
The roadway bridge over a canal is in the form of an arc of a circle of radius $20\,m$. What is the maximum speed with which a car can cross the bridge without leaving the ground at the highest point?
A) $10\,m{s^{ - 1}}$
B) $12\,m{s^{ - 1}}$
C) $14\,m{s^{ - 1}}$
D) $16\,m{s^{ - 1}}$
Answer
232.8k+ views
Hint: When the object is moving in the circular path, then the acceleration due to gravity is given by the square of the velocity of the object divided by the radius of the circular path. By using this formula, the maximum velocity or the maximum speed of the car can be determined.
Useful formula:
The acceleration of the circular motion of the object is given by,
$\dfrac{{{V^2}}}{R} = g$
Where, $V$ is the velocity of the car, $R$ is the radius of the circular path and $g$ is the acceleration due to gravity.
Complete step by step solution:
Given that,
The radius of the circular path is, $R = 20\,m$
The acceleration due to gravity is, $g = 9.81\,m{s^{ - 1}}$
Now,
The acceleration of the circular motion of the object is given by,
$\dfrac{{{V^2}}}{R} = g\,.........................\left( 1 \right)$
By keeping the velocity in one side and the other terms in the other side, then the above equation (1) is written as,
${V^2} = R \times g$
By taking square root on the both sides, then the above equation is written as,
$V = \sqrt {R \times g} \,..............\left( 2 \right)$
By substituting the radius of the circular path and the acceleration due to gravity in the above equation (2), then the above equation (2) is written as,
$V = \sqrt {20 \times 9.8} $
On multiplying the above equation, then the above equation is written as,
$V = \sqrt {196} $
By taking the square root then the above equation is written as,
$V = 14\,m{s^{ - 1}}$
Thus, the above equation shows the maximum speed or velocity of the car.
Hence, the option (C) is the correct answer.
Note: From the equation (2), the velocity is directly proportional to the radius of the circular path. As the radius of the circular path increases, the velocity of the car will increase. In real time, imagine that we are driving a vehicle, if the circular path radius is less the speed also less. If the radius of the circular path is more the velocity is more.
Useful formula:
The acceleration of the circular motion of the object is given by,
$\dfrac{{{V^2}}}{R} = g$
Where, $V$ is the velocity of the car, $R$ is the radius of the circular path and $g$ is the acceleration due to gravity.
Complete step by step solution:
Given that,
The radius of the circular path is, $R = 20\,m$
The acceleration due to gravity is, $g = 9.81\,m{s^{ - 1}}$
Now,
The acceleration of the circular motion of the object is given by,
$\dfrac{{{V^2}}}{R} = g\,.........................\left( 1 \right)$
By keeping the velocity in one side and the other terms in the other side, then the above equation (1) is written as,
${V^2} = R \times g$
By taking square root on the both sides, then the above equation is written as,
$V = \sqrt {R \times g} \,..............\left( 2 \right)$
By substituting the radius of the circular path and the acceleration due to gravity in the above equation (2), then the above equation (2) is written as,
$V = \sqrt {20 \times 9.8} $
On multiplying the above equation, then the above equation is written as,
$V = \sqrt {196} $
By taking the square root then the above equation is written as,
$V = 14\,m{s^{ - 1}}$
Thus, the above equation shows the maximum speed or velocity of the car.
Hence, the option (C) is the correct answer.
Note: From the equation (2), the velocity is directly proportional to the radius of the circular path. As the radius of the circular path increases, the velocity of the car will increase. In real time, imagine that we are driving a vehicle, if the circular path radius is less the speed also less. If the radius of the circular path is more the velocity is more.
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