Given data

• First part of motion:
  • distance $= x$
  • velocity $= v_1 = 5\,\text{m/s}$
• Second part of motion:
  • distance $= \dfrac{3x}{2}$
  • velocity $= v_2$
• Average velocity for the whole motion: $v_{\text{avg}}=\frac{50}{7}\,\text{m/s}$

Use definition of average velocity Since motion is along a straight line in the same direction, $v_{\text{avg}}=\frac{\text{total displacement}}{\text{total time}}$

Find total displacement $\text{Total displacement} = x + \frac{3x}{2} = \frac{5x}{2}$

Find total time Time in first part: $t_1 = \frac{x}{v_1} = \frac{x}{5}$

Time in second part: $t_2 = \frac{\frac{3x}{2}}{v_2} = \frac{3x}{2v_2}$

Therefore, $t_{\text{total}}=\frac{x}{5}+\frac{3x}{2v_2}$

Apply average velocity formula $\frac{\frac{5x}{2}}{\frac{x}{5}+\frac{3x}{2v_2}}=\frac{50}{7}$

Cancel $x$ $\frac{\frac{5}{2}}{\frac{1}{5}+\frac{3}{2v_2}}=\frac{50}{7}$

Solve for $v_2$ Cross-multiplying, $\frac{5}{2}=\frac{50}{7}\left(\frac{1}{5}+\frac{3}{2v_2}\right)$

Divide both sides by $50$: $\frac{5}{2}\cdot \frac{7}{50}=\frac{1}{5}+\frac{3}{2v_2}$

$\frac{35}{100}=\frac{1}{5}+\frac{3}{2v_2}$

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

Since $\frac{1}{5}=\frac{4}{20}$ we get $\frac{3}{2v_2}=\frac{7}{20}-\frac{4}{20}=\frac{3}{20}$

Hence, $\frac{1}{2v_2}=\frac{1}{20}$

$2v_2=20$

$v_2=10\,\text{m/s}$

Final answer $\boxed{10}$

Comparison with stored answer Stored correct answer = $10$

Our derived answer matches the stored answer.