A person travelling on a straight line moves with a uniform velocity v₁ for a distance x and with a uniform velocity v₂ for the next 3/2 x distance. The average velocity in this motion is 50/7 m / s. If v₁ is 5 m / s then v₂= __________ m / s.

Solution

Physics sir · 4 steps

Average velocity is total displacement over total time, so add the times for the two stretches.

1. Total displacement. The motion is along a straight line in one direction, so the displacement is x + 3/2x = 5/2x.
2. Total time. The first stretch takes x/v₁ = x/5. The second takes 3x/2v₂.

   50/7 = (5/2x)/(x/5 + 3x/2v₂)

3. Solve for v2. Cancel x and rearrange: 1/5 + 3/2v₂ = 5/2 × 7/50 = 7/20. So 3/2v₂ = 7/20 - 4/20 = 3/20, which gives 2v₂ = 20.
4. So. v₂ = 10 m s⁻¹.

The idea to keep. Never average the velocities directly. Find the total displacement, add up the time for each part, and divide. The distance x always cancels.

Answer from the NTA final answer key. Solution by PrepWiser.