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A car starting for rest, accelerates at the rate f through a distance S,then continues at constant speed for time t and then decelerates as the rate f/2 to come to rest il if the total distance traveled is 15 second,then?
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A car starting for rest, accelerates at the rate f through a distance ...
Understanding the Problem
The scenario involves a car that undergoes three phases of motion: acceleration, constant speed, and deceleration. We need to calculate the distance traveled in each phase and relate them to the total distance covered in 15 seconds.
Phase 1: Acceleration
- The car starts from rest with an acceleration of 'f'.
- The distance covered during this phase is 'S'.
- Using the equation of motion, S = (1/2) * f * t1^2, where t1 is the time taken to accelerate.
Phase 2: Constant Speed
- After reaching speed 'v', the car moves at this constant speed for time 't'.
- The speed at the end of the first phase is v = f * t1.
- The distance covered during this phase is d2 = v * t = f * t1 * t.
Phase 3: Deceleration
- The car decelerates at rate f/2 until it comes to rest.
- The time taken to decelerate from speed 'v' to rest is t3.
- Using the formula for distance during deceleration:
d3 = v * t3 - (1/2) * (f/2) * t3^2.
Combining the Phases
- The total distance covered is S + d2 + d3 = 15.
- This can be expressed as:
(1/2) * f * t1^2 + f * t1 * t + (f * t1 * t3 - (1/4) * f * t3^2) = 15.
Finding Relationships
- The total time of travel is t1 + t + t3 = 15 seconds.
- By substituting t3 in terms of t1 and t, we can simplify and solve for the variables.
Conclusion
- The key is to set up the equations correctly and relate the distances covered in each phase to solve for the unknowns.
- This methodical approach will yield the values needed for 'f', 'S', 't', and 't3'.
The scenario involves a car that undergoes three phases of motion: acceleration, constant speed, and deceleration. We need to calculate the distance traveled in each phase and relate them to the total distance covered in 15 seconds.
Phase 1: Acceleration
- The car starts from rest with an acceleration of 'f'.
- The distance covered during this phase is 'S'.
- Using the equation of motion, S = (1/2) * f * t1^2, where t1 is the time taken to accelerate.
Phase 2: Constant Speed
- After reaching speed 'v', the car moves at this constant speed for time 't'.
- The speed at the end of the first phase is v = f * t1.
- The distance covered during this phase is d2 = v * t = f * t1 * t.
Phase 3: Deceleration
- The car decelerates at rate f/2 until it comes to rest.
- The time taken to decelerate from speed 'v' to rest is t3.
- Using the formula for distance during deceleration:
d3 = v * t3 - (1/2) * (f/2) * t3^2.
Combining the Phases
- The total distance covered is S + d2 + d3 = 15.
- This can be expressed as:
(1/2) * f * t1^2 + f * t1 * t + (f * t1 * t3 - (1/4) * f * t3^2) = 15.
Finding Relationships
- The total time of travel is t1 + t + t3 = 15 seconds.
- By substituting t3 in terms of t1 and t, we can simplify and solve for the variables.
Conclusion
- The key is to set up the equations correctly and relate the distances covered in each phase to solve for the unknowns.
- This methodical approach will yield the values needed for 'f', 'S', 't', and 't3'.
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A car starting for rest, accelerates at the rate f through a distance S,then continues at constant speed for time t and then decelerates as the rate f/2 to come to rest il if the total distance traveled is 15 second,then?
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A car starting for rest, accelerates at the rate f through a distance S,then continues at constant speed for time t and then decelerates as the rate f/2 to come to rest il if the total distance traveled is 15 second,then? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A car starting for rest, accelerates at the rate f through a distance S,then continues at constant speed for time t and then decelerates as the rate f/2 to come to rest il if the total distance traveled is 15 second,then? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A car starting for rest, accelerates at the rate f through a distance S,then continues at constant speed for time t and then decelerates as the rate f/2 to come to rest il if the total distance traveled is 15 second,then?.
A car starting for rest, accelerates at the rate f through a distance S,then continues at constant speed for time t and then decelerates as the rate f/2 to come to rest il if the total distance traveled is 15 second,then? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A car starting for rest, accelerates at the rate f through a distance S,then continues at constant speed for time t and then decelerates as the rate f/2 to come to rest il if the total distance traveled is 15 second,then? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A car starting for rest, accelerates at the rate f through a distance S,then continues at constant speed for time t and then decelerates as the rate f/2 to come to rest il if the total distance traveled is 15 second,then?.
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