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A vehicle travelling at 40 kmph was stopped within 1.8 seconds after the application of brakes. The retardation (in m/s2) of vehicle is (Answer up to two decimal places)
Correct answer is '6.17'. Can you explain this answer?
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A vehicle travelling at 40 kmph was stopped within 1.8 seconds after ...
U = 40 kmph = 11.11 mps
t = 1.8 s
v = u + at
0 = 11.11 + a × 1.8
∴ a = - 6.17 m/s2
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A vehicle travelling at 40 kmph was stopped within 1.8 seconds after ...
Understanding the Problem
To calculate the retardation of the vehicle, we need to analyze the given parameters:
- Initial speed (u) = 40 km/h
- Time taken to stop (t) = 1.8 seconds
First, we need to convert the speed from kilometers per hour to meters per second.
Conversion of Speed
- 1 km/h = 1000 m / 3600 s
- Therefore, 40 km/h = (40 × 1000) / 3600 = 11.11 m/s (approximately)
Using the Equation of Motion
We can use the first equation of motion:
v = u + at
Where:
- v = final velocity (0 m/s, since the vehicle stops)
- u = initial velocity (11.11 m/s)
- a = acceleration (which will be negative in this case, as it is retardation)
- t = time (1.8 seconds)
Rearranging the equation gives:
a = (v - u) / t
Substituting the values:
a = (0 - 11.11) / 1.8
a = -11.11 / 1.8
a ≈ -6.17 m/s²
Here, the negative sign indicates that it is retardation.
Conclusion
The retardation of the vehicle is approximately:
6.17 m/s²
This means the vehicle decelerated at a rate of 6.17 meters per second squared until it came to a complete stop.
To calculate the retardation of the vehicle, we need to analyze the given parameters:
- Initial speed (u) = 40 km/h
- Time taken to stop (t) = 1.8 seconds
First, we need to convert the speed from kilometers per hour to meters per second.
Conversion of Speed
- 1 km/h = 1000 m / 3600 s
- Therefore, 40 km/h = (40 × 1000) / 3600 = 11.11 m/s (approximately)
Using the Equation of Motion
We can use the first equation of motion:
v = u + at
Where:
- v = final velocity (0 m/s, since the vehicle stops)
- u = initial velocity (11.11 m/s)
- a = acceleration (which will be negative in this case, as it is retardation)
- t = time (1.8 seconds)
Rearranging the equation gives:
a = (v - u) / t
Substituting the values:
a = (0 - 11.11) / 1.8
a = -11.11 / 1.8
a ≈ -6.17 m/s²
Here, the negative sign indicates that it is retardation.
Conclusion
The retardation of the vehicle is approximately:
6.17 m/s²
This means the vehicle decelerated at a rate of 6.17 meters per second squared until it came to a complete stop.
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A vehicle travelling at 40 kmph was stopped within 1.8 seconds after the application of brakes. The retardation (in m/s2) of vehicle is (Answer up to two decimal places)Correct answer is '6.17'. Can you explain this answer?
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A vehicle travelling at 40 kmph was stopped within 1.8 seconds after the application of brakes. The retardation (in m/s2) of vehicle is (Answer up to two decimal places)Correct answer is '6.17'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about A vehicle travelling at 40 kmph was stopped within 1.8 seconds after the application of brakes. The retardation (in m/s2) of vehicle is (Answer up to two decimal places)Correct answer is '6.17'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A vehicle travelling at 40 kmph was stopped within 1.8 seconds after the application of brakes. The retardation (in m/s2) of vehicle is (Answer up to two decimal places)Correct answer is '6.17'. Can you explain this answer?.
A vehicle travelling at 40 kmph was stopped within 1.8 seconds after the application of brakes. The retardation (in m/s2) of vehicle is (Answer up to two decimal places)Correct answer is '6.17'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about A vehicle travelling at 40 kmph was stopped within 1.8 seconds after the application of brakes. The retardation (in m/s2) of vehicle is (Answer up to two decimal places)Correct answer is '6.17'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A vehicle travelling at 40 kmph was stopped within 1.8 seconds after the application of brakes. The retardation (in m/s2) of vehicle is (Answer up to two decimal places)Correct answer is '6.17'. Can you explain this answer?.
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