A particle is starting from rest and moving with a constant acceleration ‘a’. The ratio of the distances travelled in the first, second, third and fourth seconds is

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1 : 4 : 9 : 16

1 : 1 : 1 : 3

1 : 2 : 3 : 4

1 : 3 : 5 : 7

answer is D.

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Detailed Solution

$u = 0$

$S_{1} = a (1 - \frac{1}{2}) = \frac{a}{2}$

$S_{2 d} = a (2 - \frac{1}{2}) = \frac{3}{2} a$

$S_{3 r d} = a (3 - \frac{1}{2}) = \frac{5}{2} a$

$S_{4 t h} = a (4 - \frac{1}{2}) = \frac{7}{2} a$

$S_{1} : S_{2} : S_{3} : S_{4} = \frac{a}{2} : \frac{3}{2} a : \frac{5}{2} a : \frac{7}{2} a$

$= 1 : 3 : 5 : 7$

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