JEE Main 2025 Apr 2 Shift 1, Physics Q25: Motion In One Dimension

JEE Main2025Apr 2, Shift 1Physics

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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Solution

Average velocity for piecewise-uniform motion is total distance over total time:

$v_{avg} = \frac{x _{1} + x _{2}}{t _{1} + t _{2}} = \frac{x + \frac{3}{2} x}{\frac{x}{v _{1}} + \frac{3 x /2}{v _{2}}} = \frac{\frac{5}{2} x}{\frac{x}{v _{1}} + \frac{3 x}{2 v _{2}}}$

Cancel $x$ from numerator and denominator:

$v_{avg} = \frac{5/2}{\frac{1}{v _{1}} + \frac{3}{2 v _{2}}}$

Substitute $v_{avg} = \frac{50}{7}$ and $v_{1} = 5$ :

$\frac{50}{7} = \frac{5/2}{\frac{1}{5} + \frac{3}{2 v _{2}}}$

Reciprocate and rearrange:

$\frac{1}{5} + \frac{3}{2 v _{2}} = \frac{5/2 \times 7}{50} = \frac{7}{20}$

$\frac{3}{2 v _{2}} = \frac{7}{20} - \frac{1}{5} = \frac{7 - 4}{20} = \frac{3}{20}$

$2 v_{2} = 20 \Rightarrow v_{2} = 10 m/s$

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JEE Main 2025 Apr 2 Shift 1, Physics Q25: Motion In One Dimension

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