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 $\frac{50}{7} \mathrm{~m} / \mathrm{s}$. If $v_1$ is $5 \mathrm{~m} / \mathrm{s}$ then $v_2=$ __________ $\mathrm{m} / \mathrm{s}$.

<p>To find the average velocity given the motion of a person traveling along a straight line with two different velocities, we start by calculating the average velocity ($ v_{\text{avg}} $) using the total distance traveled divided by the total time taken.</p> <p>The distances and velocities are given as follows:</p> <p><p>Distance $ x $ at velocity $ v_1 = 5 \, \text{m/s} $</p></p> <p><p>Distance $ \frac{3}{2}x $ at velocity $ v_2 $</p></p> <p>The formula for average velocity is:</p> <p>$ v_{\text{avg}} = \frac{\text{Total Distance}}{\text{Total Time}} $</p> <p>Given:</p> <p>$ v_{\text{avg}} = \frac{50}{7} \, \text{m/s} $</p> <p>The total distance traveled is:</p> <p>$ x + \frac{3x}{2} = \frac{5x}{2} $</p> <p>The time taken to cover each segment is given by:</p> <p><p>Time for $ x $: $ t_1 = \frac{x}{v_1} = \frac{x}{5} $</p></p> <p><p>Time for $ \frac{3x}{2} $: $ t_2 = \frac{\frac{3x}{2}}{v_2} = \frac{3x}{2v_2} $</p></p> <p>Now, using the formula for average velocity:</p> <p>$ \frac{50}{7} = \frac{\frac{5x}{2}}{\frac{x}{5} + \frac{3x}{2v_2}} $</p> <p>Simplifying the equation:</p> <p>$ \frac{50}{7} = \frac{5/2}{\frac{1}{5} + \frac{3}{2v_2}} $</p> <p>Cross-multiplying gives:</p> <p>$ \frac{1}{5} + \frac{3}{2v_2} = \frac{7}{20} $</p> <p>Solving for $ \frac{3}{2v_2} $:</p> <p>$ \frac{3}{2v_2} = \frac{7}{20} - \frac{1}{5} = \frac{7-4}{20} = \frac{3}{20} $</p> <p>Finally, solving for $ v_2 $:</p> <p>$ \frac{3}{2v_2} = \frac{3}{20} $</p> <p>$ v_2 = 10 \, \text{m/s} $</p> <p>Thus, the value of $ v_2 $ is $ 10 \, \text{m/s} $.</p>