The driver of a train moving with a constant speed $v_1$ along a straight track sights another train at a distance d ahead of him on the same track moving in the same direction with a constant speed $v_2$. He at once applies the brakes and gives his train a constant retardation f. There will be a collision of the trains if:
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$\displaystyle v_1 > v_2$ and $\displaystyle \frac{(v_1 - v_2)^2}{2f} < d$
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$\displaystyle v_1 < v_2$ and $\displaystyle \frac{(v_1 + v_2)^2}{2f} > d$
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$\displaystyle v_1 > v_2$ and $\displaystyle \frac{(v_1 - v_2)^2}{2f} > d$
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$\displaystyle v_1 > v_2$ and $\displaystyle \frac{(v_1^2 - v_2^2)}{2f} > d$
C
Correct answer
Explanation
The relative speed is (v1 - v2). The distance covered by the first train to stop is v1^2 / (2f). The distance covered by the second train in that time is v2 * (v1/f) - 0.5 * f * (v1/f)^2? No, simpler: relative velocity v_rel = v1 - v2. Relative acceleration a_rel = -f. Collision occurs if the relative stopping distance (v1-v2)^2 / (2f) is greater than the initial distance d.