Question

Two objects moving along the same straight line are leaving point A with an accelerations $a$ and $2a$, and velocities $2u$ and $u$ respectively at time $t=0$. The distance moved by the objects with respect to point A when one object overtakes the other is

Answer 1

Step 1: Motion in a straight line

Consider a point B, at a distance $x$ from point A, where both the objects meet each other.

Step 2: Calculation of distance $x$ for object 1

The distance covered by object 1 can be calculated by the following expression.

$x=2ut+\frac{1}{2}a{t}^2$

Here, $x$ is the distance covered by object 1, $2u$ is the initial velocity, $a$ is the acceleration, and $t$ is the time taken by object 1.

Step 3: Calculation of distance $x$ for object 2

The distance covered by object 2 can be calculated by the following expression.

$x=ut+\frac{1}{2}\times 2a{t}^2$

Here, $u$ is the initial velocity, $2a$ is the acceleration for object 2.

Step 4: Calculation of time taken by both the objects to reach point B

Since, the distance covered by both the objects is the same, time taken by the objects can be calculated by equating the equations.

$$\begin{gathered} 2ut+\frac{1}{2}a{t}^2=ut+a{t}^2 \\ ut=\frac{1}{2}a{t}^2 \\ t=\frac{2u}{a} \end{gathered}$$

Step 5: Calculation of distance covered by the objects

The distance covered by the objects can be determined by putting the value of time taken by any object in the respective equation for displacement.

$$\begin{gathered} x=u\times \frac{2u}{a}+\frac{1}{2}\times 2a{\left(\frac{2u}{a}\right)}^2 \\ x=\frac{2u^2}{a}+\frac{4u^2}{a} \\ x=\frac{6u^2}{a} \end{gathered}$$

Hence, the distance moved by the objects with respect to point A when one object overtakes the other is $\frac{6u^2}{a}$.