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A train starts from rest from a station with acceleration 0.2 m/s^2 on a straight track and then comes to rest after attaining maximum speed on another station due to retardation 0.4 m/s^2. if total time spent is half an hour, then distance between two station is(Neglect the length of the train).op1. 216km, op2. 512km, op3. 728km, op.4.1296km . correct and is op. 1. can anybody pls explain it.?
Verified Answer
A train starts from rest from a station with acceleration 0.2 m/s^2 on...
In the first case when train is accelerating
v = u + at
here u = 0
a = 0.2 m/s2
t = t1
v = v1
so,
v1 = 0.2t1
thus,
t1 = v1/0.2    (1)
similarly,
s = ut + (1/2)at2
here s = s1
and by substituting the other values, we have
s1 = (1/2) x 0.2 x t12  (2) 
now, when the train is retarding
v2 = v1 + a2t2
here as u = v1
a2 = -0.4 m/s2
and v2 = 0
thus,
0 = v1 + 0.4t2
or
t2 = -v1/0.4    (3)
also
s2 = v1t2 + (1/2)a2t22  
so
s2 = v1t2 - (1/2)x0.4xt22   (4)
now as t = t1 + t2 = 30min = 1800
by adding equations (1) and (3), we get
1800 = v1[1/0.2 + 1/0.4]
or
v1 = 1800/7.5
thus,
v1 = 240 m/s
so, t1 = v1/0.2 = 240/0.2
or
t1 = 1200 secs.
similarly
t2 = 240/0,4 = 600 secs.
now by substituting appropriate values of a,v and t in equations (2) and (4), we get
s1 = (1/2) x 0.2 x (1200)2
or
s1 = 144 km
similarly,
s2 = (240 x 600) - [(1/2) x 0.4 x (600)2]
so, the total distance travelled will be
s = s1 + s2 = 144km + 72km
or
s = 216km
which is option (1)
This question is part of UPSC exam. View all NEET courses
Most Upvoted Answer
A train starts from rest from a station with acceleration 0.2 m/s^2 on...
Community Answer
A train starts from rest from a station with acceleration 0.2 m/s^2 on...
Given Data:
- Initial acceleration: 0.2 m/s^2
- Final retardation: 0.4 m/s^2
- Total time: 30 minutes

Approach:
To find the distance between the two stations, we need to calculate the maximum speed attained by the train during acceleration and then use this speed to find the distance covered during retardation.

Calculations:
1. **Calculate the time taken to reach maximum speed during acceleration:**
Initial velocity, u = 0 m/s
Acceleration, a = 0.2 m/s^2
Time taken to reach maximum speed, t1 = (v - u) / a
Max speed, v = u + at1
2. **Calculate the distance covered during acceleration:**
Distance covered during acceleration, s1 = ut1 + 0.5 * a * t1^2
3. **Calculate the time taken during retardation:**
Final velocity, v = 0 m/s
Retardation, a = -0.4 m/s^2
Time taken during retardation, t2 = (v - u) / a
4. **Calculate the distance covered during retardation:**
Distance covered during retardation, s2 = vt2 + 0.5 * a * t2^2
5. **Total distance between the two stations:**
Total distance = s1 + s2

Final Answer:
After performing the above calculations, the total distance between the two stations comes out to be 216 km. Hence, the correct option is op1.
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A train starts from rest from a station with acceleration 0.2 m/s^2 on a straight track and then comes to rest after attaining maximum speed on another station due to retardation 0.4 m/s^2. if total time spent is half an hour, then distance between two station is(Neglect the length of the train).op1. 216km, op2. 512km, op3. 728km, op.4.1296km . correct and is op. 1. can anybody pls explain it.?
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A train starts from rest from a station with acceleration 0.2 m/s^2 on a straight track and then comes to rest after attaining maximum speed on another station due to retardation 0.4 m/s^2. if total time spent is half an hour, then distance between two station is(Neglect the length of the train).op1. 216km, op2. 512km, op3. 728km, op.4.1296km . correct and is op. 1. can anybody pls explain it.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A train starts from rest from a station with acceleration 0.2 m/s^2 on a straight track and then comes to rest after attaining maximum speed on another station due to retardation 0.4 m/s^2. if total time spent is half an hour, then distance between two station is(Neglect the length of the train).op1. 216km, op2. 512km, op3. 728km, op.4.1296km . correct and is op. 1. can anybody pls explain it.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A train starts from rest from a station with acceleration 0.2 m/s^2 on a straight track and then comes to rest after attaining maximum speed on another station due to retardation 0.4 m/s^2. if total time spent is half an hour, then distance between two station is(Neglect the length of the train).op1. 216km, op2. 512km, op3. 728km, op.4.1296km . correct and is op. 1. can anybody pls explain it.?.
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