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Question: A particle moves from \(A\) to \(B\) diametrically opposite in a circle of radius \(5\,\,m\) with a ...

A particle moves from AA to BB diametrically opposite in a circle of radius 5m5\,\,m with a velocity 10ms110\,\,m{s^{ - 1}}. Find the average acceleration.
(A) Zero.
(B) 40πms2\dfrac{{40}}{\pi }\,\,m{s^{ - 2}}.
(C) 20πms2\dfrac{{20}}{\pi }\,\,m{s^{ - 2}}.
(D) none.

Explanation

Solution

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 AA to BB is;
a=vta = \dfrac{v}{t}
Where, aa denotes the acceleration on the particle, vv denotes the velocity of the particle, tt denotes the time taken by the particle from the point AA to BB.

Complete step-by-step solution:
The data given in the problem are;
Radius of the circle, r=5mr = 5\,\,m,
Velocity of the particle, v=10ms1v = 10\,\,m{s^{ - 1}}.
Time taken:
In order to find the time taken by the particle to move from the point AA to BB;
t=πrvt = \dfrac{{\pi r}}{v}
Where, rr denotes the radius of the circle.
Substitute the values of rr and vv;
t=π×5ms110ms1t = \dfrac{{\pi \times 5\,\,m{s^{ - 1}}}}{{10\,\,m{s^{ - 1}}}}
By simplifying the above equation, we get;
t=π2st = \dfrac{\pi }{2}\,\,s
Therefore, the time taken by the particle to move from the point AA to BB is t=π2st = \dfrac{\pi }{2}\,\,s .
Average acceleration:
The average acceleration of the particle that moves from AA to BB is;
a=vta = \dfrac{v}{t}
Hence the change in velocity is v=20ms1v = 20\,\,m{s^{ - 1}}
Substitute the values of tt and vv;
a=20ms1π2sa = \dfrac{{20\,\,m{s^{ - 1}}}}{{\dfrac{\pi }{2}\,\,s}}
By simplifying the above equation, we get;
a=40πms2a = \dfrac{{40}}{\pi }\,\,m{s^{ - 2}}
The average acceleration of the particle that moves from AA to BB is a=40πms2a = \dfrac{{40}}{\pi }\,\,m{s^{ - 2}}.
Hence the option (B) a=40πms2a = \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.