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A train started from rest from a station and accelerated at 2 ms -2 for 10 s. Then, it ran at constant speed for 30 s and thereafter it decelerated at 4 ms -2 until it stopped at the next station. The distance between two stations is :
  • a)
    650 m
  • b)
    700 m
  • c)
    750 m ​
  • d)
    800 m
Correct answer is option 'D'. Can you explain this answer?
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A train started from rest from a station and accelerated at 2 ms -2 fo...
Given data:
Initial velocity (u) = 0 m/s
Acceleration (a1) = 2 m/s²
Time taken (t1) = 10 s
Velocity at the end of the acceleration (v1) = u + a1t1 = 0 + 2×10 = 20 m/s
Time taken for constant speed (t2) = 30 s
Distance covered during constant speed (d) = v1t2 = 20×30 = 600 m
Deceleration (a2) = -4 m/s² (negative sign indicates deceleration)
Final velocity (v2) = 0 m/s

Calculation:
Using the third equation of motion, we can find the distance covered during acceleration and deceleration as below:
Distance covered during acceleration (d1) = ut1 + ½a1t1² = 0 + ½×2×10² = 100 m
Distance covered during deceleration (d2) = ½(v1 + v2)t3 = ½(20 + 0)×(v1/a2) = 5v1/a2 = 5×20/4 = 25 m
Total distance covered = d1 + d + d2 = 100 + 600 + 25 = 725 m
However, the question asks for the distance between two stations, which is the total distance covered minus the distance covered during constant speed. Therefore,
Distance between two stations = Total distance covered - Distance covered during constant speed = 725 - 600 = 125 m

Therefore, the correct answer is option (C) 750 m.
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A train started from rest from a station and accelerated at 2 ms -2 fo...
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A train started from rest from a station and accelerated at 2 ms -2 for 10 s. Then, it ran at constant speed for 30 s and thereafter it decelerated at 4 ms -2 until it stopped at the next station. The distance between two stations is :a)650 mb)700 mc)750 m ​d)800 mCorrect answer is option 'D'. Can you explain this answer?
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A train started from rest from a station and accelerated at 2 ms -2 for 10 s. Then, it ran at constant speed for 30 s and thereafter it decelerated at 4 ms -2 until it stopped at the next station. The distance between two stations is :a)650 mb)700 mc)750 m ​d)800 mCorrect answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A train started from rest from a station and accelerated at 2 ms -2 for 10 s. Then, it ran at constant speed for 30 s and thereafter it decelerated at 4 ms -2 until it stopped at the next station. The distance between two stations is :a)650 mb)700 mc)750 m ​d)800 mCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A train started from rest from a station and accelerated at 2 ms -2 for 10 s. Then, it ran at constant speed for 30 s and thereafter it decelerated at 4 ms -2 until it stopped at the next station. The distance between two stations is :a)650 mb)700 mc)750 m ​d)800 mCorrect answer is option 'D'. Can you explain this answer?.
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