A body moving with uniform acceleration covers a distance of $14 m$ in first $2$ second and $88 m$ in next $4$ second. How much distance is travelled by it in $5^{t h}$ second?

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Solution

Distance in $2 s e c = 14 m$
$S = U (2) + \frac{1}{2} a (2)^{2}$
$14 = 24 + 29 \to 4 + 9 = 7$ ....(1)
$(88 + 14) = 64 + \frac{1}{2} a (36)$
$102 = 64 + 189$
$4 + 3 a = 17$ ....(2)
$2 a = 10 ⟹ \frac{9 = 5 m / s}{4 = 2 m / s}$
Distance in $5$ sec.
$S_{1} = 2 (5) + \frac{1}{2} (5) (25)$
$= 10 + 62.5 = 72.5 m$
Distance in $4$ sec
$S_{2} = 2 (4) + \frac{1}{2} (5) (16)$
$= 8 + 40 = 48 m$
Distance in $5^{t h} s e c = 72.5 - 48$
$= 24.5 m$

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