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Relating Circular Motion to SHM

A body starts...

Question

A body starts from rest and travels a distance x with uniform acceleration then it travels a distance 2x with uniform speed, finally it travels a distance 3x with uniform retardation and comes to rest.If the complete motion of particle is along a straight line,then the ratio of its average velocity to maximum velocity during its motion is -
a )( 2 / 5 ) b )( 3 / 5 ) c )( 4 / 5 ) d ) (6 / 7 )

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Solution

$F o r t h e f i r s t p a r t : x = 1 / 2 * a * t^{2} t_{1} = \sqrt{2 x / a} v = a t_{1} = a \sqrt{2 x / a} F o r t h e s e c o n d p a r t : t_{2} = 2 x / a \sqrt{2 x / a} = 2 x / a \sqrt{2 x / a} F o r t h e t h i r d p a r t : v^{2} - u^{2} = 2 a' s - u^{2} = 2 a' s - 4 x^{2} / - a^{2} 2 x / a = 2 a' s a' = - 2 a / 3 s o, v = u + a t t = - u / a = a \sqrt{2 x / a} / - 2 a / 3 = 3 \sqrt{2 x / a} / 2 A v e r a g e v e l o c i t y = t o t a l d i s \tan c e / t o t a l t i m e t a k e n = x + 3 x + 2 x / \sqrt{2 x / a} + 2 x / a \sqrt{2 x / a} + 3 \sqrt{2 x / a} / 2 = 6 x / 7 x * a * \sqrt{2 x / a} M a x i m u m v e l o c i t y = a \sqrt{2 x / a} a n s w e r = 6 / 7$

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