Question

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

Answer 1

Distance in $2sec=14m$
$S=U\left(2\right)+\frac{1}{2}a(2{)}^2$
$14=24+29\to 4+9=7$ ....(1)
$(88+14)=64+\frac{1}{2}a\left(36\right)$
$102=64+189$
$4+3a=17$ ....(2)
$2a=10⟹\frac{9=5m/s}{4=2m/s}$
Distance in $5$ sec.
$S_1=2\left(5\right)+\frac{1}{2}\left(5\right)\left(25\right)$
$=10+62.5=72.5m$
Distance in $4$ sec
$S_2=2\left(4\right)+\frac{1}{2}\left(5\right)\left(16\right)$
$=8+40=48m$
Distance in $5^{th}sec=72.5-48$
$=24.5m$