 ms^-1.

Hence the correct choices are (a), (b) and (c)