Question

A particle is falling freely under gravity from rest. In first $t\,\sec $ it covers distance ${x_1}$ and in the next $t\,\sec $ it covers distance ${x_2}$, then $t$ is given by:
(A) $\sqrt {\dfrac{{{x_2} - {x_1}}}{g}} $
(B) $\sqrt {\dfrac{{{x_2} + {x_1}}}{{2g}}} $
(C) $\sqrt {\dfrac{{{x_2} - {x_1}}}{{2g}}} $
(D) $\sqrt {\dfrac{{{x_2} + {x_1}}}{g}} $

Answer 1

Hint The time is given by the formula of the acceleration equation of the motion, by using this formula and the given information in the question, the time taken can be determined. By creating the two equations and subtracting the two equations, then the solution is determined.

Useful formula
The acceleration equation of the motion is given by,
$s = ut + \dfrac{1}{2}a{t^2}$
Where, $s$ is the distance travelled by the object, $u$ is the initial velocity, $t$ is the time taken and $a$ is the acceleration of the object.

Complete step by step solution
Given that,
The distance travelled by the object in the first $t\,\sec $ is ${x_1}$,
The distance travelled by the object in the next $t\,\sec $ is ${x_2}$.
Now,
The acceleration equation of the motion is given by,
$s = ut + \dfrac{1}{2}a{t^2}\,...................\left( 1 \right)$
By substituting the distance and taking the acceleration as the acceleration due to gravity and the initial velocity is zero in the above equation, then the above equation is written as,
${x_1} = \left( {0 \times t} \right) + \dfrac{1}{2}g{t^2}$
By multiplying the terms in the above equation, then the above equation is written as,
${x_1} = \dfrac{1}{2}g{t^2}\,................\left( 2 \right)$
Now, the object travelled after the $t\,\sec $, so it will have some velocity, so the velocity is given as product of the acceleration and time, it is derived from the acceleration equation $a = \dfrac{v}{t}$, so the distance after $t\,\sec $ is written as,
${x_2} = \left( {gt} \right)t + \dfrac{1}{2}g{t^2}$
By multiplying the terms in the above equation, then the above equation is written as,
${x_2} = g{t^2} + \dfrac{1}{2}g{t^2}$
By adding the terms in the above equation, then the above equation is written as,
${x_2} = \dfrac{3}{2}g{t^2}\,................\left( 3 \right)$
Now subtracting the equation (3) by equation (2), then
${x_2} - {x_1} = \dfrac{3}{2}g{t^2} - \dfrac{1}{2}g{t^2}$
By subtracting the terms in the above equation, then the above equation is written as,
${x_2} - {x_1} = \dfrac{2}{2}g{t^2}$
By cancelling the terms in the above equation, then the above equation is written as,
${x_2} - {x_1} = g{t^2}$
By rearranging the terms in the above equation, then the above equation is written as,
${t^2} = \dfrac{{{x_2} - {x_1}}}{g}$
By taking the square root on both side in the above equation, then the above equation is written as,
$t = \sqrt {\dfrac{{{x_2} - {x_1}}}{g}} $

Hence, the option (A) is the correct answer.

Note The acceleration of the object is directly proportional to the velocity of the object and the acceleration of the object is inversely proportional to the time. As the velocity of the object increases, the acceleration of the object also increases.