so initial velocity of body is $10 \mathrm{m} / \mathrm{s}$ and acceleration of the object is $-5 \mathrm{m} / \mathrm{s}^{2}$

so time taken to stop once is 2 sec. using $v=u+a t$ displacement at that time is $10 \mathrm{m}$. using $S=u t+\frac{1}{2} a t^{2}$ or by $v^{2}-u^{2}=2 a s$

after this the body will start moving in reverse direction so maximum displacement is $10 \mathrm{m}$

and distance covered in next $1\, sec$ id calculate by $S=u t+\frac{1}{2} a t^{2}$

so $D_{2}=2.5 m$ so total distance covered in first 3 sec is $10+2.5=12.5$

so best possible answers are $A,C.$