Question

distance Travelled by an object in 4th sec is twice the distance travelled by it in 2nd second. If the acceleration of the object $3m{s}^{-2}$n is calculate (i) its initial velocity (ii) the distance travelled in 7th second.

Answer 1

Let $s_n$ be the distance travelled in n_{th} sec

$s_n=u+\frac{a}{2}(2n-1)$

$s_4=2s_2$

$u+3.5a=2u+3a$

$u-0.5a=0$

$u=0.5a$

$u=1.5$

using:$s_n=u+\frac{a}{2}(2n-1)$

$s_7=1.5+\frac{3}{2}(2\times 7-1)$

$s_7=21m$