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Standard IX

Physics

Derivation of Position-Velocity Relation by Graphical Method

The driver of...

Question

The driver of a train travelling at $115 k m h^{- 1}$ seen on the same track, $100 m$ in front of him, a slow train travelling in the same direction at $25 k m h^{- 1}$ . The least retardation that must be applied to faster train to avoid a collision is

25 m s^{- 2}

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50 m s^{- 2}

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75 m s^{- 2}

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3.125 m s^{- 2}

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Solution

The correct option is D $3.125 m s^{- 2}$
Relative velocity of faster train w.r.t. slower train $u = 115 - 25 = 90 k m h^{- 1}$
$u = 90 \times \frac{5}{18} m s^{- 1} = 25 m s^{- 1}$
Using $v^{2} - u^{2} = 2 a S$
$0^{2} - 25^{2} = 2 \times a \times 100$
or $200 a = - 625 \Rightarrow a = - \frac{625}{200} m s^{- 2}$
or $a = - 3.125 m s^{- 2}$ .

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The driver of a train travelling at 115 km h−1 seen on the same track, 100 m in front of him, a slow train travelling in the same direction at 25 km h−1. The least retardation that must be applied to faster train to avoid a collision is

The correct option is D 3.125 ms−2Relative velocity of faster train w.r.t. slower train u=115−25=90 km h−1u=90×518ms−1=25 ms−1Using v2−u2=2aS02−252=2×a×100or 200a=−625⇒a=−625200ms−2or a=−3.125 ms−2.

The driver of a train travelling at $115 k m h^{- 1}$ seen on the same track, $100 m$ in front of him, a slow train travelling in the same direction at $25 k m h^{- 1}$ . The least retardation that must be applied to faster train to avoid a collision is