\\\\(v=u+at\\\\)\\\\(1.5=u+5a...........(i)\\\\)\\\\(0=u+6a.............(ii)\\\\)From (i) and (ii)\\\\(0=1.5+a(6-5)\\\\)\\\\(a=-1.5m/s^2\\\\)Om substituting \\\\(a=-1.5m/s^2\\\\) in equation (ii)\\\\(0=u+6(-1.5)\\\\)\\\\(0=u-9\\\\)\\\\(u=9m/s\\\\)The distance traveled by the body in its forward journey is \\\\(s_1\\\\) can be calculated as follows,\\\\(s_1=ut+\\frac{1}{2}at^2\\\\)\\\\(s_1=9\ imes6+\\frac{1}{2}\ imes(-1.5)\ imes6^2\\\\)\\\\(s_1=54+(-1.5\ imes18)\\\\)\\\\(s_1=27m\\\\)The distance traveled by the body in its forward journey \\\\((s_1)\\\\)= the distance traveled by the body in its backward journey\\\\((s_2)=27m\\\\)So, the total distance traveled by the body in forward journey and its backward journey\\\\(=s_1+s_2=27+27=54m\\\\)