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A car starts from rest and accelerates uniformly at the rate of 1m/s sq. for 5sec itthen maintains a constant velocity for 30sec . Then brakes are applied and the car is uniformly retarded to the rest in 10sec . Find the maxinum velocity attained by the car and the total distance travelled by it?
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A car starts from rest and accelerates uniformly at the rate of 1m/s s...
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A car starts from rest and accelerates uniformly at the rate of 1m/s s...
I solve it. Is the answer as follows: === maximum velocity =5m/s and distance travelled=212.5meter.
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A car starts from rest and accelerates uniformly at the rate of 1m/s s...
Given data:
- Acceleration (a) = 1 m/s²
- Time taken to accelerate (t₁) = 5 seconds
- Time taken to maintain constant velocity (t₂) = 30 seconds
- Time taken to retard (t₃) = 10 seconds
To find:
- Maximum velocity attained by the car (v_max)
- Total distance travelled by the car (d_total)
Calculating maximum velocity (v_max):
- The car starts from rest, so the initial velocity (u) is 0 m/s.
- Using the equation: v = u + at, where v is the final velocity attained after time t, we can find the final velocity after the acceleration phase.
- Substituting the values, v = 0 + (1 m/s²)(5 s) = 5 m/s.
- Therefore, the maximum velocity attained by the car is 5 m/s.
Calculating distance travelled during acceleration:
- Using the equation: s = ut + (1/2)at², where s is the distance travelled, u is the initial velocity, a is the acceleration, and t is the time taken.
- Substituting the values, s₁ = (0)(5 s) + (1/2)(1 m/s²)(5 s)² = 12.5 m.
Calculating distance travelled during constant velocity:
- During this phase, the velocity remains constant at 5 m/s for 30 seconds.
- Using the equation: s = vt, where s is the distance travelled, v is the constant velocity, and t is the time taken.
- Substituting the values, s₂ = (5 m/s)(30 s) = 150 m.
Calculating distance travelled during retardation:
- The car decelerates uniformly with a negative acceleration to come to rest.
- Using the same equation, s = ut + (1/2)at², but with a negative acceleration (-1 m/s²) and the final velocity (v) as 0 m/s.
- Substituting the values, s₃ = (5 m/s)(10 s) + (1/2)(-1 m/s²)(10 s)² = 25 m.
Calculating total distance travelled:
- The total distance travelled is the sum of the distances travelled during the acceleration, constant velocity, and retardation phases.
- Therefore, d_total = s₁ + s₂ + s₃ = 12.5 m + 150 m + 25 m = 187.5 m.
Summary:
The maximum velocity attained by the car is 5 m/s, and the total distance travelled by the car is 187.5 m.
- Acceleration (a) = 1 m/s²
- Time taken to accelerate (t₁) = 5 seconds
- Time taken to maintain constant velocity (t₂) = 30 seconds
- Time taken to retard (t₃) = 10 seconds
To find:
- Maximum velocity attained by the car (v_max)
- Total distance travelled by the car (d_total)
Calculating maximum velocity (v_max):
- The car starts from rest, so the initial velocity (u) is 0 m/s.
- Using the equation: v = u + at, where v is the final velocity attained after time t, we can find the final velocity after the acceleration phase.
- Substituting the values, v = 0 + (1 m/s²)(5 s) = 5 m/s.
- Therefore, the maximum velocity attained by the car is 5 m/s.
Calculating distance travelled during acceleration:
- Using the equation: s = ut + (1/2)at², where s is the distance travelled, u is the initial velocity, a is the acceleration, and t is the time taken.
- Substituting the values, s₁ = (0)(5 s) + (1/2)(1 m/s²)(5 s)² = 12.5 m.
Calculating distance travelled during constant velocity:
- During this phase, the velocity remains constant at 5 m/s for 30 seconds.
- Using the equation: s = vt, where s is the distance travelled, v is the constant velocity, and t is the time taken.
- Substituting the values, s₂ = (5 m/s)(30 s) = 150 m.
Calculating distance travelled during retardation:
- The car decelerates uniformly with a negative acceleration to come to rest.
- Using the same equation, s = ut + (1/2)at², but with a negative acceleration (-1 m/s²) and the final velocity (v) as 0 m/s.
- Substituting the values, s₃ = (5 m/s)(10 s) + (1/2)(-1 m/s²)(10 s)² = 25 m.
Calculating total distance travelled:
- The total distance travelled is the sum of the distances travelled during the acceleration, constant velocity, and retardation phases.
- Therefore, d_total = s₁ + s₂ + s₃ = 12.5 m + 150 m + 25 m = 187.5 m.
Summary:
The maximum velocity attained by the car is 5 m/s, and the total distance travelled by the car is 187.5 m.
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A car starts from rest and accelerates uniformly at the rate of 1m/s sq. for 5sec itthen maintains a constant velocity for 30sec . Then brakes are applied and the car is uniformly retarded to the rest in 10sec . Find the maxinum velocity attained by the car and the total distance travelled by it?
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A car starts from rest and accelerates uniformly at the rate of 1m/s sq. for 5sec itthen maintains a constant velocity for 30sec . Then brakes are applied and the car is uniformly retarded to the rest in 10sec . Find the maxinum velocity attained by the car and the total distance travelled by it? 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 car starts from rest and accelerates uniformly at the rate of 1m/s sq. for 5sec itthen maintains a constant velocity for 30sec . Then brakes are applied and the car is uniformly retarded to the rest in 10sec . Find the maxinum velocity attained by the car and the total distance travelled by it? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A car starts from rest and accelerates uniformly at the rate of 1m/s sq. for 5sec itthen maintains a constant velocity for 30sec . Then brakes are applied and the car is uniformly retarded to the rest in 10sec . Find the maxinum velocity attained by the car and the total distance travelled by it?.
A car starts from rest and accelerates uniformly at the rate of 1m/s sq. for 5sec itthen maintains a constant velocity for 30sec . Then brakes are applied and the car is uniformly retarded to the rest in 10sec . Find the maxinum velocity attained by the car and the total distance travelled by it? 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 car starts from rest and accelerates uniformly at the rate of 1m/s sq. for 5sec itthen maintains a constant velocity for 30sec . Then brakes are applied and the car is uniformly retarded to the rest in 10sec . Find the maxinum velocity attained by the car and the total distance travelled by it? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A car starts from rest and accelerates uniformly at the rate of 1m/s sq. for 5sec itthen maintains a constant velocity for 30sec . Then brakes are applied and the car is uniformly retarded to the rest in 10sec . Find the maxinum velocity attained by the car and the total distance travelled by it?.
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