First slide

Rectilinear Motion

Question

A freely falling body moves from A to B during 2sec and from B to C during 1sec. If AB=BC the distance between A&B is

Easy

A

4g
B

3g
C

2g
D

g

Solution

Let ‘O’ be the starting point of the body and let ‘t’ be the time taken by it to move from O to A .
Given that from A to B it takes 2sec
And from B to C it takes 1sec
$\begin{array}{l} A B = B C \\ \Rightarrow (O B - O A) = O C - O B \\ \Rightarrow \frac{1}{2} g [(t + 2)^{2} - t^{2}] = \frac{1}{2} g [(t + 3)^{2} - (t + 2)^{2}] \\ \Rightarrow 4 t + t = 2 t + 5 \\ \Rightarrow 2 t = 1 \\ \Rightarrow t = \frac{1}{2} s . \end{array}$
The distance between A&B is
$\begin{array}{l} \frac{1}{2} g [(t + 2)^{2} - t^{2}] \\ = \frac{1}{2} g [4 t + 4] \\ = \frac{1}{2} g [2 + 4] \\ = \frac{6 g}{2} \\ = 3 g \end{array}$

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