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Question

If a body starts from rest and travels $$120$$ cm in the $$6^{th}$$ second then what is the acceleration ?


A
0.20m/s2
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B
0.027m/s2
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C
0.218m/s2
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D
0.003m/s2
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Solution

The correct option is B $$0.218m/s^{2}$$
$$S = ut + 0.5 at^2$$
displacement in n-1 sec is given by:
$$S_{n-1} = u(n-1) + 0.5 a(n-1)^2$$
$$S_{n -1} = un - u + 0.5an^2 +0.5a(1-2n)$$  
displacement in n seconds is given by:
$$S_{n} = un + 0.5an^2$$
Subtract above equations to get the displacement in "nth" second:
$$S = u +0.5a(2n-1)$$
Therefore, displacement in n = 6s
$$1.2 = 0 + 0.5a(12-1)$$
$$a = \dfrac{1.2}{5.5} = 0.218 \ m/s^2$$

Physics

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