Answer
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Hint: The acceleration of an object moving on a circular path is the vector sum of tangential acceleration and centripetal acceleration. The acceleration in circular motion, which changes the magnitude of velocity, is the tangential acceleration.
Complete step by step solution:
The radius of the circular path along which the car is moving is 500 m. Its velocity is 30 m/s. Since the car is moving on a circular path, it has two components of acceleration. One is tangential which acts tangent to the circular path and another is centripetal which acts towards the center and allows the car to move on a circular path. It is given that the speed is increasing at the rate 0f 2 m/sec$^2$. This is the tangential acceleration of the car. Let us denote it as a$_t$.
${a_t} = 2$ m/sec$^2$
The centripetal component of acceleration can be found by using the following formula,
${a_c} = \dfrac{{{v^2}}}{r}$ …equation (1)
On substituting the values of the velocity and the radius of circular path in equation (1), we obtain,
${a_c} = \dfrac{{{{\left( {30} \right)}^2}}}{{500}} = \dfrac{{900}}{{500}} = 1.8$ m/sec$^2$
Now, the total acceleration can be found by taking the vector sum of both the components of acceleration. The tangential and centripetal acceleration are at right angles to each other because the tangential acceleration is along the tangent to the circular path and centripetal acceleration is along the radius. Therefore the cosine of the angle between the two components of acceleration is 0. The total acceleration is as follows,
$a = \sqrt {a_t^2 + a_c^2} = \sqrt {{{\left( 2 \right)}^2} + {{\left( {1.8} \right)}^2}} = \sqrt {7.24} = 2.7$ m/sec$^2$
Hence, the total acceleration of the car is 2.7 m/sec$^2$.
Therefore, the correct answer is option (D).
Note: When the magnitude of velocity is not changing while moving on a circular path, its tangential acceleration is zero. Since the direction of velocity is changing at every point, the acceleration is never non-zero, even with constant magnitude.
Complete step by step solution:
The radius of the circular path along which the car is moving is 500 m. Its velocity is 30 m/s. Since the car is moving on a circular path, it has two components of acceleration. One is tangential which acts tangent to the circular path and another is centripetal which acts towards the center and allows the car to move on a circular path. It is given that the speed is increasing at the rate 0f 2 m/sec$^2$. This is the tangential acceleration of the car. Let us denote it as a$_t$.
${a_t} = 2$ m/sec$^2$
The centripetal component of acceleration can be found by using the following formula,
${a_c} = \dfrac{{{v^2}}}{r}$ …equation (1)
On substituting the values of the velocity and the radius of circular path in equation (1), we obtain,
${a_c} = \dfrac{{{{\left( {30} \right)}^2}}}{{500}} = \dfrac{{900}}{{500}} = 1.8$ m/sec$^2$
Now, the total acceleration can be found by taking the vector sum of both the components of acceleration. The tangential and centripetal acceleration are at right angles to each other because the tangential acceleration is along the tangent to the circular path and centripetal acceleration is along the radius. Therefore the cosine of the angle between the two components of acceleration is 0. The total acceleration is as follows,
$a = \sqrt {a_t^2 + a_c^2} = \sqrt {{{\left( 2 \right)}^2} + {{\left( {1.8} \right)}^2}} = \sqrt {7.24} = 2.7$ m/sec$^2$
Hence, the total acceleration of the car is 2.7 m/sec$^2$.
Therefore, the correct answer is option (D).
Note: When the magnitude of velocity is not changing while moving on a circular path, its tangential acceleration is zero. Since the direction of velocity is changing at every point, the acceleration is never non-zero, even with constant magnitude.
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