In motion \(BC\)

\(x = d _1+ d _2\)

where \(d _1 , d _2\) we the distance travelled with \(10\,m / s\) and \(15\,m / s\) respectively in equal time intervals

\(\frac{' t^{\prime}}{2} \text { each }\)

\(d_1=\frac{10 t}{2}, d_2=\frac{15 t }{2}\)

\(d_1+d_2=x=\frac{t}{2}(10+15)=\frac{25 t }{2}\)

\(< v >=\frac{2 x }{\frac{ x }{5}+\frac{2 x }{25}}=\frac{2 \times 25}{5+2}=\frac{50}{7} m / s\)

Ans. : \(50\)