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`        a train starts from rest from a station with acceleration 0.2ms-2 on astraight track and then comes to rest after attaining maximum speed on another station due to retardation 0.4ms-2 . If the total time spent is 30 minutes, then the distance betwen the two stations is(neglect the length of train):1)216km2)512km3)728km4)1296km`
5 years ago Vasantha Kumari
38 Points
```							Case I:Train
accelerates with a = 0.2 m/s2 , t = t1 and
v= v1v=u + at
and v1=0.2t1 or t1= 0.2 v1 /2 …(1)s=ut+(1/2)at2Case II:v2=v1+a2t2(train
is retarding and s =s1)s1=1/2
x 0.2 x t12 v2=v1+a2t20 = v1+0.4t2t2=–v1/0.4 …(2)s2=v1t2
+(1 /2) x a2t22 t=t1+t2
= 30 min = 1800t1=1200
secs and t2 = 600 secssubstituting
a, v and t in (1) and (3), we get s1
= 144 km and s2= 72 kmTotal
distance s= s2+ s2s=216kmThanks
```
5 years ago
```							Thank you sir for this answer. Thank you very much.
```
4 years ago
```							You can
```
one year ago
```							S=abT^2/(2a+b) where a and be are ACC and retardation respectively and T is total time
.On putting the given values gives S=216m.this formula is driven by graphical approach to physics
basic mechanics.
```
one year ago
```								Then I will get the correct answerSimple tricks for this is distance = 1/2(AB/A+B)t^2. Where A and B are the value of acceleration and t is the time so plz try believe on Mahbub Azhari
```
one year ago
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