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A particle is starting from rest and moves 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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