A body starting from rest moves with constant acceleration. The ratio of distance covered by the body during the $5^{th}$ sec to that covered in $5\, sec$ is
A body starting from rest moves with constant acceleration. The ratio of distance covered by the body during the $5^{th}$ sec to that covered in $5\, sec$ is
$9/25$
$3/5$
$25/9$
$1/25$
A body starting from rest moves with constant acceleration. The ratio of distance covered by the body during the $5^{th}$ sec to that covered in $5\, sec$ is
Distance covered in $5^{th}$ second,
${S_{{5^{th}}}} = u + \frac{a}{2}(2n - 1) = 0 + \frac{a}{2}(2 \times 5 - 1) = \frac{{9a}}{2}$
and distance covered in $5$ second,
${S_5} = ut + \frac{1}{2}a{t^2} = 0 + \frac{1}{2} \times a \times 25 = \frac{{25a}}{2}$
$\therefore \frac{{{S_{{5^{th}}}}}}{{{S_5}}} = \frac{9}{{25}}$
Other Language