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A train starts from rest from a station with acceleration 0.2m/s^2 on a straight track and then comes to rest after attaining maximum speed on another station due to retardation 0.4m/s^2. If total time spent is half an hour, then distance between two stations is (neglect length of train)?
Verified Answer
A train starts from rest from a station with acceleration 0.2m/s^2 on ...
If A is the acceleration of the train and B is the retardation of the train , then the total distance travelled is given by [(AB)/(A+B)].(T^2/2
Here A = 0.2m/s^2 and B = 0.4m/s^2 and T = 1/2 hr = 1800 sec
Therefore x=[ (0.2)(0.4)/0.2+0.4)]x1800x1800/2 = 216000m = 216km
Hence the ans is 216km
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Most Upvoted Answer
A train starts from rest from a station with acceleration 0.2m/s^2 on ...
Given:
Acceleration (a) = 0.2 m/s^2
Retardation (r) = 0.4 m/s^2
Total time spent (t) = 0.5 hours
To Find:
Distance between two stations
Explanation:
Let's break down the problem into different phases:
Phase 1: Acceleration
During this phase, the train starts from rest and accelerates until it reaches its maximum speed. We need to find the time taken and the distance covered during this phase.
Using the equation:
v = u + at
Where:
v = final velocity
u = initial velocity (which is 0 as the train starts from rest)
a = acceleration
t = time taken
We can rearrange the equation to find the time taken:
t = (v - u) / a
Substituting the given values:
t = (v - 0) / 0.2
t = v / 0.2
Phase 2: Retardation
During this phase, the train decelerates with a retardation of 0.4 m/s^2 until it comes to rest. We need to find the time taken and the distance covered during this phase.
Using the same equation as in Phase 1, but with the negative value of retardation:
t = (0 - v) / -r
t = v / r
Total Time:
The total time spent is given as 0.5 hours. Since we have two phases, the total time is the sum of the times taken in each phase.
t_total = t_phase1 + t_phase2
0.5 = v/0.2 + v/0.4
Calculating the Maximum Speed:
To determine the maximum speed, we can use the equation:
v_max = a * t_phase1
Substituting the value of t_phase1:
v_max = 0.2 * (v / 0.2)
v_max = v
Solving the Equations:
Let's solve the equation 0.5 = v/0.2 + v/0.4 for v.
0.5 = 2.5v/1
0.5 = 2.5v
v = 0.5/2.5
v = 0.2 m/s
Calculating the Distance:
Now that we have the maximum speed, we can calculate the distance covered during each phase.
Distance in Phase 1:
Using the equation:
s_phase1 = u * t_phase1 + (1/2) * a * t_phase1^2
Since u = 0, the equation simplifies to:
s_phase1 = (1/2) * a * t_phase1^2
s_phase1 = (1/2) * 0.2 * (v/0.2)^2
s_phase1 = 0.1 * (0.2)^2
s_phase1 = 0.004 m
Distance in Phase 2:
Using the equation:
s_phase2 = v * t_phase2 + (1/2) * (-r) * t_phase2^2
s_phase2 = v * (v/r)
Acceleration (a) = 0.2 m/s^2
Retardation (r) = 0.4 m/s^2
Total time spent (t) = 0.5 hours
To Find:
Distance between two stations
Explanation:
Let's break down the problem into different phases:
Phase 1: Acceleration
During this phase, the train starts from rest and accelerates until it reaches its maximum speed. We need to find the time taken and the distance covered during this phase.
Using the equation:
v = u + at
Where:
v = final velocity
u = initial velocity (which is 0 as the train starts from rest)
a = acceleration
t = time taken
We can rearrange the equation to find the time taken:
t = (v - u) / a
Substituting the given values:
t = (v - 0) / 0.2
t = v / 0.2
Phase 2: Retardation
During this phase, the train decelerates with a retardation of 0.4 m/s^2 until it comes to rest. We need to find the time taken and the distance covered during this phase.
Using the same equation as in Phase 1, but with the negative value of retardation:
t = (0 - v) / -r
t = v / r
Total Time:
The total time spent is given as 0.5 hours. Since we have two phases, the total time is the sum of the times taken in each phase.
t_total = t_phase1 + t_phase2
0.5 = v/0.2 + v/0.4
Calculating the Maximum Speed:
To determine the maximum speed, we can use the equation:
v_max = a * t_phase1
Substituting the value of t_phase1:
v_max = 0.2 * (v / 0.2)
v_max = v
Solving the Equations:
Let's solve the equation 0.5 = v/0.2 + v/0.4 for v.
0.5 = 2.5v/1
0.5 = 2.5v
v = 0.5/2.5
v = 0.2 m/s
Calculating the Distance:
Now that we have the maximum speed, we can calculate the distance covered during each phase.
Distance in Phase 1:
Using the equation:
s_phase1 = u * t_phase1 + (1/2) * a * t_phase1^2
Since u = 0, the equation simplifies to:
s_phase1 = (1/2) * a * t_phase1^2
s_phase1 = (1/2) * 0.2 * (v/0.2)^2
s_phase1 = 0.1 * (0.2)^2
s_phase1 = 0.004 m
Distance in Phase 2:
Using the equation:
s_phase2 = v * t_phase2 + (1/2) * (-r) * t_phase2^2
s_phase2 = v * (v/r)
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A train starts from rest from a station with acceleration 0.2m/s^2 on a straight track and then comes to rest after attaining maximum speed on another station due to retardation 0.4m/s^2. If total time spent is half an hour, then distance between two stations is (neglect length of train)?
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A train starts from rest from a station with acceleration 0.2m/s^2 on a straight track and then comes to rest after attaining maximum speed on another station due to retardation 0.4m/s^2. If total time spent is half an hour, then distance between two stations is (neglect length of train)? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A train starts from rest from a station with acceleration 0.2m/s^2 on a straight track and then comes to rest after attaining maximum speed on another station due to retardation 0.4m/s^2. If total time spent is half an hour, then distance between two stations is (neglect length of train)? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A train starts from rest from a station with acceleration 0.2m/s^2 on a straight track and then comes to rest after attaining maximum speed on another station due to retardation 0.4m/s^2. If total time spent is half an hour, then distance between two stations is (neglect length of train)?.
A train starts from rest from a station with acceleration 0.2m/s^2 on a straight track and then comes to rest after attaining maximum speed on another station due to retardation 0.4m/s^2. If total time spent is half an hour, then distance between two stations is (neglect length of train)? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A train starts from rest from a station with acceleration 0.2m/s^2 on a straight track and then comes to rest after attaining maximum speed on another station due to retardation 0.4m/s^2. If total time spent is half an hour, then distance between two stations is (neglect length of train)? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A train starts from rest from a station with acceleration 0.2m/s^2 on a straight track and then comes to rest after attaining maximum speed on another station due to retardation 0.4m/s^2. If total time spent is half an hour, then distance between two stations is (neglect length of train)?.
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