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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)216km 2)512km 3)728km 4)1296km

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

Grade:11

6 Answers

Vasantha Kumari
askIITians Faculty 38 Points
9 years ago
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

KAPIL MANDAL
132 Points
8 years ago
Thank you sir for this answer. Thank you very much.
Joy
11 Points
5 years ago
You can
Narendra Sharma
35 Points
5 years 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.
Mahbub
15 Points
5 years ago
  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
Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the solution to your problem below.
 
Let the peak velocity be v
Let the time of acceleration bee t1 and time of deacceleration be t2
Now, 0.2 = (v – 0)/t1
 – 0.4 = (0 – v)/t2
Hence, t1 = v/0.2 ; t2 = v/0.4
t1 + t2 = 3v/0.4 = 30 mins = 30 x 60 s
Hence, v = 240 m/s
Now, if we plot the velocity time graph it would look like a triangle having coordinates: (0,0), (t1,v), (t1+t2,0)
The area of this triangle will give us the displacement.
Hence, the distance between two stations = ½ x (t1 + t2) x v
                                                             = ½ x 30 x 60 x 240
                                                             = 216000 m
                                                             = 216 Km
 
Hope it helps.
Thanks and regards,
Kushagra

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