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JEE Main 2025 (2 Apr, Shift 1) · Distance, Displacement & Average Speed

Distance, Displacement & Average Speed question from JEE Main 2025 (2 Apr, Shift 1)

Q5JEE Main 2025 · 2 APR · SHIFT 1Easy

Distance, Displacement & Average Speed

A person travelling on a straight line moves with a uniform velocity v1v_1 for a distance xx and with a uniform velocity v2v_2 for the next 32x\frac{3}{2}x distance. The average velocity in this motion is 507 m/s\frac{50}{7}\text{ m/s}. If v1v_1 is 5 m/s5\text{ m/s} then v2=v_2 = __________ m/s\text{m/s}.

Numerical answer type. Work it out, then open the answer.

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Answer: 10

  1. Total distance =x+32x=5x2= x + \frac{3}{2}x = \frac{5x}{2}; total time =x5+3x2v2= \frac{x}{5} + \frac{3x}{2v_2}.
  2. 5x/2x/5+3x/(2v2)=507⇒15+32v2=52×750=0.35\frac{5x/2}{x/5 + 3x/(2v_2)} = \frac{50}{7} \Rightarrow \frac{1}{5} + \frac{3}{2v_2} = \frac{5}{2} \times \frac{7}{50} = 0.35
  3. 32v2=0.15⇒v2=10\frac{3}{2v_2} = 0.15 \Rightarrow v_2 = 10
  4. So v2=10v_2 = 10 m/s.
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