Multiple choice

The driver of a train moving with a speed $v_{ 1 }$ sights another train at a distance $s$, ahead of him moving in the same direction with a slower speed $v_{ 2 }$. He applies the brakes and gives a constant deceleration $a$ to his train. For no collision, $s$ is

  1. $=\cfrac { { \left( v_{ 2 }-v_{ 1 } \right) }^{ 2 } }{ 2a } $
  2. $>{ \cfrac{ \left ( v_{ 1 }-v_{ 2 } \right )^2 }{ 2a } }$
  3. $<\cfrac { { \left( v_{ 1 }-v_{ 2 } \right) }^{ 2 } }{ 2a } $
  4. $<\cfrac { v_{ 1 }-v_{ 2 } }{ 2a } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For no collision, the stopping distance of the first train must be less than the initial distance s plus the distance covered by the second train during the braking time. The relative velocity approach shows s > (v1 - v2)^2 / 2a.

AI explanation

To avoid a collision, the distance s must strictly accommodate the relative deceleration of the faster train as it slows to the slower train's speed. Using the kinematic equation v^2 = u^2 + 2as and substituting the relative initial velocity (v1 - v2) and final velocity of zero, the required stopping distance is (v1 - v2)^2 / (2a). Therefore, to ensure no collision ever occurs, the initial distance s must be strictly greater than (v1 - v2)^2 / (2a).

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