A particle moves from $A$ to $B$ diametrically opposite in a circle of radius $5\,\,m$ with a velocity $10\,\,m{s^{ - 1}}$. Find the average acceleration.
(A) Zero.
(B) $\dfrac{{40}}{\pi }\,\,m{s^{ - 2}}$.
(C) $\dfrac{{20}}{\pi }\,\,m{s^{ - 2}}$.
(D) none.

Hint:The particle that travels in a diametrically opposite direction, the velocity of the velocity particle is opposite to initial velocity. Average acceleration is the asses at which the velocity changes or it is the vary in velocity divided by an progressed time.

Formulae Used:
The average acceleration of the particle that moves from $A$ to $B$ is;
$a = \dfrac{v}{t}$
Where, $a$ denotes the acceleration on the particle, $v$ denotes the velocity of the particle, $t$ denotes the time taken by the particle from the point $A$ to $B$.

Complete step-by-step solution:
The data given in the problem are;
Radius of the circle, $r = 5\,\,m$,
Velocity of the particle, $v = 10\,\,m{s^{ - 1}}$.
Time taken:
In order to find the time taken by the particle to move from the point $A$ to $B$;
$t = \dfrac{{\pi r}}{v}$
Where, $r$ denotes the radius of the circle.
Substitute the values of $r$ and $v$;
$t = \dfrac{{\pi \times 5\,\,m{s^{ - 1}}}}{{10\,\,m{s^{ - 1}}}}$
By simplifying the above equation, we get;
$t = \dfrac{\pi }{2}\,\,s$
Therefore, the time taken by the particle to move from the point $A$ to $B$ is $t = \dfrac{\pi }{2}\,\,s$ .
Average acceleration:
The average acceleration of the particle that moves from $A$ to $B$ is;
$a = \dfrac{v}{t}$
Hence the change in velocity is $v = 20\,\,m{s^{ - 1}}$
Substitute the values of $t$ and $v$;
$a = \dfrac{{20\,\,m{s^{ - 1}}}}{{\dfrac{\pi }{2}\,\,s}}$
By simplifying the above equation, we get;
$a = \dfrac{{40}}{\pi }\,\,m{s^{ - 2}}$
The average acceleration of the particle that moves from $A$ to $B$ is $a = \dfrac{{40}}{\pi }\,\,m{s^{ - 2}}$.
Hence the option (B) $a = \dfrac{{40}}{\pi }\,\,m{s^{ - 2}}$ is the correct answer.

Note:- The lines that pass between the middle circle cross the area in two points, these two points are called antipodal. This is known as the diametrically opposing points of the circle.