# Question #cdc14

Mar 4, 2016

(B) $= \left(2 + \sqrt{2}\right) s$

#### Explanation:

Let $s$ be the total distance covered in time $t$.
We know from general expression that $s = u t + \frac{1}{2} g . {t}^{2}$
Since the object falls from rest, $u = 0$
Equation becomes $s = \frac{1}{2} g . {t}^{2}$ ......(1)

It is given that object covers half of the distance in the last second. It implies that it covers first half of the distance in $\left(t - 1\right)$ seconds.
We obtain
$\frac{s}{2} = \frac{1}{2} g . {\left(t - 1\right)}^{2}$.......(2)
Dividing equation (1) by (2)
We obtain
$2 = {t}^{2} / {\left(t - 1\right)}^{2}$, Solving for $t$
$2 \times {\left(t - 1\right)}^{2} = {t}^{2}$
or $2 \times {\left({t}^{2} - 2 t + 1\right)}^{2} = {t}^{2}$
or ${t}^{2} - 4 t + 2 = 0$
Using the quadratic formula for finding roots
$t = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
or $t = \frac{4 \pm \sqrt{{\left(- 4\right)}^{2} - 4 \times 1 \times 2}}{2}$
or $t = \frac{4 \pm \sqrt{8}}{2}$
or $t = \left(2 \pm \sqrt{2}\right)$

Considering $t = \left(2 - \sqrt{2}\right)$ root.

We observe that $t \approx 0.6 s$. For this root $\left(t - 1\right)$ is negative. Therefore, the condition for obtaining equation (2) can not be physically fulfilled. Hence, this root is to be ignored.
$\therefore t = \left(2 + \sqrt{2}\right) s$