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A train starts from rest and accelerates uniformly for 20 seconds to attain a velocity of 90 km per hour. it travels at constant velocity for 15 minutes .the driver applies break to stop the train in that 5 seconds find the displacement of the train? solve this?
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A train starts from rest and accelerates uniformly for 20 seconds to a...
Given information:
Finding acceleration:
We know that, v = u + at
Here, u = 0, v = 25 m/s and t = 20 seconds
25 = 0 + a * 20
a = 1.25 m/s^2
Finding displacement during acceleration:
We know that, s = ut + (1/2)at^2
Here, u = 0, t = 20 seconds and a = 1.25 m/s^2
s = 0 + (1/2) * 1.25 * (20)^2
s = 500 meters
Finding displacement during constant velocity:
We know that, s = vt
Here, v = 25 m/s and t = 900 seconds
s = 25 * 900
s = 22500 meters
Finding acceleration during deceleration:
We know that, v = u + at
Here, u = 25 m/s, v = 0 and t = 5 seconds
0 = 25 + a * 5
a = -5 m/s^2
Finding displacement during deceleration:
We know that, s = ut + (1/2)at^2
Here, u = 25 m/s, t = 5 seconds and a = -5 m/s^2
s = 25 * 5 + (1/2) * (-5) * (5)^2
s = 62.5 meters
Total displacement:
Total displacement = Displacement during acceleration + Displacement during constant velocity + Displacement during deceleration
Total displacement = 500 + 22500 + 62.5
Total displacement = 23162.5 meters
Therefore, the displacement of the train is 23162.5 meters.
- Initial velocity of the train, u = 0
- Time taken to attain velocity, t1 = 20 seconds
- Final velocity of the train, v = 90 km/hr = 25 m/s
- Time for which the train travels at constant velocity, t2 = 15 minutes = 900 seconds
- Time taken to stop the train, t3 = 5 seconds
Finding acceleration:
We know that, v = u + at
Here, u = 0, v = 25 m/s and t = 20 seconds
25 = 0 + a * 20
a = 1.25 m/s^2
Finding displacement during acceleration:
We know that, s = ut + (1/2)at^2
Here, u = 0, t = 20 seconds and a = 1.25 m/s^2
s = 0 + (1/2) * 1.25 * (20)^2
s = 500 meters
Finding displacement during constant velocity:
We know that, s = vt
Here, v = 25 m/s and t = 900 seconds
s = 25 * 900
s = 22500 meters
Finding acceleration during deceleration:
We know that, v = u + at
Here, u = 25 m/s, v = 0 and t = 5 seconds
0 = 25 + a * 5
a = -5 m/s^2
Finding displacement during deceleration:
We know that, s = ut + (1/2)at^2
Here, u = 25 m/s, t = 5 seconds and a = -5 m/s^2
s = 25 * 5 + (1/2) * (-5) * (5)^2
s = 62.5 meters
Total displacement:
Total displacement = Displacement during acceleration + Displacement during constant velocity + Displacement during deceleration
Total displacement = 500 + 22500 + 62.5
Total displacement = 23162.5 meters
Therefore, the displacement of the train is 23162.5 meters.
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A train starts from rest and accelerates uniformly for 20 seconds to a...
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A train starts from rest and accelerates uniformly for 20 seconds to attain a velocity of 90 km per hour. it travels at constant velocity for 15 minutes .the driver applies break to stop the train in that 5 seconds find the displacement of the train? solve this?
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A train starts from rest and accelerates uniformly for 20 seconds to attain a velocity of 90 km per hour. it travels at constant velocity for 15 minutes .the driver applies break to stop the train in that 5 seconds find the displacement of the train? solve this? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about A train starts from rest and accelerates uniformly for 20 seconds to attain a velocity of 90 km per hour. it travels at constant velocity for 15 minutes .the driver applies break to stop the train in that 5 seconds find the displacement of the train? solve this? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A train starts from rest and accelerates uniformly for 20 seconds to attain a velocity of 90 km per hour. it travels at constant velocity for 15 minutes .the driver applies break to stop the train in that 5 seconds find the displacement of the train? solve this?.
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