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

= \frac{{(v_{2} - v_{1})}^{2}}{2 a}

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> \frac{{(v_{1} - v_{2})}_{1}^{2}}{2 a}

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< \frac{{(v_{1} - v_{2})}^{2}}{2 a}

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< \frac{v_{1} - v_{2}}{2 a}

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Solution

The correct option is D $> \frac{{(v_{1} - v_{2})}_{1}^{2}}{2 a}$
Let $d$ be the distance traveled by the fast train
Then, according to the slower train
$V_{r e l}^{2} - U_{r e l}^{2} = 2 a_{r e l} d \Rightarrow d = \frac{{(v_{1} - v_{2})}^{2}}{2 a}$
Thus, $s > d$

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The driver of a train moving with a constant velocity v₁along a straing track sights another train at a distance d ahrad of him on the same track moving in a same direction with a constant speed V2. He at once applies a brakes and gives his train a constant retardation f. There will be a collosioc of yhe trains if......?

a) v1 > V2 amd (V1 -V2)²/2f < d

b) V1<V2 and (V1+V2)² /2f > d

c) V1 > V2 and ( V1-V2)²/2f > d

d) V1>V2 and (V1²- V2²)/2f > d

Q. The driver of a train moving with a constant speed