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The engine of a motorcycle can produce a maximum acceleration 5 m/s^2 .it's break can produce a maximum retardation of 10 m/s ^2 . What is the minimum time in which it can cover a distance of of 1.5 km ?
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Calculation of Minimum Time Required for a Motorcycle to Cover 1.5 km Distance
Given:
Maximum acceleration = 5 m/s^2
Maximum retardation = 10 m/s^2
Distance to be covered = 1.5 km = 1500 m
Approach:
To calculate the minimum time required to cover a distance of 1.5 km, we need to consider the following scenarios:
1. Motorcycle achieves maximum acceleration until it covers half of the distance and then applies maximum retardation to stop.
2. Motorcycle achieves maximum acceleration until it covers the entire distance and then applies maximum retardation to stop.
Scenario 1:
In this scenario, the motorcycle will achieve maximum acceleration until it covers half the distance, i.e., 750 m. Then it will apply maximum retardation to stop.
Let's calculate the time taken to cover the first 750 m:
Initial velocity, u = 0
Final velocity, v = ?
Acceleration, a = 5 m/s^2
Distance, s = 750 m
Using the equation of motion, v^2 = u^2 + 2as, we get:
v^2 = 0 + 2(5)(750)
v = 54.77 m/s
Time taken, t1 = (v - u)/a
t1 = (54.77 - 0)/5
t1 = 10.95 s
Now, the motorcycle needs to stop. It will apply maximum retardation, i.e., a = -10 m/s^2, to stop.
Using the equation of motion, v = u + at, we get:
0 = 54.77 - 10t2
t2 = 5.48 s
Total time taken in scenario 1, t = t1 + t2
t = 10.95 + 5.48
t = 16.43 s
Scenario 2:
In this scenario, the motorcycle will achieve maximum acceleration until it covers the entire distance of 1.5 km, i.e., 1500 m. Then it will apply maximum retardation to stop.
Let's calculate the time taken to cover the entire distance:
Initial velocity, u = 0
Final velocity, v = ?
Acceleration, a = 5 m/s^2
Distance, s = 1500 m
Using the equation of motion, v^2 = u^2 + 2as, we get:
v^2 = 0 + 2(5)(1500)
v = 122.47 m/s
Time taken, t3 = (v - u)/a
t3 = (122.47 - 0)/5
t3 = 24.49 s
Now, the motorcycle needs to stop. It will apply maximum retardation, i.e., a = -10 m/s^2, to stop.
Using the equation of motion, v = u + at, we get:
0 = 122.47 - 10t4
t4 = 12.25 s
Total time taken in scenario 2, t = t3 + t4
t = 24.49 + 12.25
t = 36.74 s
Conclusion:
The minimum time required for the
Given:
Maximum acceleration = 5 m/s^2
Maximum retardation = 10 m/s^2
Distance to be covered = 1.5 km = 1500 m
Approach:
To calculate the minimum time required to cover a distance of 1.5 km, we need to consider the following scenarios:
1. Motorcycle achieves maximum acceleration until it covers half of the distance and then applies maximum retardation to stop.
2. Motorcycle achieves maximum acceleration until it covers the entire distance and then applies maximum retardation to stop.
Scenario 1:
In this scenario, the motorcycle will achieve maximum acceleration until it covers half the distance, i.e., 750 m. Then it will apply maximum retardation to stop.
Let's calculate the time taken to cover the first 750 m:
Initial velocity, u = 0
Final velocity, v = ?
Acceleration, a = 5 m/s^2
Distance, s = 750 m
Using the equation of motion, v^2 = u^2 + 2as, we get:
v^2 = 0 + 2(5)(750)
v = 54.77 m/s
Time taken, t1 = (v - u)/a
t1 = (54.77 - 0)/5
t1 = 10.95 s
Now, the motorcycle needs to stop. It will apply maximum retardation, i.e., a = -10 m/s^2, to stop.
Using the equation of motion, v = u + at, we get:
0 = 54.77 - 10t2
t2 = 5.48 s
Total time taken in scenario 1, t = t1 + t2
t = 10.95 + 5.48
t = 16.43 s
Scenario 2:
In this scenario, the motorcycle will achieve maximum acceleration until it covers the entire distance of 1.5 km, i.e., 1500 m. Then it will apply maximum retardation to stop.
Let's calculate the time taken to cover the entire distance:
Initial velocity, u = 0
Final velocity, v = ?
Acceleration, a = 5 m/s^2
Distance, s = 1500 m
Using the equation of motion, v^2 = u^2 + 2as, we get:
v^2 = 0 + 2(5)(1500)
v = 122.47 m/s
Time taken, t3 = (v - u)/a
t3 = (122.47 - 0)/5
t3 = 24.49 s
Now, the motorcycle needs to stop. It will apply maximum retardation, i.e., a = -10 m/s^2, to stop.
Using the equation of motion, v = u + at, we get:
0 = 122.47 - 10t4
t4 = 12.25 s
Total time taken in scenario 2, t = t3 + t4
t = 24.49 + 12.25
t = 36.74 s
Conclusion:
The minimum time required for the
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The engine of a motorcycle can produce a maximum acceleration 5 m/s^2 .it's break can produce a maximum retardation of 10 m/s ^2 . What is the minimum time in which it can cover a distance of of 1.5 km ?
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The engine of a motorcycle can produce a maximum acceleration 5 m/s^2 .it's break can produce a maximum retardation of 10 m/s ^2 . What is the minimum time in which it can cover a distance of of 1.5 km ? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The engine of a motorcycle can produce a maximum acceleration 5 m/s^2 .it's break can produce a maximum retardation of 10 m/s ^2 . What is the minimum time in which it can cover a distance of of 1.5 km ? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The engine of a motorcycle can produce a maximum acceleration 5 m/s^2 .it's break can produce a maximum retardation of 10 m/s ^2 . What is the minimum time in which it can cover a distance of of 1.5 km ?.
The engine of a motorcycle can produce a maximum acceleration 5 m/s^2 .it's break can produce a maximum retardation of 10 m/s ^2 . What is the minimum time in which it can cover a distance of of 1.5 km ? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The engine of a motorcycle can produce a maximum acceleration 5 m/s^2 .it's break can produce a maximum retardation of 10 m/s ^2 . What is the minimum time in which it can cover a distance of of 1.5 km ? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The engine of a motorcycle can produce a maximum acceleration 5 m/s^2 .it's break can produce a maximum retardation of 10 m/s ^2 . What is the minimum time in which it can cover a distance of of 1.5 km ?.
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