A train of length a, traveling at a constant velocity, : PS Archive
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# A train of length a, traveling at a constant velocity,

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A train of length a, traveling at a constant velocity, [#permalink]

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03 Jun 2007, 06:43
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A train of length a, traveling at a constant velocity, passes a pole in t seconds. If the same train traveling at the same velocity passes a platform in 3t seconds, then the length of the platform is

A) 0.5a
B) a
C) 1.5a
D) 2a
E) 3a
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03 Jun 2007, 07:49
(D)

Train passes pole in t secs => distance = a

say velocity = r

rt=a ---------- (1)

now say length of platform is X

hence distance = X + a

r(3t) = X + a => 3a = X+a

=> X=2a
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03 Jun 2007, 08:04
(D)

Train passes pole in t secs => distance = a

say velocity = r

rt=a ---------- (1)

now say length of platform is X

hence distance = X + a

r(3t) = X + a => 3a = X+a

=> X=2a

I dont understand why you did x+a...aren't the pole and platform scenarios separate from each other?

I get a = r(t) where a is the length of the train (distance), r is the velocity, and t is the time.

we know that r = a/t, so if the same train travelling at the same velocity passes the platform in 3t seconds, then p, the lenght of the platform is just p=a/t*3t

which simplifies to 3a...

I guess that'd be too easy huh...what's the asnwer?
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03 Jun 2007, 08:32
E. 3a

Pole:
Length of train = distance = a
time = t
speed = a/t

Station:
Length = x
time = 3t
speed = a/t
x = (a/t)*3t = 3a
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03 Jun 2007, 11:59
(D)

Pole:
Length of train = distance = a
time = t
speed = a/t

Station:
Length = x
time = 3t
speed = a/t
x = (a/t)*3t = 3a

this is true but this is the length which the train has totally traveled , also including its own length so by subtracting its own length we get length of platform as 2a which I suppose should be the correct answer.
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03 Jun 2007, 20:00
plaguerabbit wrote:
(D)

Train passes pole in t secs => distance = a

say velocity = r

rt=a ---------- (1)

now say length of platform is X

hence distance = X + a

r(3t) = X + a => 3a = X+a

=> X=2a

I dont understand why you did x+a...aren't the pole and platform scenarios separate from each other?

I get a = r(t) where a is the length of the train (distance), r is the velocity, and t is the time.

we know that r = a/t, so if the same train travelling at the same velocity passes the platform in 3t seconds, then p, the lenght of the platform is just p=a/t*3t

which simplifies to 3a...

I guess that'd be too easy huh...what's the asnwer?

the Q says that the train passes the platform in 3t seconds

passes as in - till the tail of the train crosses the end point of the platform

hence total distance = length of the train + lenght of the platform

= X + a
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03 Jun 2007, 20:16

Train has to cover its own length and the length of the platform
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04 Jun 2007, 04:27
Plug in to solve:

make a=10
and t = 2

so S = 10/2 = 5

so 3t = 6
and 5 = x/6
x = 30

but as the train has already passed the plataform the lenght of it is 3a-a what makes 30-10 = 20 or 2a!!!!
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04 Jun 2007, 10:23
The OA is 2a

Rule: don't forget to deduct the lenght of the train
04 Jun 2007, 10:23
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