A body travels $200 c m$ in the first two seconds and $220 c m$ in the next four seconds. What will be the velocity at the end of $7^{t h}$ second from the start?

10 c m s^{- 1}

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20 c m s^{- 1}

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15 c m s^{- 1}

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5 c m s^{- 1}

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Solution

The correct option is C $10 c m s^{- 1}$
$s = u t + \frac{1}{2} a t^{2}$
For $c a s e 1.$
$200 = 2 u + 2 a$
$u + a = 100 . . . . (1)$
For $c a s e 2.$ .i.e. next 4 seconds, we have distance $= 420 c m$ and time $= 6 s$
$420 = 6 u + 18 a$
$u + 3 a = 70 . . . . (2)$
Solving equation $1$ and $2,$ we get,
$u = 115 c m / s$ and $a = - 15 c m / s^{2}$
Now by
$v = u + a t$
$v = 115 - 15 \times 7$
$v = 10 c m / s$
Hence answer is A.

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