Question 14 Marks
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 $v _2=$ __________ $m / s$.
Answer
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Explanation: $v_{\text {avg }}=\frac{x_1+x_2}{t_1+t_2}$
$\Rightarrow \frac{50}{7}=\frac{x+\frac{3 x}{2}}{\frac{x}{5}+\frac{3 x}{2 v_2}}$
$\Rightarrow \frac{50}{7}=\frac{5 / 2}{\frac{1}{5}+\frac{3}{2 v _2}}$
$\Rightarrow \frac{1}{5}+\frac{3}{2 v _2}=\frac{7}{20}$
$\Rightarrow \frac{3}{2 v _2}=\frac{7}{20}-\frac{1}{5}=\frac{7-4}{20}$
$\Rightarrow \frac{3}{2 v _2}=\frac{3}{20}$
$\Rightarrow v _2=10 m / s$
Explanation: $v_{\text {avg }}=\frac{x_1+x_2}{t_1+t_2}$
$\Rightarrow \frac{50}{7}=\frac{x+\frac{3 x}{2}}{\frac{x}{5}+\frac{3 x}{2 v_2}}$
$\Rightarrow \frac{50}{7}=\frac{5 / 2}{\frac{1}{5}+\frac{3}{2 v _2}}$
$\Rightarrow \frac{1}{5}+\frac{3}{2 v _2}=\frac{7}{20}$
$\Rightarrow \frac{3}{2 v _2}=\frac{7}{20}-\frac{1}{5}=\frac{7-4}{20}$
$\Rightarrow \frac{3}{2 v _2}=\frac{3}{20}$
$\Rightarrow v _2=10 m / s$

