A horse rider covers half the distance with 5m s ^ - 1 speed. The...

Let total distance travelled by horse rider be $2x$.

Now, the time taken to cover first half distance is ${t}_{AB} = \dfrac{x}{5}$

Second half distance $x$ between $BC$ is covered in two parts, $x = {d}_{1} + {d}_{2}$

here ${d}_{1}$ is the distance travelled with $10m{s}^{- 1}$ and ${d}_{2}$ is the distance travelled with $15m{s}^{- 1}$ in equal time intervals $\dfrac{t}{2}$ each.

Then, ${d}_{1} = \dfrac{10t}{2}, {d}_{2} = \dfrac{15t}{2}$

Second half distance $x = \dfrac{t}{2} ( 10 + 15 ) = \dfrac{25t}{2}$

Now, the mean speed of rider is $v = \dfrac{2x}{\dfrac{x}{5} + \dfrac{2x}{25}} = \dfrac{2 \times 25}{5 + 2} = \dfrac{50}{7}m{s}^{- 1}$

Thus, the value of $x = 50$.