First slide

Rectilinear Motion

Question

A body falls from a height h=300m, the ratio of distances travelled in each 2s, during t=0 to t=6s in the journey is:…….

Moderate

A

$1 : 5 : 9$
B

$1 : 3 : 5$
C

$1 : 5 : 3$
D

$1 : 2 : 3$

Solution

$For a freely body s = \frac{1}{2} g t^{2}$
$\begin{matrix} For first 2 s : s_{1} = \frac{1}{2} g (2)^{2} = \frac{1}{2} g (4) \dots \dots \dots . . (i) \\ For the next 2 s : s_{2} = \frac{1}{2} g (4)^{2} - \frac{1}{2} g (2)^{2} \\ \begin{aligned} = \frac{1}{2} g [16 - 4] \\ = \frac{1}{2} g [12] \dots \dots . . . (i i) \end{aligned} \end{matrix}$
$\begin{array}{l} For last 2 s : s_{3} = \frac{1}{2} g [(6)^{2} - 4^{4}] \\ = \frac{1}{2} g [36 - 16] = \frac{1}{2} g (20) \dots \dots . . . (i i i) \end{array}$
The ratio of all these distance is
$\begin{array}{l} s_{1} : s_{2} : s_{3} = \frac{1}{2} g (4) : \frac{1}{2} g (12) : \frac{1}{2} g (20) \\ = 4 : 12 : 20 \\ = 1 : 3 : 5 \end{array}$

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