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# Two objects moving along the same straight are leaving point A with an acceleration $a,\,2a$ and velocity $2u,\,u$ respectively at time $t = 0$. The distance moved by the object with respect to the point A when one overtakes the other is:A) $\dfrac{{6{u^2}}}{a}$B) $\dfrac{{2{u^2}}}{a}$C) $\dfrac{{4{u^2}}}{a}$D) None of these

Last updated date: 10th Sep 2024
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Hint: When one object overtakes the other, the distance travelled by them will be equal. Use the first equation of motion to calculate the distance travelled by each of them, equate them as they will be equal and solve to get value of time. After that, use the first motion of the equation again and put the value of time as calculated to get the value of distance travelled by the object.

Formula Used:
According to the first equation of motion, $s = ut + \dfrac{1}{2}a{t^2}$ where, $s$ is the distance travelled by the object, $u$ is its initial velocity, $t$ is the time period for which distance is being calculated, $a$ is the acceleration of the object.

Complete Step by Step Solution:
The two objects start moving at $t = 0$ with an acceleration $a,\,2a$ and velocity $2u,\,u$ respectively for object 1 and 2. At the time where one of them is overtaking the other, both of them would have covered the same distance from the initial point. Suppose it takes them time $t$ for one of them to be at this same place.
If distance covered by object 1 is ${s_1}$ and distance covered by object 2 is ${s_2}$ , then ${s_1} = {s_2}$
By using the first equation of motion, we get ${s_1} = 2ut + \dfrac{1}{2}a{t^2}$
(given velocity and acceleration of object 1 is $2u$ and $a$ respectively)
Also, ${s_1} = ut + \dfrac{1}{2} \times 2a{t^2} = ut + a{t^2}$
Since these two distances are equal at time $t$ , we can write $2ut + \dfrac{1}{2}a{t^2} = ut + a{t^2}$
On further simplifying, we get $ut = \dfrac{1}{2}a{t^2}$
Or, $t = \dfrac{{2u}}{a}$ this will be the time taken by them to get at a position where they have covered equal distances from the initial point.
Now, we have ${s_1} = 2ut + \dfrac{1}{2}a{t^2}$
Substituting the value $t = \dfrac{{2u}}{a}$ , we get
$\Rightarrow$ ${s_1} = 2u(\dfrac{{2u}}{a}) + \dfrac{1}{2}a{(\dfrac{{2u}}{a})^2}$
Which gives, ${s_1} = \dfrac{{4{u^2}}}{a} + \dfrac{{2{u^2}}}{a} = \dfrac{{6{u^2}}}{a}$
This is the distance covered by them at that point and this will be the final answer.

Hence, option (A) is the correct answer.

Note: There are a total three equations of motion in physics, all relating velocity, time, acceleration and distance. All equations are different. However, more than one equation can be used to solve a question. Choose the appropriate equation by checking whether the value of variables in an equation have been given to you in question or not.