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14 Oct 2009 09:45:40 IST

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A train STARTING FROM REST ACCELERATES UNIFORMLY FOR 100s ,RUNS AT A CONSTANT SPEED FOR 5MINUTES AND

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A train STARTING FROM REST ACCELERATES UNIFORMLY FOR 100s ,RUNS AT A CONSTANT SPEED FOR 5MINUTES AND THEN COMES TO A STOP WITH UNIFORM RETARDATION IN THE NEXT 150 SECONDS. DURING THIS MOTION IT COVERS A DISTANCE OF 4.25 KM. FIND ITS CONSTANT SPEED,ACCELERATION AND RETARDATION.

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Deepak Aggarwal

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14 Oct 2009 10:36:13 IST

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1. Consider the accelerated motion. Body starts from rest and after 100 sec. let its velocity be v. Also, let the acceleration be a.

As, v = u + at₁ = 0 + a ( 100 ) = 100 a -----(1)

Distance covered during accelerated motion, S₁ = (1/2) a (100)² = (1/2) v (100) [ Using (1) ]

So, S₁ = 50v

2. Consider the motion when body moves with constant speed.

S₂ = v t₂ = 300v

3. Considered the motion when body undergoes uniform retardation (-A). Initial velocity in this motion is v and final velocity is zero.

So, v = At₃ = 150A -------- (2)

DIstance covered, S₃ = v² / 2A = 75v [ Using (2) ]

Now, according to question, S₁ + S₂ + S₃ = 4.25 Km = 4250m

So, ( 50 + 300 + 75 ) v = 4250

v = 4250 / 425

v = 10 m/s

So, a = 10 / 100 = 0.1 m/s² and A = 10 / 150 = 0.067 m/s²

PLEASE RATE !

VARUN RAJ

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14 Oct 2009 11:34:09 IST

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see dude always attemt these qustions by drawing graphs its goin to be much easier

Sagar Saxena

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Joined: 8 Oct 2008

Posts: 7221

14 Oct 2009 12:50:48 IST

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hello dear

S = a/2 at^2

= 1/2 a 10000 = 5000a

V = at = 100a

d = 300*100a = 30000a

let retardation be x

v = u + xt

0 = 100a - x 150

x = 100a / 150 = 2a /3

now when it gets retardation

D = 100a (150) - 1/2 x t^2 =

D+d+S = 4250 m

500a + 30000a + 7500a = 4250

Anubhav

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Joined: 4 Feb 2009

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14 Oct 2009 12:52:08 IST

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ya deepak is correct

Deepak Aggarwal

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Joined: 9 May 2009

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14 Oct 2009 21:36:45 IST

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Varun, the solution i did is not that much lengthy. Its looking so bcoz i have explained it fully to make it understand better. It took me only 30 sec. to solve dis ques. which i beleive is a very gud time.

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