Answer
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Hint: Average speed is calculated by dividing the total distance that a body has travelled by the total amount of time it took it to travel that particular distance whereas the average velocity of an object is its total displacement divided by the total time taken. Average speed is a scalar quantity whereas average velocity is vector quantity.
Complete step by step solution:
Since, distance is always greater than or equal to displacement, so speed will always be greater than or equal to velocity. This concludes that the average speed of a particle in a given time is never less than the magnitude of the average velocity. So, option (A) is a correct statement.
We know that in a uniform circular motion,
$ \left| {\dfrac{{d\overrightarrow v }}{{dt}}} \right| = {\omega ^2}R \ne 0 $
Since, $ \left| {\overrightarrow v } \right| $ is constant,
$ \dfrac{d}{{dt}}\left| v \right| = 0 $
So, option (A) is the correct statement.
It is true that if the average velocity of a particle is zero in a time interval, then It is possible that the instantaneous velocity is never zero in the interval. This happens in the case of uniform circular motion, instantaneous velocity is never zero but average velocity in the time period of rotation is zero. So, option (C) is the correct statement.
If the average velocity of a particle moving in a straight line is zero, then it means that the net displacement is also zero. Now this can happen in a straight line motion only in two ways. Firstly, it is possible if the particle is at rest. Secondly, it is also possible if the particle reverses its direction at least once in the time interval. Both the conditions demand that the instantaneous velocity of the particle is zero at least once in the time interval. So, option (D) is not the correct statement.
So, (A), (B) and (C) are the correct statements.
Note:
An object undergoing uniform circular motion is always accelerating, even though the magnitude of its velocity is constant. This is because we know that acceleration is a vector quantity and in a uniform circular motion, the direction of the particle changes at every point.
Complete step by step solution:
Since, distance is always greater than or equal to displacement, so speed will always be greater than or equal to velocity. This concludes that the average speed of a particle in a given time is never less than the magnitude of the average velocity. So, option (A) is a correct statement.
We know that in a uniform circular motion,
$ \left| {\dfrac{{d\overrightarrow v }}{{dt}}} \right| = {\omega ^2}R \ne 0 $
Since, $ \left| {\overrightarrow v } \right| $ is constant,
$ \dfrac{d}{{dt}}\left| v \right| = 0 $
So, option (A) is the correct statement.
It is true that if the average velocity of a particle is zero in a time interval, then It is possible that the instantaneous velocity is never zero in the interval. This happens in the case of uniform circular motion, instantaneous velocity is never zero but average velocity in the time period of rotation is zero. So, option (C) is the correct statement.
If the average velocity of a particle moving in a straight line is zero, then it means that the net displacement is also zero. Now this can happen in a straight line motion only in two ways. Firstly, it is possible if the particle is at rest. Secondly, it is also possible if the particle reverses its direction at least once in the time interval. Both the conditions demand that the instantaneous velocity of the particle is zero at least once in the time interval. So, option (D) is not the correct statement.
So, (A), (B) and (C) are the correct statements.
Note:
An object undergoing uniform circular motion is always accelerating, even though the magnitude of its velocity is constant. This is because we know that acceleration is a vector quantity and in a uniform circular motion, the direction of the particle changes at every point.
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