Physics, asked by udhbhav38, 11 months ago

A particle moves in a circular path of radius R such that its speed v varies with distance x as v=a√x, where a is positive constant. The acceleration of the particle after traversing a distance √2R is..............
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Answers

Answered by jacksondave610
4

Answer: alpha square / 2

Explanation:

We know that v² = u² + 2as

ATQ, v = alpha√x

v² = alpha²x

Comparing both the eqns,

'2as' is equivalent to 'alpha²x'

2ax = alpha²x

a = alpha²/2

Thank you

Answered by farruminoo
1

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