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Blazing goIITian

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8 Dec 2007 21:55:43 IST

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Kinematics-Train

None

A train can reach its maximum speed of 60kmph over 1km and can stop over 0.5km.One part of the track on its path is damaged.It can travel at a maximum speed of 20kmph here. Find the time by which it is delayed between two stations(it travels across damaged part)

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Learner .

Blazing goIITian

Joined: 5 Nov 2007

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8 Dec 2007 21:56:47 IST

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i got the answer as 2min40s.

Please give your solutions;

Rates assured to all those who attempt..

abc xyz

Blazing goIITian

Joined: 3 Aug 2007

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8 Dec 2007 23:37:52 IST

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Vmax = 60000/3600 = 16.67m/s
v^2 = 2as
gives acc =0.139m/s^2
also deceleration = 0.278m/s^2

but give length of damaged section man ???!!!!!!!!!

Learner .

Blazing goIITian

Joined: 5 Nov 2007

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9 Dec 2007 00:00:29 IST

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VERY SORRY GUYS-LENGTH IS 1 KM!!

Learner .

Blazing goIITian

Joined: 5 Nov 2007

Posts: 507

9 Dec 2007 00:00:32 IST

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VERY SORRY GUYS-LENGTH IS 1 KM!!

Learner .

Blazing goIITian

Joined: 5 Nov 2007

Posts: 507

9 Dec 2007 00:00:42 IST

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VERY SORRY GUYS;LENGTH IS 1 KM!!

Learner .

Blazing goIITian

Joined: 5 Nov 2007

Posts: 507

9 Dec 2007 00:01:01 IST

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VERY SORRY GUYS;LENGTH IS 1 KM!!

Indzee ...

Hot goIITian

Joined: 2 Dec 2007

Posts: 124

9 Dec 2007 00:11:27 IST

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Case I (max speed = 60km/hr)

a) Acceleration to reach the max speed:

u = 0m/s; v = 60km/hr = 16.67m/s; t = ?;

s = 1km = 1000m; a = ?

v² = u² + 2as

16.6² = 0² + 2 (a) (1000)

Therefore,

a = 0.13m/s

v = u + at

16.6 = 0 + 0.13 (t)

Therefore,

t = 127.6 sec

b) Declaration at the end of the station:

u = 60km/hr = 16.67m/s; v = 0m/s; t = ?;

s = 0.5km = 500m; a = ?

v² = u² + 2as

0² = 16.67² + 2 (a) (500)

Therefore,

a = - 0.27m/s

v = u + at

0 = 16.67 + (-0.27) (t)

Therefore,

t = 60.18 sec

c) Total time taken

Let the distance between the villages be x meters

Therefore, total time taken

T₁ = 127.6 + 60.18 + (x - 1000 ? 500) / 16.67

T₁ = 187.78 + (x - 1000 ? 500) /16.67

Case 2 (max speed = 20km/hr)

Follow the same procedure,

a) Acceleration to reach the max speed:

b) Declaration at the end of the station:

c) Total time taken

Then when you get T₂

Just subtract T₁ from T₂, to get the final time elapsed.

Now I don?t know how you got (2 min 40 sec) without (x). Some information is also lacking in Case 2. (ie. the distance covered to reach 20km/hr and declarate)

abc xyz

Blazing goIITian

Joined: 3 Aug 2007

Posts: 603

9 Dec 2007 00:15:26 IST

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5.56^2 = 16.67^2 - 2a2*s1
s1 comes 444 meters. t1 comes 40 sec.
s2 = 1000m. t2=60 sec.

16.67^2 = 5.56^2 + 2a2*s3
s3 = 888m. t3 = 80 sec.

Stotal = 2332m
t original = 140sec.
but t actual = 180 sec.

time lost is 40 secs.

Learner .

Blazing goIITian

Joined: 5 Nov 2007

Posts: 507

9 Dec 2007 11:58:32 IST

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Umm..Pannaguma,have you taken the time taken to decelerate to 20m/s from 60 m/s and vice-e-versa?

You need to do that,right?

First find the time it takes to do that and compare it to time taken if nothing was damaged...

I'll just post my solution in a little while-its pretty long.Please correct me if I've made a mistake.

Btw,I typed that previous post just once;why did it come four times?

Learner .

Blazing goIITian

Joined: 5 Nov 2007

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9 Dec 2007 16:16:15 IST

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See what I did :

Max. speed =50/3mps

Speed on damaged track =50/9mps

Using v²=u²+2as,

acceleration from 0 to 50/3 =5/36 ms^-2

Thus,distance covered in acceleration from 50/9 to 50/3=(8000/9) m

Using v²=u²+2as,

acceleration from 50/3 to 0 = 5/18 ms^-2

Thus,distance covered in retarding from 50/3 to 50/9=4000/9 m

If track is undamaged,it covers 8000/9+1000+4000/9 is covered at 50/3mps.

Learner .

Blazing goIITian

Joined: 5 Nov 2007

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9 Dec 2007 16:19:51 IST

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But,time taken to cover the 1000 damaged metres is =9000/50

=180s

Time consumed in accelerating and retarding =80+40 =120s

Total=300s

Otherwise(if there was no damage),it would have taken time=140s

Thus,delay=300-140=160s

=2min40s

What say guys?

abc xyz

Blazing goIITian

Joined: 3 Aug 2007

Posts: 603

9 Dec 2007 17:59:01 IST

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read my post carefully, what i have calculated is the T actual,(where t1, t2, t3 are the parts) T original is simply 2332/16.67=180 secs.

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