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the brake of a train which is travelling at 30m /sec are appliedas train passse from point A. the brake produces a constant retardation of magnitude 3 Xm/sec^2 until the speed of train is reduced to 10 m / sec. the train travels at this speed for a distance and then uniformly accelerated at X m /sec^2 until it again reaches a speed of 30 m / sec as it passes from point B. the time taken by the train in travelling from A to B, a distance of 4 km is 4 min.
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the brake of a train which is travelling at 30m /sec are appliedas tra...
**Braking from Point A to 10 m/s**
To calculate the time taken by the train to travel from point A to B, we need to break down the journey into different segments.
First, let's calculate the time taken by the train to slow down from its initial speed of 30 m/s to 10 m/s. We know that the train has a constant retardation of 3 m/s^2 during this phase.
Using the equation of motion v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance, we can find the distance traveled during this deceleration.
v^2 = u^2 + 2as
(10)^2 = (30)^2 + 2(-3)s
100 = 900 - 6s
6s = 800
s = 800/6
s = 133.33 m
So, the train travels a distance of 133.33 meters while slowing down from 30 m/s to 10 m/s.
The time taken to cover this distance can be calculated using the equation of motion v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
v = u + at
10 = 30 + (-3)t
-3t = -20
t = 20/3
t ≈ 6.67 seconds
**Traveling at 10 m/s**
After reaching a speed of 10 m/s, the train travels at this constant speed for a certain distance. Let's denote this distance as d1.
We know that the time taken to travel at a constant speed is given by the formula t = d/v, where d is the distance and v is the velocity.
t = d/v
t = d1/10
**Accelerating from 10 m/s to 30 m/s**
Next, the train accelerates uniformly from 10 m/s to 30 m/s. Let's denote the acceleration during this phase as a1. We are given that the train accelerates at X m/s^2.
Using the equation of motion v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance, we can find the distance traveled during this acceleration.
v^2 = u^2 + 2as
(30)^2 = (10)^2 + 2(a1)s
900 = 100 + 2(a1)s
2(a1)s = 800
s = 400/a1
The time taken to cover this distance can be calculated using the equation of motion v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
v = u + at
30 = 10 + a1t
a1t = 20
t = 20/a1
**Total Time Taken**
The total time taken by the train to travel from point A to B can be calculated by adding the time taken for each segment.
Total time = time taken to slow down + time taken to travel at a constant speed + time taken to accelerate
Total time = 6.67 + d1/10 +
To calculate the time taken by the train to travel from point A to B, we need to break down the journey into different segments.
First, let's calculate the time taken by the train to slow down from its initial speed of 30 m/s to 10 m/s. We know that the train has a constant retardation of 3 m/s^2 during this phase.
Using the equation of motion v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance, we can find the distance traveled during this deceleration.
v^2 = u^2 + 2as
(10)^2 = (30)^2 + 2(-3)s
100 = 900 - 6s
6s = 800
s = 800/6
s = 133.33 m
So, the train travels a distance of 133.33 meters while slowing down from 30 m/s to 10 m/s.
The time taken to cover this distance can be calculated using the equation of motion v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
v = u + at
10 = 30 + (-3)t
-3t = -20
t = 20/3
t ≈ 6.67 seconds
**Traveling at 10 m/s**
After reaching a speed of 10 m/s, the train travels at this constant speed for a certain distance. Let's denote this distance as d1.
We know that the time taken to travel at a constant speed is given by the formula t = d/v, where d is the distance and v is the velocity.
t = d/v
t = d1/10
**Accelerating from 10 m/s to 30 m/s**
Next, the train accelerates uniformly from 10 m/s to 30 m/s. Let's denote the acceleration during this phase as a1. We are given that the train accelerates at X m/s^2.
Using the equation of motion v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance, we can find the distance traveled during this acceleration.
v^2 = u^2 + 2as
(30)^2 = (10)^2 + 2(a1)s
900 = 100 + 2(a1)s
2(a1)s = 800
s = 400/a1
The time taken to cover this distance can be calculated using the equation of motion v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
v = u + at
30 = 10 + a1t
a1t = 20
t = 20/a1
**Total Time Taken**
The total time taken by the train to travel from point A to B can be calculated by adding the time taken for each segment.
Total time = time taken to slow down + time taken to travel at a constant speed + time taken to accelerate
Total time = 6.67 + d1/10 +
Community Answer
the brake of a train which is travelling at 30m /sec are appliedas tra...
calculate 1.) value of X2.)distance travelled at 10 m / sec
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the brake of a train which is travelling at 30m /sec are appliedas train passse from point A. the brake produces a constant retardation of magnitude 3 Xm/sec^2 until the speed of train is reduced to 10 m / sec. the train travels at this speed for a distance and then uniformly accelerated at X m /sec^2 until it again reaches a speed of 30 m / sec as it passes from point B. the time taken by the train in travelling from A to B, a distance of 4 km is 4 min.
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the brake of a train which is travelling at 30m /sec are appliedas train passse from point A. the brake produces a constant retardation of magnitude 3 Xm/sec^2 until the speed of train is reduced to 10 m / sec. the train travels at this speed for a distance and then uniformly accelerated at X m /sec^2 until it again reaches a speed of 30 m / sec as it passes from point B. the time taken by the train in travelling from A to B, a distance of 4 km is 4 min. 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 the brake of a train which is travelling at 30m /sec are appliedas train passse from point A. the brake produces a constant retardation of magnitude 3 Xm/sec^2 until the speed of train is reduced to 10 m / sec. the train travels at this speed for a distance and then uniformly accelerated at X m /sec^2 until it again reaches a speed of 30 m / sec as it passes from point B. the time taken by the train in travelling from A to B, a distance of 4 km is 4 min. covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for the brake of a train which is travelling at 30m /sec are appliedas train passse from point A. the brake produces a constant retardation of magnitude 3 Xm/sec^2 until the speed of train is reduced to 10 m / sec. the train travels at this speed for a distance and then uniformly accelerated at X m /sec^2 until it again reaches a speed of 30 m / sec as it passes from point B. the time taken by the train in travelling from A to B, a distance of 4 km is 4 min. .
the brake of a train which is travelling at 30m /sec are appliedas train passse from point A. the brake produces a constant retardation of magnitude 3 Xm/sec^2 until the speed of train is reduced to 10 m / sec. the train travels at this speed for a distance and then uniformly accelerated at X m /sec^2 until it again reaches a speed of 30 m / sec as it passes from point B. the time taken by the train in travelling from A to B, a distance of 4 km is 4 min. 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 the brake of a train which is travelling at 30m /sec are appliedas train passse from point A. the brake produces a constant retardation of magnitude 3 Xm/sec^2 until the speed of train is reduced to 10 m / sec. the train travels at this speed for a distance and then uniformly accelerated at X m /sec^2 until it again reaches a speed of 30 m / sec as it passes from point B. the time taken by the train in travelling from A to B, a distance of 4 km is 4 min. covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for the brake of a train which is travelling at 30m /sec are appliedas train passse from point A. the brake produces a constant retardation of magnitude 3 Xm/sec^2 until the speed of train is reduced to 10 m / sec. the train travels at this speed for a distance and then uniformly accelerated at X m /sec^2 until it again reaches a speed of 30 m / sec as it passes from point B. the time taken by the train in travelling from A to B, a distance of 4 km is 4 min. .
Solutions for the brake of a train which is travelling at 30m /sec are appliedas train passse from point A. the brake produces a constant retardation of magnitude 3 Xm/sec^2 until the speed of train is reduced to 10 m / sec. the train travels at this speed for a distance and then uniformly accelerated at X m /sec^2 until it again reaches a speed of 30 m / sec as it passes from point B. the time taken by the train in travelling from A to B, a distance of 4 km is 4 min. in English & in Hindi are available as part of our courses for Class 11.
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the brake of a train which is travelling at 30m /sec are appliedas train passse from point A. the brake produces a constant retardation of magnitude 3 Xm/sec^2 until the speed of train is reduced to 10 m / sec. the train travels at this speed for a distance and then uniformly accelerated at X m /sec^2 until it again reaches a speed of 30 m / sec as it passes from point B. the time taken by the train in travelling from A to B, a distance of 4 km is 4 min. , a detailed solution for the brake of a train which is travelling at 30m /sec are appliedas train passse from point A. the brake produces a constant retardation of magnitude 3 Xm/sec^2 until the speed of train is reduced to 10 m / sec. the train travels at this speed for a distance and then uniformly accelerated at X m /sec^2 until it again reaches a speed of 30 m / sec as it passes from point B. the time taken by the train in travelling from A to B, a distance of 4 km is 4 min. has been provided alongside types of the brake of a train which is travelling at 30m /sec are appliedas train passse from point A. the brake produces a constant retardation of magnitude 3 Xm/sec^2 until the speed of train is reduced to 10 m / sec. the train travels at this speed for a distance and then uniformly accelerated at X m /sec^2 until it again reaches a speed of 30 m / sec as it passes from point B. the time taken by the train in travelling from A to B, a distance of 4 km is 4 min. theory, EduRev gives you an
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