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Grade: 11
        
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)216km
2)512km
3)728km
4)1296km
5 years ago

Answers : (5)

Vasantha Kumari
askIITians Faculty
38 Points
							Case I:

Train accelerates with a = 0.2 m/s2 , t = t1 and v= v1

v=u + at and v1=0.2t1 or t1= 0.2 v1 /2 …(1)

s=ut+(1/2)at2

Case II:

v2=v1+a2t2(train is retarding and s =s1)

s1=1/2 x 0.2 x t12

v2=v1+a2t2

0 = v1+0.4t2

t2=–v1/0.4 …(2)

s2=v1t2 +(1 /2) x a2t22

t=t1+t2 = 30 min = 1800

t1=1200 secs and t2 = 600 secs

substituting a, v and t in (1) and (3), we get

s1 = 144 km and s2= 72 km

Total distance s= s2+ s2

s=216km

Thanks & Regards,

Vasantha Sivaraj,

askIITians faculty

5 years ago
KAPIL MANDAL
132 Points
							
Thank you sir for this answer. Thank you very much.
4 years ago
Joy
11 Points
							You can
						
one year ago
Narendra Sharma
35 Points
							
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
Mahbub
15 Points
							
  1. 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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