Question

Find ratio avg.speed to the avg. Velocity in motion
from A to B In a circle

Answer 1

Answer:

$\frac{\pi}{2 \sqrt{2} }$

Explanation:

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•

$avg.speed = \frac{total \: distance}{total \: time} \\ avg.velocity = \frac{x2 - x1}{t2 - t1}$

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avg.speed:

• Total distance travelled = (1/4)th the perimeter of circle.
• perimeter of the circle=2πR
• 1/4th perimeter= (2πR)/4 = πR/2
• let the total time be 't'
• therefore,

$avg.speed = \frac{ \frac{\pi r}{2} }{t} \\ = \frac{\pi r}{2t}$

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avg.velocity:

• consider ∆ABO in the circle, it is a right angled triangle, so acc.to Pythagorean theorem,

${hypt}^{2} = {base}^{2} + {height}^{2} \\ {hypt}^{2} = {r}^{2} + {r}^{2} \\ hypt = \sqrt{2 {r}^{2} } \\ hypt = \sqrt{2} r$

• let the time taken be 't'
• therefore,
• $avg.speed = \frac{ \sqrt{2} r}{t}$

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Then, the ratio is given by,

$\frac{avg.speed}{avg.velocity} = \frac{ \frac{\pi\:r}{2t} }{ \frac{ \sqrt{2}r }{t} } \\ = \frac{\pi \: r}{2t} \times \frac{t}{ \sqrt{2} r} \\ = \frac{\pi}{2 \sqrt{2} }$