Multiple choice

The two ends of a train moving with a constant acceleration pass a certain pole with velocities $u$ and $v$. The velocity with which the middle point of the train passes the same pole is

  1. $\sqrt{\displaystyle\frac{{u}^{2} + {v}^{2}}{2}}$
  2. $\displaystyle\frac{u + v}{2}$
  3. $\sqrt{uv}$
  4. $\displaystyle\frac{{u}^{2} + {v}^{2}}{2}$
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A Correct answer
Explanation

For a train with constant acceleration a and length L, the velocity v at any point x is given by v^2 = u^2 + 2ax. At the front, x=0 and v=u. At the back, x=L and v=v. For the middle point, x=L/2. Thus, v_mid^2 = u^2 + 2a(L/2) = u^2 + aL. Since v^2 = u^2 + 2aL, we have aL = (v^2 - u^2)/2. Substituting this, v_mid^2 = u^2 + (v^2 - u^2)/2 = (u^2 + v^2)/2.

AI explanation

Using the third equation of motion, $v^2 = u^2 + 2as$, the length of the train is the distance traveled, so the middle point is at $L/2$. For the midpoint, its velocity $w$ satisfies $w^2 = u^2 + 2a(L/2)$, which simplifies to $w^2 = u^2 + aL$. From the full train length, $v^2 = u^2 + 2aL$, so $aL = (v^2 - u^2)/2$. Substituting this into the midpoint equation gives $w^2 = u^2 + (v^2 - u^2)/2 = (u^2 + v^2)/2$. Taking the square root, the velocity of the middle point is the square root of $(u^2 + v^2)/2$.

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