Question 99: Motion in 1D

Question

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 $v_2=$ __________ $\mathrm{m} / \mathrm{s}$.

Accepted Answer

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_{	ext{avg}} $) using the total distance traveled divided by the total time taken.
The distances and velocities are given as follows:
Distance $ x $ at velocity $ v_1 = 5 \, 	ext{m/s} $
Distance $ \frac{3}{2}x $ at velocity $ v_2 $
The formula for average velocity is:
$ v_{	ext{avg}} = \frac{	ext{Total Distance}}{	ext{Total Time}} $
Given:
$ v_{	ext{avg}} = \frac{50}{7} \, 	ext{m/s} $
The total distance traveled is:
$ x + \frac{3x}{2} = \frac{5x}{2} $
The time taken to cover each segment is given by:
Time for $ x $: $ t_1 = \frac{x}{v_1} = \frac{x}{5} $
Time for $ \frac{3x}{2} $: $ t_2 = \frac{\frac{3x}{2}}{v_2} = \frac{3x}{2v_2} $
Now, using the formula for average velocity:
$ \frac{50}{7} = \frac{\frac{5x}{2}}{\frac{x}{5} + \frac{3x}{2v_2}} $
Simplifying the equation:
$ \frac{50}{7} = \frac{5/2}{\frac{1}{5} + \frac{3}{2v_2}} $
Cross-multiplying gives:
$ \frac{1}{5} + \frac{3}{2v_2} = \frac{7}{20} $
Solving for $ \frac{3}{2v_2} $:
$ \frac{3}{2v_2} = \frac{7}{20} - \frac{1}{5} = \frac{7-4}{20} = \frac{3}{20} $
Finally, solving for $ v_2 $:
$ \frac{3}{2v_2} = \frac{3}{20} $
$ v_2 = 10 \, 	ext{m/s} $
Thus, the value of $ v_2 $ is $ 10 \, 	ext{m/s} $.

Motion in 1D

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