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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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
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