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(D) For uniformly accelerated motion, the distance covered in the $n^{th}$ second is given by $S_n = u + \frac{a}{2}(2n - 1)$.For the fourth second $(

Vedclass · Accepted Answer

$92$

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(D) For uniformly accelerated motion, the distance covered in the $n^{th}$ second is given by $S_n = u + \frac{a}{2}(2n - 1)$.For the fourth second $(n=4)$: $25 = u + \frac{a}{2}(2 	imes 4 - 1) \implies 25 = u + 3.5a$  --- $(i)$For the sixth second $(n=6)$: $37 = u + \frac{a}{2}(2 	imes 6 - 1) \implies 37 = u + 5.5a$  --- (ii)Subtracting $(i)$ from (ii): $(37 - 25) = (5.5a - 3.5a) \implies 12 = 2a \implies a = 6 \ m/s^2$.Substituting $a=6$ in $(i)$: $25 = u + 3.5(6) \implies 25 = u + 21 \implies u = 4 \ m/s$.The distance covered in the next two seconds (from $t=6 \ s$ to $t=8 \ s$) is calculated using $S = ut + \frac{1}{2}at^2$ where $u$ is the velocity at $t=6 \ s$.Velocity at $t=6 \ s$: $v = u + at = 4 + (6 	imes 6) = 40 \ m/s$.Distance covered in $t=2 \ s$: $S = (40 	imes 2) + \frac{1}{2} 	imes 6 	imes (2)^2 = 80 + 12 = 92 \ m$.

$A$ body moving with uniform acceleration travels a distance of $25 \ m$ in the fourth second and $37 \ m$ in the sixth second. The distance covered by the body in the next two seconds is (in $m$)

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