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A body is revolving in a circle of diameter 0.40m at the rate of 1000 revolution/minute. Calculate its linear velocity and centripetal accelerationA car accelerates from rest at constant rate A for some time after which it retards at a constant rate B to come to rest. If the time elapsed is T seconds, Calculate the maximum velocity reached and the total distance travelled in terms A,B, and T.?
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A body is revolving in a circle of diameter 0.40m at the rate of 1000 ...
Linear Velocity:
The linear velocity of a body moving in a circle can be calculated using the formula:
v = πdN
where v is the linear velocity, d is the diameter of the circle, and N is the number of revolutions per unit time.
Given that the diameter of the circle is 0.40m and the body is making 1000 revolutions per minute, we can calculate the linear velocity as follows:
v = π(0.40m)(1000 rev/min)
v = 3.14(0.40m)(1000/60 min/s)
v ≈ 20.93 m/s
Therefore, the linear velocity of the body is approximately 20.93 m/s.
Centripetal Acceleration:
The centripetal acceleration of a body moving in a circle can be calculated using the formula:
a = (v^2)/r
where a is the centripetal acceleration, v is the linear velocity, and r is the radius of the circle.
Since the diameter of the circle is given, we need to calculate the radius first:
r = d/2
r = 0.40m/2
r = 0.20m
Now we can calculate the centripetal acceleration:
a = (20.93 m/s)^2 / 0.20m
a ≈ 216.84 m/s^2
Therefore, the centripetal acceleration of the body is approximately 216.84 m/s^2.
Maximum Velocity and Total Distance Traveled:
To calculate the maximum velocity reached by the car, we need to determine the time taken to accelerate and decelerate. Let's assume the car accelerates for t1 seconds and decelerates for t2 seconds.
During acceleration, the velocity of the car can be calculated using the equation:
v1 = at1
During deceleration, the velocity of the car can be calculated using the equation:
v2 = -Bt2
The maximum velocity is reached when the acceleration becomes zero, so:
at1 = -Bt2
Simplifying the equation, we get:
t2 = (A/B)t1
The total time is given as T seconds, so:
T = t1 + t2
T = t1 + (A/B)t1
T = t1(1 + A/B)
Solving for t1, we get:
t1 = T / (1 + A/B)
Substituting this value back into the equation for t2, we get:
t2 = (A/B)(T / (1 + A/B))
t2 = (A/B)T / (1 + A/B)
The maximum velocity is given by:
vmax = at1 = A(T / (1 + A/B))
The total distance traveled can be calculated using the equation:
s = (1/2)at1^2 + v1t1 + (1/2)(-B)t2^2 + v2t2
Substituting the values of v1, v2, t1, and t2, we get:
s = (1/2)A(T / (1 + A/B))^2 + A(T / (1 + A/B))(T / (1 + A/B)) + (1/2)(-B)(A/BT / (1
The linear velocity of a body moving in a circle can be calculated using the formula:
v = πdN
where v is the linear velocity, d is the diameter of the circle, and N is the number of revolutions per unit time.
Given that the diameter of the circle is 0.40m and the body is making 1000 revolutions per minute, we can calculate the linear velocity as follows:
v = π(0.40m)(1000 rev/min)
v = 3.14(0.40m)(1000/60 min/s)
v ≈ 20.93 m/s
Therefore, the linear velocity of the body is approximately 20.93 m/s.
Centripetal Acceleration:
The centripetal acceleration of a body moving in a circle can be calculated using the formula:
a = (v^2)/r
where a is the centripetal acceleration, v is the linear velocity, and r is the radius of the circle.
Since the diameter of the circle is given, we need to calculate the radius first:
r = d/2
r = 0.40m/2
r = 0.20m
Now we can calculate the centripetal acceleration:
a = (20.93 m/s)^2 / 0.20m
a ≈ 216.84 m/s^2
Therefore, the centripetal acceleration of the body is approximately 216.84 m/s^2.
Maximum Velocity and Total Distance Traveled:
To calculate the maximum velocity reached by the car, we need to determine the time taken to accelerate and decelerate. Let's assume the car accelerates for t1 seconds and decelerates for t2 seconds.
During acceleration, the velocity of the car can be calculated using the equation:
v1 = at1
During deceleration, the velocity of the car can be calculated using the equation:
v2 = -Bt2
The maximum velocity is reached when the acceleration becomes zero, so:
at1 = -Bt2
Simplifying the equation, we get:
t2 = (A/B)t1
The total time is given as T seconds, so:
T = t1 + t2
T = t1 + (A/B)t1
T = t1(1 + A/B)
Solving for t1, we get:
t1 = T / (1 + A/B)
Substituting this value back into the equation for t2, we get:
t2 = (A/B)(T / (1 + A/B))
t2 = (A/B)T / (1 + A/B)
The maximum velocity is given by:
vmax = at1 = A(T / (1 + A/B))
The total distance traveled can be calculated using the equation:
s = (1/2)at1^2 + v1t1 + (1/2)(-B)t2^2 + v2t2
Substituting the values of v1, v2, t1, and t2, we get:
s = (1/2)A(T / (1 + A/B))^2 + A(T / (1 + A/B))(T / (1 + A/B)) + (1/2)(-B)(A/BT / (1
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A body is revolving in a circle of diameter 0.40m at the rate of 1000 revolution/minute. Calculate its linear velocity and centripetal accelerationA car accelerates from rest at constant rate A for some time after which it retards at a constant rate B to come to rest. If the time elapsed is T seconds, Calculate the maximum velocity reached and the total distance travelled in terms A,B, and T.?
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A body is revolving in a circle of diameter 0.40m at the rate of 1000 revolution/minute. Calculate its linear velocity and centripetal accelerationA car accelerates from rest at constant rate A for some time after which it retards at a constant rate B to come to rest. If the time elapsed is T seconds, Calculate the maximum velocity reached and the total distance travelled in terms A,B, and T.? 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 A body is revolving in a circle of diameter 0.40m at the rate of 1000 revolution/minute. Calculate its linear velocity and centripetal accelerationA car accelerates from rest at constant rate A for some time after which it retards at a constant rate B to come to rest. If the time elapsed is T seconds, Calculate the maximum velocity reached and the total distance travelled in terms A,B, and T.? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A body is revolving in a circle of diameter 0.40m at the rate of 1000 revolution/minute. Calculate its linear velocity and centripetal accelerationA car accelerates from rest at constant rate A for some time after which it retards at a constant rate B to come to rest. If the time elapsed is T seconds, Calculate the maximum velocity reached and the total distance travelled in terms A,B, and T.?.
A body is revolving in a circle of diameter 0.40m at the rate of 1000 revolution/minute. Calculate its linear velocity and centripetal accelerationA car accelerates from rest at constant rate A for some time after which it retards at a constant rate B to come to rest. If the time elapsed is T seconds, Calculate the maximum velocity reached and the total distance travelled in terms A,B, and T.? 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 A body is revolving in a circle of diameter 0.40m at the rate of 1000 revolution/minute. Calculate its linear velocity and centripetal accelerationA car accelerates from rest at constant rate A for some time after which it retards at a constant rate B to come to rest. If the time elapsed is T seconds, Calculate the maximum velocity reached and the total distance travelled in terms A,B, and T.? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A body is revolving in a circle of diameter 0.40m at the rate of 1000 revolution/minute. Calculate its linear velocity and centripetal accelerationA car accelerates from rest at constant rate A for some time after which it retards at a constant rate B to come to rest. If the time elapsed is T seconds, Calculate the maximum velocity reached and the total distance travelled in terms A,B, and T.?.
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