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Q.
A body covers $100 \,cm$ in first $2$ second and $128\, cm$ in next four seconds moving with constant acceleration. Find the velocity of the body at the end of $8$ sec?
Solution:
The distance in first two seconds
$s_{1}=u t_{1}+\frac{1}{2} a t_{1}^{2}$
$100=2 u+\frac{a}{2} 4 \ldots \ldots .(1)$
$200=4 u+4 a$
The distance in $6$ seconds
$s_{1}+s_{2}=6 u+\frac{a}{2} 36$
$2\left(s_{1}+s_{2}\right)=12 u+36 \ldots \ldots(2)$
$2(228)=12 u+36$
From equation $1$ and $2$ We get
$a=-6\, cm / sec ^{2}$
$100=2 u-\frac{6}{2} \times 4$
$2 u=112 \,\,\,V=u+a t$
$v=56-6 \times 8=8 \,cm / sec$