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(B) The distance travelled by a particle in the $n^{	ext{th}}$ second is given by $S_n = u + \frac{a}{2}(2n - 1)$.Given that the sum of the distance

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$50$

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(B) The distance travelled by a particle in the $n^{	ext{th}}$ second is given by $S_n = u + \frac{a}{2}(2n - 1)$.Given that the sum of the distance travelled in the $t^{	ext{th}}$ and $(t+1)^{	ext{th}}$ seconds is $100 	ext{ cm}$.$S_t + S_{t+1} = 100$$[u + \frac{a}{2}(2t - 1)] + [u + \frac{a}{2}(2(t+1) - 1)] = 100$$2u + \frac{a}{2}(2t - 1 + 2t + 2 - 1) = 100$$2u + \frac{a}{2}(4t) = 100$$2u + 2at = 100$$u + at = 50$Since the velocity after $t$ seconds is given by $v = u + at$,we have $v = 50 	ext{ cm/s}$.

$A$ particle is moving in a straight line with initial velocity $u$ and uniform acceleration $a$. If the sum of the distance travelled in the $t^{\text{th}}$ and $(t+1)^{\text{th}}$ seconds is $100 \text{ cm}$,then its velocity after $t$ seconds,in $\text{cm/s}$,is:

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