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
B
0.027m/s2
C
0.218m/s2
D
0.003m/s2

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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