A person travelling on a straight line moves with a uniform velocity $$v_1$$ for a distance x and with a uniform velocity $$v_2$$ for the next $$\frac{3}{2}x$$ distance. The average velocity in this motion is $$\frac{50}{7}$$ m/s. If $$v_1$$ is 5 m/s then $$v_2$$ = ______ m/s.

Total distance covered  $$S = x + \frac{3}{2}x = \frac{5}{2}x$$.

Time taken in the first part  $$t_1 = \frac{\text{distance}}{\text{speed}} = \frac{x}{v_1}$$.

Given  $$v_1 = 5 \text{ m/s}$$, so  $$t_1 = \frac{x}{5}$$.

Time taken in the second part  $$t_2 = \frac{\frac{3}{2}x}{v_2} = \frac{3x}{2v_2}$$.

Total time  $$T = t_1 + t_2 = \frac{x}{5} + \frac{3x}{2v_2}$$.

The average velocity formula is
$$\text{Average velocity} = \frac{\text{Total distance}}{\text{Total time}}$$.

According to the question,
$$\frac{\frac{5}{2}x}{\frac{x}{5} + \frac{3x}{2v_2}} = \frac{50}{7}$$.

Cancel the common factor $$x$$ from numerator and denominator:
$$\frac{\frac{5}{2}}{\frac{1}{5} + \frac{3}{2v_2}} = \frac{50}{7}$$.

Cross-multiply:
$$\frac{5}{2} \times \frac{7}{50} = \frac{1}{5} + \frac{3}{2v_2}$$.

Simplify the left side:
$$\frac{5}{2} \times \frac{7}{50} = \frac{35}{100} = \frac{7}{20}$$.

Thus,
$$\frac{7}{20} = \frac{1}{5} + \frac{3}{2v_2}$$.

Convert $$\frac{1}{5}$$ to a denominator of 20:
$$\frac{1}{5} = \frac{4}{20}$$.

Subtract $$\frac{4}{20}$$ from both sides:
$$\frac{7}{20} - \frac{4}{20} = \frac{3}{2v_2}$$
$$\frac{3}{20} = \frac{3}{2v_2}$$.

Cancel the common factor 3:
$$\frac{1}{20} = \frac{1}{2v_2}$$.

Cross-multiply to find $$v_2$$:
$$2v_2 = 20 \; \Longrightarrow \; v_2 = 10 \text{ m/s}.$$

Hence, the required velocity $$v_2$$ is 10 m/s.