
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
576.6k+ 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.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

