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Standard XII

Physics

Relative Acceleration

The driver of...

Question

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

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

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

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

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a > \frac{v_{1}^{2} - v_{2}^{2}}{2 d}

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Solution

The correct option is A $a > \frac{(v_{1} - v_{2})^{2}}{2 d}$
Relative speed $= V_{1} - V_{2}$
$\Rightarrow 0 > (V_{1} - V_{2})^{2} - 2 a d$
$\Rightarrow a > [\frac{(V_{1} - V_{2})^{2}}{2 d}]$

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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......?

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The driver of an express train travelling at a speed of v1 sees on the same track at distance d in front of him a goods train travelling in the same direction at a speed v2 such that v1>v2. Immediately he applies brakes to his express train producing retardation a to avoid collision. Then

The correct option is A a>(v1−v2)22dRelative speed =V1−V2⇒0>(V1−V2)2−2ad⇒a>[(V1−V2)22d]

The correct option is A $a > \frac{(v_{1} - v_{2})^{2}}{2 d}$
Relative speed $= V_{1} - V_{2}$
$\Rightarrow 0 > (V_{1} - V_{2})^{2} - 2 a d$
$\Rightarrow a > [\frac{(V_{1} - V_{2})^{2}}{2 d}]$