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`        A ggod train accelarates uniformly on a straight railway track approaches an electric pole standing on the side of track It’s engine passes the pole with velocity u and guards room with velocity v the middle wagon passes the pole with velocity: 	root(u2+v2 /2)      b. ½ root(u2+v2)      c. root uv                                                                                             I know the answer but i want full solution on this.`
2 years ago

erra akhil
208 Points
```										Dear Ananya,From the laws of motion we know,v2 - u2 = 2 a S -----eqn.1v-final velocityu-intial velocitya-acceleartionS-dispacementLet us assume that the train length is L;When the train front part passes from the pole let the velocity of the train be u;Since the train is accelerating,  let the velocity of the train be v when the back part passes besides the pole.When the back part or ending part of train passes besides the pole it implies that the train front part has travelled by a distance L  from the pole.So we can use the eqn1;v2 - u2 = 2 a L ----------eqn.aWhen the middle part of the train passes by the pole it implies that the front part of the train has travelled by a distance L/2 from the pole. Let the middle part of the train travel with a velocity (v`);(v`)2  -  (u)2 = 2 a (L/2) -----------eqn.b Divide both the eqn a & b;  So that a ans L get cancelled;v2  -  u2 = 2{(v`2) – (u2)}on solving we get ;root{ (v2+u2) / 2 } = v`Approve if my answer helped you.
```
2 years ago
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