
A car travelling at $20m/s$ on a circular road of radius 100m. It is increasing in speed at the rate of $3m/{s^2}$. Its acceleration is
(a) $3m/{s^2}$
(b) $4m/{s^2}$
(c) $5m/{s^2}$
(d) $7m/{s^2}$
Answer
588.3k+ views
Hint: This is a case of circular motion. Total acceleration will be the resultant of centripetal and tangential accelerations for an object moving on a circular path.
Formula used:
1. Centripetal acceleration of a body in circular motion: ${a_c} = \dfrac{{{v^2}}}{r}$ ……(1)
Where,
$v$ is the initial speed of the body
$r$ is the radius of the circle on which the body is moving.
2. Total acceleration of the body: ${a_{total}} = \sqrt {{a_r}^2 + {a_t}^2} $ ……(2)
Where,
${a_r}$ is the acceleration in the radial direction
${a_t}$ is the acceleration in the tangential direction
Complete step by step answer:
Given:
1. Initial speed of the car $v = 20m/s$
2. Radius of circular road $r = 100m$
3. Acceleration of the car in the tangential direction $a = 3m/{s^2}$
To find: The acceleration of the car.
Step 1 of 3:
Use eq (1) to find the centripetal acceleration experienced by the car:
${a_c} = \dfrac{{{{20}^2}}}{{100}} $
${a_c} = 4m/{s^2} $
Since, centripetal acceleration acts towards the center of the circle, it will point in the radial direction for each point on the circle. We can write:
${a_c} = {a_r} $
${a_r} = 4m/{s^2} $
Step 2 of 3:
When the car accelerates in its direction of motion, it will always be tangential its point on the circle. Hence, we can write:
${a_t} = a $
${a_t} = 3m/{s^2} $
Step 3 of 3:
The 2 components of acceleration ${a_t}$ and ${a_r}$ are perpendicular to each other. So, we can use eq (2) to find the magnitude of the resultant acceleration:
${a_{total}} = \sqrt {{4^2} + {3^2}} $
${a_{total}} = 5m/{s^2} $
The acceleration of the car is $5m/{s^2}$. Hence, option (c) is correct.
Note:
Remember to distinguish between the radial and tangential acceleration. As they are perpendicular, their resultant can be found somewhere close to circumference pointing in the direction of motion. If there is non-zero tangential acceleration, then there would be an increase in velocity so stronger the centripetal force required to hold the motion to the same path is needed.
Formula used:
1. Centripetal acceleration of a body in circular motion: ${a_c} = \dfrac{{{v^2}}}{r}$ ……(1)
Where,
$v$ is the initial speed of the body
$r$ is the radius of the circle on which the body is moving.
2. Total acceleration of the body: ${a_{total}} = \sqrt {{a_r}^2 + {a_t}^2} $ ……(2)
Where,
${a_r}$ is the acceleration in the radial direction
${a_t}$ is the acceleration in the tangential direction
Complete step by step answer:
Given:
1. Initial speed of the car $v = 20m/s$
2. Radius of circular road $r = 100m$
3. Acceleration of the car in the tangential direction $a = 3m/{s^2}$
To find: The acceleration of the car.
Step 1 of 3:
Use eq (1) to find the centripetal acceleration experienced by the car:
${a_c} = \dfrac{{{{20}^2}}}{{100}} $
${a_c} = 4m/{s^2} $
Since, centripetal acceleration acts towards the center of the circle, it will point in the radial direction for each point on the circle. We can write:
${a_c} = {a_r} $
${a_r} = 4m/{s^2} $
Step 2 of 3:
When the car accelerates in its direction of motion, it will always be tangential its point on the circle. Hence, we can write:
${a_t} = a $
${a_t} = 3m/{s^2} $
Step 3 of 3:
The 2 components of acceleration ${a_t}$ and ${a_r}$ are perpendicular to each other. So, we can use eq (2) to find the magnitude of the resultant acceleration:
${a_{total}} = \sqrt {{4^2} + {3^2}} $
${a_{total}} = 5m/{s^2} $
The acceleration of the car is $5m/{s^2}$. Hence, option (c) is correct.
Note:
Remember to distinguish between the radial and tangential acceleration. As they are perpendicular, their resultant can be found somewhere close to circumference pointing in the direction of motion. If there is non-zero tangential acceleration, then there would be an increase in velocity so stronger the centripetal force required to hold the motion to the same path is needed.
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