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The maximum possible acceleration of a train starting from rest and moving on straight track is 10m/s² and maximum possible retardation is 5m/s². The maximum speed that train can achieve is 85m/s. Minimum time in which train can complete a journey of 1000m ending at rest is n(2/3)^1/2 sec. Where n is integer . Find n.?
Verified Answer
The maximum possible acceleration of a train starting from rest and mo...
Time for acceleration :
85 / 10 = 8.5 s
Time for deceleration :
85 / 5= 17s
Total time for retardation and acceleration :
8.5 + 17 = 25.5 s
This motion forms a triangle of height 85 and length 25.5
Distance covered in retardation and acceleration is :
0.5 x 25.5 x 85 = 1083 m
The distance moved here is 1000m
We therefore look for the value of the that gives 1000
0.5 x t x 85 = 1000
t = 1000 / 42.5 = 23.53 s
This is equal to the total time :
23.53 = n√2/3
23.53 = 0.816n
n = 23.53 / 0.816 = 28.84
n = 28.84
This question is part of UPSC exam. View all JEE courses
Most Upvoted Answer
The maximum possible acceleration of a train starting from rest and mo...
Given data:
- Maximum acceleration (a) = 10 m/s²
- Maximum retardation (r) = 5 m/s²
- Maximum speed (v) = 85 m/s
- Distance (s) = 1000 m
Calculating time using acceleration:
When the train is accelerating, the time taken to reach the maximum speed can be calculated using the formula:
v = u + at
Here, u = 0 (initial velocity) and v = 85 m/s (maximum speed)
So, 85 = 0 + 10t
t = 8.5 seconds
Calculating time using retardation:
When the train is decelerating, the time taken to come to rest can be calculated using the formula:
v² = u² + 2as
Here, u = 85 m/s (maximum speed), v = 0 (final velocity), and a = -5 m/s² (negative because it is deceleration)
So, 0 = 85² + 2(-5)s
0 = 7225 - 10s
10s = 7225
s = 722.5 meters
The distance covered during deceleration is 722.5 meters.
Calculating time for the remaining distance:
The remaining distance to be covered is 1000 - 722.5 = 277.5 meters.
Let's assume the time taken to cover this remaining distance is t2. During this time, the train is moving at a constant speed of 85 m/s.
So, t2 = distance/speed = 277.5/85 = 3.265 seconds
Calculating total time:
The total time taken for the journey can be calculated by adding the time taken during acceleration, deceleration, and the time for the remaining distance.
Total time = t1 + t2 = 8.5 + 3.265 = 11.765 seconds
Calculating n:
Given that the minimum time required is n(2/3)^1/2 seconds, we need to find the value of n.
n(2/3)^1/2 = 11.765
n = 11.765/(2/3)^1/2
n = 11.765/√(2/3)
Simplifying further, we get:
n = 11.765/√(2/3) * √(3/2) * √(3/2)
n = 11.765 * √(9/4)
n = 11.765 * (3/2)
n = 17.648
Therefore, the value of n is 17.
- Maximum acceleration (a) = 10 m/s²
- Maximum retardation (r) = 5 m/s²
- Maximum speed (v) = 85 m/s
- Distance (s) = 1000 m
Calculating time using acceleration:
When the train is accelerating, the time taken to reach the maximum speed can be calculated using the formula:
v = u + at
Here, u = 0 (initial velocity) and v = 85 m/s (maximum speed)
So, 85 = 0 + 10t
t = 8.5 seconds
Calculating time using retardation:
When the train is decelerating, the time taken to come to rest can be calculated using the formula:
v² = u² + 2as
Here, u = 85 m/s (maximum speed), v = 0 (final velocity), and a = -5 m/s² (negative because it is deceleration)
So, 0 = 85² + 2(-5)s
0 = 7225 - 10s
10s = 7225
s = 722.5 meters
The distance covered during deceleration is 722.5 meters.
Calculating time for the remaining distance:
The remaining distance to be covered is 1000 - 722.5 = 277.5 meters.
Let's assume the time taken to cover this remaining distance is t2. During this time, the train is moving at a constant speed of 85 m/s.
So, t2 = distance/speed = 277.5/85 = 3.265 seconds
Calculating total time:
The total time taken for the journey can be calculated by adding the time taken during acceleration, deceleration, and the time for the remaining distance.
Total time = t1 + t2 = 8.5 + 3.265 = 11.765 seconds
Calculating n:
Given that the minimum time required is n(2/3)^1/2 seconds, we need to find the value of n.
n(2/3)^1/2 = 11.765
n = 11.765/(2/3)^1/2
n = 11.765/√(2/3)
Simplifying further, we get:
n = 11.765/√(2/3) * √(3/2) * √(3/2)
n = 11.765 * √(9/4)
n = 11.765 * (3/2)
n = 17.648
Therefore, the value of n is 17.
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The maximum possible acceleration of a train starting from rest and moving on straight track is 10m/s² and maximum possible retardation is 5m/s². The maximum speed that train can achieve is 85m/s. Minimum time in which train can complete a journey of 1000m ending at rest is n(2/3)^1/2 sec. Where n is integer . Find n.?
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The maximum possible acceleration of a train starting from rest and moving on straight track is 10m/s² and maximum possible retardation is 5m/s². The maximum speed that train can achieve is 85m/s. Minimum time in which train can complete a journey of 1000m ending at rest is n(2/3)^1/2 sec. Where n is integer . Find n.? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The maximum possible acceleration of a train starting from rest and moving on straight track is 10m/s² and maximum possible retardation is 5m/s². The maximum speed that train can achieve is 85m/s. Minimum time in which train can complete a journey of 1000m ending at rest is n(2/3)^1/2 sec. Where n is integer . Find n.? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The maximum possible acceleration of a train starting from rest and moving on straight track is 10m/s² and maximum possible retardation is 5m/s². The maximum speed that train can achieve is 85m/s. Minimum time in which train can complete a journey of 1000m ending at rest is n(2/3)^1/2 sec. Where n is integer . Find n.?.
The maximum possible acceleration of a train starting from rest and moving on straight track is 10m/s² and maximum possible retardation is 5m/s². The maximum speed that train can achieve is 85m/s. Minimum time in which train can complete a journey of 1000m ending at rest is n(2/3)^1/2 sec. Where n is integer . Find n.? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The maximum possible acceleration of a train starting from rest and moving on straight track is 10m/s² and maximum possible retardation is 5m/s². The maximum speed that train can achieve is 85m/s. Minimum time in which train can complete a journey of 1000m ending at rest is n(2/3)^1/2 sec. Where n is integer . Find n.? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The maximum possible acceleration of a train starting from rest and moving on straight track is 10m/s² and maximum possible retardation is 5m/s². The maximum speed that train can achieve is 85m/s. Minimum time in which train can complete a journey of 1000m ending at rest is n(2/3)^1/2 sec. Where n is integer . Find n.?.
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