A body covers $200 c m$ in the first $2 s e c$ and $220 c m$ in next $4 s e c$ . Assuming constant acceleration, what is the velocity of the body at the end of $7 t h$ second?

40 c m / s e c

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20 c m / s e c

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10 c m / s e c

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5 c m / s e c

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Solution

The correct option is A $10 c m / s e c$
Distance travelled in first 2 seconds $= 200 c m$
$s = u t + \frac{1}{2} a t^{2}$
$200 = 2 u + \frac{1}{2} a \times 4$
$u + a = 100 \to (1)$
Distance travelled in next 4 sec $=$ 220 cm
Distance travelled in first 6 sec $=$ 420 cm
$420 = u \times 6 + \frac{1}{2} a \times 36$
$70 = u + 3 a \to (2)$
Solving equation 1 and 2
$2 a = - 30$
$a = - 15 c m / s^{2}$
From equation 1 , $u = 115 c m / s$
$v = u + a t$
$v = 115 + (- 15) (7)$
$v = 10 c m / s$

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