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(A) The average velocity is defined as the total displacement divided by the total time taken.$v_{	ext{avg}} = \frac{x_1 + x_2}{t_1 + t_2}$Given $x_1

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(A) The average velocity is defined as the total displacement divided by the total time taken.$v_{	ext{avg}} = \frac{x_1 + x_2}{t_1 + t_2}$Given $x_1 = x$,$x_2 = \frac{3}{2}x$,$v_1 = 5 \ m/s$,and $v_{	ext{avg}} = \frac{50}{7} \ m/s$.The time taken for the first part is $t_1 = \frac{x}{v_1} = \frac{x}{5}$.The time taken for the second part is $t_2 = \frac{x_2}{v_2} = \frac{3x}{2v_2}$.Substituting these into the average velocity formula:$\frac{50}{7} = \frac{x + \frac{3}{2}x}{\frac{x}{5} + \frac{3x}{2v_2}}$$\frac{50}{7} = \frac{\frac{5}{2}x}{x(\frac{1}{5} + \frac{3}{2v_2})}$$\frac{50}{7} = \frac{2.5}{\frac{1}{5} + \frac{3}{2v_2}}$$\frac{1}{5} + \frac{3}{2v_2} = \frac{2.5 	imes 7}{50} = \frac{17.5}{50} = \frac{7}{20}$$\frac{3}{2v_2} = \frac{7}{20} - \frac{1}{5} = \frac{7-4}{20} = \frac{3}{20}$$\frac{3}{2v_2} = \frac{3}{20} \Rightarrow 2v_2 = 20 \Rightarrow v_2 = 10 \ m/s$.

$A$ 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 = $ . . . . . . $m/s$.

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