Question

A car is travelling with a linear velocity v on a circular road of radius r. If it is increasing its speed at the rate of $a\,\,{\text{m/}}{{\text{s}}^{\text{2}}}$, then the resultant acceleration will be
A. $\left( {\dfrac{{{v^2}}}{r} + a} \right)$
B. ${\left( {\dfrac{{{v^2}}}{{{r^2}}} + a} \right)^{1/2}}$
C. ${\left( {\dfrac{{{v^2}}}{{{r^2}}} + {a^2}} \right)^{1/2}}$
D. ${\left( {\dfrac{{{v^2}}}{{{r^2}}} - {a^2}} \right)^{1/2}}$

Answer 1

The motion of the car is non-uniform circular motion because its speed is increasing. In the non-uniform circular motion, the total acceleration of the car is the sum of its tangential acceleration vector and radial acceleration vector. Recall the expression for the centripetal acceleration to determine the total acceleration of the car.

Formula used:
${\vec a_T} = {\vec a_C} + {\vec a_t}$
Here, ${\vec a_T}$ is the total acceleration vector, ${\vec a_C}$ is the centripetal acceleration vector and ${\vec a_t}$ is the tangential acceleration vector.
Centripetal acceleration, ${a_C} = \dfrac{{{v^2}}}{r}$
Here, v is the velocity of the car on the circular track and r is the radius of the circular track.

Complete step by step answer:
We have given that the velocity of the car is increasing. Therefore, the motion of the car is non-uniform circular motion. In the non-uniform circular motion, the total acceleration of the car is the sum of its tangential acceleration and radial acceleration (centripetal acceleration). That is,
${\vec a_T} = {\vec a_C} + {\vec a_t}$
Here, ${\vec a_T}$ is the total acceleration vector, ${\vec a_C}$ is the centripetal acceleration vector and ${\vec a_t}$ is the tangential acceleration vector.

In terms of magnitude, we can write the above equation as,
$a_T^2 = a_C^2 + a_t^2$
$\Rightarrow {a_T} = \sqrt {a_C^2 + a_t^2}$

We have given that the tangential acceleration of the car is $a\,\,{\text{m/}}{{\text{s}}^{\text{2}}}$. Therefore, the above equation becomes,
${a_T} = \sqrt {a_C^2 + {a^2}}$ …… (1)

We have the expression for centripetal acceleration of the car,
${a_C} = \dfrac{{{v^2}}}{r}$
Here, v is the velocity of the car on the circular track and r is the radius of the circular track.

Using the above equation in equation (1), we get,
${a_T} = \sqrt {\dfrac{{{v^4}}}{{{r^2}}} + {a^2}}$
$\therefore {a_T} = {\left( {\dfrac{{{v^4}}}{{{r^2}}} + {a^2}} \right)^{1/2}}$

So, the correct answer is option C.

Note: If the car has the uniform motion on the circular track, the tangential acceleration of the car will be zero and the total acceleration of the car will be the centripetal acceleration towards the centre. Note that both the centripetal acceleration and tangential acceleration are vector quantities. Therefore, the total acceleration is not the sum of magnitude of the centripetal acceleration and the tangential acceleration.