Question
Derive an equation for the distance covered by a uniformly accelerated body in nth second of its motion. A body travels half its total path in the last second of its fall from rest. Calculate the time of its fall.
✓
Answer
For a body having a uniform acceleration ‘a’ in a straight line, starting with an initial velocity u, the displacement in ‘n’ seconds is given by,
$\text{S}_\text{n}=\text{nu}+\frac{1}{2}\text{an}^2$
In (n - 1) seconds,
$\text{S}_{\text{n}-1}=(\text{n}-1)\text{u}+\frac{1}{2}\text{a}(\text{n}-1)^2$
$\therefore$ Displacement in nth secound $=\text{S}_\text{n}-\text{S}_{\text{n}-1}$
Let S be the complere length of fall and t be the time taken for it. Then,
$\text{S}=\frac{1}{2}\text{gt}^2$ $(\because \text{u}=0)\ \dots(\text{i})$
Also, $\frac{\text{S}}{2}$ is covered in the last second.
$\therefore \frac{\text{S}}{2}=0+\frac{\text{g}}{2}(2\text{t}-1)\ \dots(\text{ii})$
Using (i) and (ii), solve for t to be,
$\text{S}=\text{g}(2\text{t}-1)=\frac{1}{2}\text{gt}^2,$
i.e., $4\text{tg}-2\text{g}=\text{gt}^2$
$\text{gt}^2-4\text{tg}+2\text{g}=0$
$\Rightarrow \text{t}^2-4\text{t}+2=0$
i.e., $\text{t}=\frac{4\pi\sqrt{16-8}}2{}=\frac{4\pm2\sqrt{2}}{2}$
$\text{t}=2\pm\sqrt{2}$
$\text{S}_\text{n}=\text{nu}+\frac{1}{2}\text{an}^2$
In (n - 1) seconds,
$\text{S}_{\text{n}-1}=(\text{n}-1)\text{u}+\frac{1}{2}\text{a}(\text{n}-1)^2$
$\therefore$ Displacement in nth secound $=\text{S}_\text{n}-\text{S}_{\text{n}-1}$
Let S be the complere length of fall and t be the time taken for it. Then,
$\text{S}=\frac{1}{2}\text{gt}^2$ $(\because \text{u}=0)\ \dots(\text{i})$
Also, $\frac{\text{S}}{2}$ is covered in the last second.
$\therefore \frac{\text{S}}{2}=0+\frac{\text{g}}{2}(2\text{t}-1)\ \dots(\text{ii})$
Using (i) and (ii), solve for t to be,
$\text{S}=\text{g}(2\text{t}-1)=\frac{1}{2}\text{gt}^2,$
i.e., $4\text{tg}-2\text{g}=\text{gt}^2$
$\text{gt}^2-4\text{tg}+2\text{g}=0$
$\Rightarrow \text{t}^2-4\text{t}+2=0$
i.e., $\text{t}=\frac{4\pi\sqrt{16-8}}2{}=\frac{4\pm2\sqrt{2}}{2}$
$\text{t}=2\pm\sqrt{2}$
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