Multiple choice

The driver of an express train travelling at a speed of $v_{1}$ sees on the same track at distance $d$ in front of him a goods train travelling in the same direction at a speed $v_{2}$ such that $v_1>v_2$. Immediately he applies brakes to his express train producing retardation $a$ to avoid collision. Then

  1. $a < \dfrac{v^{2}_{1}-v^{2}_{2}}{2d}$
  2. $a < \dfrac{(v_{1}-v_{2})^{2}}{2d}$
  3. $a > \dfrac{(v_{1}-v_{2})^{2}}{2d}$
  4. $a > \dfrac{v^{2}_{1}-v^{2}_{2}}{2d}$
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C Correct answer
Explanation

To avoid collision, the relative velocity (v1 - v2) must be reduced to zero before the distance d is covered. Using the kinematic equation v_final^2 = v_initial^2 - 2ad, where v_final = 0, v_initial = v1 - v2, we get 0 = (v1 - v2)^2 - 2ad. Thus, a = (v1 - v2)^2 / (2d). To avoid collision, the retardation must be greater than this value.

AI explanation

To avoid a collision, the express train must decelerate to the speed of the goods train before covering the distance d. Using the kinematic equation v squared equals u squared minus 2as, the relative final velocity is zero and the initial relative velocity is v1 minus v2. Substituting these values gives zero equals (v1 minus v2) squared minus 2ad. Therefore, the required retardation is a equals (v1 minus v2) squared divided by 2d, so the applied retardation must be greater than this value.

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