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A particle moving with uniform retardation covers distances 18 m. 14 m and 10 m in successive seconds. It comes to rest after travelling a further distance of

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42m

50m

12m

answer is B.

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Detailed Solution

$s = ut + \frac{1}{2} {at}^{2} \Rightarrow 18 = u (1) + \frac{1}{2} a (1)^{2}$

$18 + 14 = u (2) + \frac{1}{2} a (2)^{2}$

u = 20ms^–1; a= 4ms^–2 ; v = u + at

$\Rightarrow$ v = 20 – (4)(3)

$\Rightarrow$ v = 8ms^–1

v² - u² = 2aS

$\Rightarrow$ 0 - 64 = 2(-4)S

S = 8m

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