Question

A body covers $12\text{m}$ in $2^{nd}$ second and $20\text{m}$ in $4^{th}$ second. How much distance will it cover in 4 second after the $5^{th}$ second.

Answer 1

$s_n=u+a/2(2n-1)$

$s_2=12$

$\Rightarrow 12=u+a/2(2\times 2–1)$

$\Rightarrow 12=u+3a/2$ ………. (1)

$s_4=20$

$\Rightarrow 20=u+a/2(2\times 4–1)$

$\Rightarrow 20=u+7a/2$ ………. (2)

Subtract equation (1) from (2)

$s=4a/2$

$\Rightarrow a=4m{s}^{-2}$

$12=u+3/2\times 4$

$\Rightarrow u=6m{s}^{-1}$

Distance covered in the $4$ seconds after $5^{th}$ second,

$=S_9–S_5$

$=u\left(9\right)+1/2a(9{)}^2–[u\left(5\right)+1/2a\left(5{)}^2\right]$

$=4u+a/2(81–25)$

$=4\times 6+4/2\times 56=136m.$