A body starts with initial velocity u and moves with uniform acceleration. When the velocity was increased to 5u, the acceleration is reversed in direction, the magnitude remains constant. Find its velocity when it returns to the starting point?

-7u

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-5u

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-u

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Solution

The correct option is C -u
I. By III equation of motion,
$v^{2} = u^{2} + 2 a s$
$(5 v)^{2} = u^{2} + 2 a s$
$25 v^{2} - u^{2} = 2 a s$
$24 u^{2} = s$
$2 a$
$∴$ Distance travelled $= S = \frac{12 u^{2}}{a}$
II. $v^{2} = u^{2} + 2 a s$
$v^{2} = (5 u)^{2} + 2 (- a) (\frac{12 v^{2}}{a})$
$v^{2} = 25 v^{2} - 24 u^{2}$
$v^{2} = u^{2}$
$v = 1 u$
$\Rightarrow v = - u$ .

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A body starts with initial velocity u and moves with uniform acceleration. When the velocity was increased to 5u, the acceleration is reversed in direction, the magnitude remains constant. Find its velocity when it returns to the starting point?

The correct option is C -uI. By III equation of motion,v2=u2+2as(5v)2=u2+2as25v2−u2=2as24u2=s2a∴ Distance travelled =S=12u2aII. v2=u2+2asv2=(5u)2+2(−a)(12v2a)v2=25v2−24u2v2=u2v=1u⇒v=−u.