A body starts from rest and moves with uniform acceleration. What is the ratio of kinetic energies at the end of $1 s t, 2 n d a n d 3 r d s e c o n d s$ of its journey?

1 : 8 : 27

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1 : 2 : 3

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1 : 4 : 9

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3 : 2 : 1

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Solution

The correct option is C $1 : 4 : 9$
$C a s e 1 : \frac{V_{1} - V_{0}}{t} = a \Rightarrow V_{1} = a \Rightarrow K E_{1} = \frac{1}{2} m a^{2}$ (since the body starts from rest)

$C a s e 2 : \frac{V_{2} - V_{0}}{2} = a \Rightarrow V_{2} = 2 a \Rightarrow K E_{2} = \frac{4}{2} m a^{2}$ (since the body starts from rest)

$C a s e 3 : \frac{V_{3} - V_{0}}{3} = a \Rightarrow V_{3} = 3 a \Rightarrow K E_{3} = \frac{9}{2} m a^{2}$ (since the body starts from rest)

$\Rightarrow K E_{1} : K E_{2} : K E_{3} = 1 : 4 : 9$

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