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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(u 2 +v 2 /2) b. ½ root(u 2 +v 2 ) c. root uv I know the answer but i want full solution on this.

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:
 
  1. root(u2+v2 /2)      b. ½ root(u2+v2)      c. root uv                                                                                             I know the answer but i want full solution on this.

Grade:11

1 Answers

erra akhil
208 Points
8 years ago
Dear Ananya,
From the laws of motion we know,
v2 - u2 = 2 a S -----eqn.1
v-final velocity
u-intial velocity
a-acceleartion
S-dispacement
Let 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.a
When 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.

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