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Question

Physics Question on Motion in a plane

A particle moves along a circle of radius (20π)m\left(\frac{20}{\pi}\right) m with constant tangential acceleration. If the velocity of the particle is 80m/s80\, m/s at the end of the second revolution after motion has begun, the tangential acceleration is

A

40 m/s2 m/s^2

B

640 πm/s2 \pi \, m/s^2

C

160 πm/s2 \pi \, m/s^2

D

40πm/s240\, \pi \, m/s^2

Answer

40 m/s2 m/s^2

Explanation

Solution

r=20πr =\frac{20}{\pi}
In two revolution, distance is s=2×2πr=80ms =2 \times 2 \pi r =80 m
Now, v2=2v ^{2}=2 as for zero initial speed.
a=v22s=80×802×80\Longrightarrow a =\frac{ v ^{2}}{2 s }=\frac{80 \times 80}{2 \times 80}
a=40m/s2\Longrightarrow a =40 m / s ^{2}