A person travelling in a straight line moves with uniform velocity V₁ for a distance x and with a uniform velocity V₂ for the next 3x/2 distance. The average velocity in this motion is 50/7 m/s. If V₁ is 5 m/s then V₂ (in m/s) is (a) 5 (b) 15 (c) 10 (d) 8

Answer: (c). Total distance = x + 3x/2 = 5x/2; total time = x/5 + (3x)/(2V₂). (5x/2)/(x/5 + 3x/(2V₂)) = 50/7 ⇒ 1/5 + 3/2V₂ = 0.35 3/2V₂ = 0.15 ⇒ V₂ = 10 m/s So the answer is (c) 10.

A person travelling in a straight line moves with uniform velocity V₁… | MHT CET 2026 (15 May, Shift 2)

Q2MHT CET 2026 · 15 MAY · SHIFT 2Medium

Distance, Displacement & Average Speed

A person travelling in a straight line moves with uniform velocity $V_1$ for a distance $x$ and with a uniform velocity $V_2$ for the next $\frac{3x}{2}$ distance. The average velocity in this motion is $\frac{50}{7}$ m/s. If $V_1$ is $5$ m/s then $V_2$ (in m/s) is

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Answer: (c)

Total distance $= x + \frac{3x}{2} = \frac{5x}{2}$ ; total time $= \frac{x}{5} + \frac{3x}{2V_2}$ .
$\frac{5x/2}{x/5 + 3x/(2V_2)} = \frac{50}{7} \Rightarrow \frac{1}{5} + \frac{3}{2V_2} = 0.35$
$\frac{3}{2V_2} = 0.15 \Rightarrow V_2 = 10$ m/s
So the answer is (c) $10$ .

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Distance, Displacement & Average Speed question from MHT CET 2026 (15 May, Shift 2)

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