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# A particle falling from rest under gravity covers a distance $x$ in 4$s$. If it continues failing then next $2x$ distance will be covered in approximatelyA. $1.41s$B. $1.73s$C. $2.05s$D. $2.92s$

Last updated date: 13th Jun 2024
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Hint:
- We have to use $S = ut + \dfrac{1}{2}g{t^2}$.
- First we have to calculate the value of $x$ using the equation. Then using that value of$x$, we can calculate the time taken to cover the $2x$ distance.

Complete step by step solution:
If the particle falls with an initial velocity $u$ and acceleration $g$. And, after time t, it travels a distance $s$.
Then, this equation can be used,
$S = ut + \dfrac{1}{2}a{t^2}$ - (equation 1)
We will solve this problem in two parts.
First, the particle falling from rest under gravity covers a distance $x$ in $4s$.

So, here
$s = x \\ g = 9.8 \\ u = 0 \\ t = 4 \\$
Putting this value on equation 1,
$x = 0 \times 4 + \dfrac{1}{2} \times 9.8 \times {4^2} \\ \Rightarrow x = \dfrac{1}{2} \times 9.8 \times 16 \\ \Rightarrow x = 78.4unit \\$

Second, we will calculate the time taken by the particle to cover $2x$ distance. After covering $x$distance, the particle covers $2x$.
So, here
$s = 3x \\ g = 9.8 \\ u = 0 \\ t = ? \\$
By putting this values on equation 1,

$3x = 0 \times t + \dfrac{1}{2} \times 9.8 \times {t^2} \\ \Rightarrow 3 \times 78.4 = \dfrac{1}{2} \times 9.8 \times {t^2} \\ \Rightarrow 235.2 = 4.9 \times {t^2} \\ \Rightarrow {t^2} = \dfrac{{235.5}}{{4.9}} \\ \Rightarrow t = \sqrt {\dfrac{{235.5}}{{4.9}}} \\ \Rightarrow t = \sqrt {48.06} \\ \Rightarrow t = 6.93s \\$
Now, to cover $3x$ distance it takes $6.93$. By subtracting the times covered by the particle from this time, we will get the time taken by the particle for the next $2x$ distance.
So, the next $2x$ distance will be covered in approximately = 6.93 – 4.0 = 2.93s.

The Correct option is D. $2.93s$

Note: We should use the equation of motion to solve this kind of problem.