
It has been found that a particle is in motion along a circular path of radius $5m$ with a uniform speed of $5m{{s}^{-1}}$. Calculate the magnitude of the average acceleration in the time interval at which the particle is completing a half revolution?
\[\begin{align}
& A.zero \\
& B.10m{{s}^{-1}} \\
& C.10\pi m{{s}^{-1}} \\
& D.\dfrac{10}{\pi }m{{s}^{-1}} \\
\end{align}\]
Answer
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Hint: Time taken by the particle in order to complete the mentioned half revolution can be found by taking the ratio of the distance covered by the particle and the velocity of the travel. The average acceleration can be found by taking the ratio of the total velocity to the total time taken. This will help you in answering this question.
Complete step by step answer:
Time taken by the particle in order to complete the mentioned half revolution can be found by taking the ratio of the distance covered by the particle and the velocity of the travel. This can be written as,
\[T=\dfrac{x}{v}\]
Where \[x\] be the distance of the travel and \[v\] be the velocity of the particle. The distance of the travel can be found by taking the half of the circumference of the circular path. The circumference can be found by taking the product of twice the pi of the radius of the circular path. This can be shown as,
\[C=2\pi r\]
Therefore the distance travelled will be,
\[x=\dfrac{2\pi r}{2}=\pi r\]
The radius of the path has been mentioned as,
\[r=5m\]
Substituting this in the equation can be shown as,
\[x=5\pi \]
The velocity of travel has been mentioned as,
\[v=5m{{s}^{-1}}\]
Substituting these in the equation of time taken can be shown as,
\[T=\dfrac{5\pi }{5}=\pi s\]
The average acceleration can be found by taking the ratio of the total velocity to the total time taken. Therefore we can write that,
\[{{a}_{avg}}=\dfrac{\Delta v}{\Delta T}\]
The change in velocity will be,
\[\Delta v=5-\left( -5 \right)\]
Substituting this in the equation will give,
\[{{a}_{avg}}=\dfrac{5-\left( -5 \right)}{\pi }=\dfrac{10}{\pi }m{{s}^{-2}}\]
So, the correct answer is “Option A”.
Note: The acceleration of a body can be defined as the rate of variation of velocity with respect to the time taken. The velocity of a body can be defined as the rate of variation of displacement with respect to the time taken. Both are found to be vector quantities.
Complete step by step answer:
Time taken by the particle in order to complete the mentioned half revolution can be found by taking the ratio of the distance covered by the particle and the velocity of the travel. This can be written as,
\[T=\dfrac{x}{v}\]
Where \[x\] be the distance of the travel and \[v\] be the velocity of the particle. The distance of the travel can be found by taking the half of the circumference of the circular path. The circumference can be found by taking the product of twice the pi of the radius of the circular path. This can be shown as,
\[C=2\pi r\]
Therefore the distance travelled will be,
\[x=\dfrac{2\pi r}{2}=\pi r\]
The radius of the path has been mentioned as,
\[r=5m\]
Substituting this in the equation can be shown as,
\[x=5\pi \]
The velocity of travel has been mentioned as,
\[v=5m{{s}^{-1}}\]
Substituting these in the equation of time taken can be shown as,
\[T=\dfrac{5\pi }{5}=\pi s\]
The average acceleration can be found by taking the ratio of the total velocity to the total time taken. Therefore we can write that,
\[{{a}_{avg}}=\dfrac{\Delta v}{\Delta T}\]
The change in velocity will be,
\[\Delta v=5-\left( -5 \right)\]
Substituting this in the equation will give,
\[{{a}_{avg}}=\dfrac{5-\left( -5 \right)}{\pi }=\dfrac{10}{\pi }m{{s}^{-2}}\]
So, the correct answer is “Option A”.
Note: The acceleration of a body can be defined as the rate of variation of velocity with respect to the time taken. The velocity of a body can be defined as the rate of variation of displacement with respect to the time taken. Both are found to be vector quantities.
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