Question

find the value of alpha

Answer 1

Object 1

Acceleration = a

Initial Velocity = 2u

Distance Covered = $\mathsf{s_1}$ (say)

Then

$\mathsf{s_1=(2u)t+\dfrac{1}{2}at^2}$

Object 2

Acceleration = 2a

Initial Velocity = u

Distance Covered = $\mathsf{s_2}$ (say)

Then

$\mathsf{s_2=ut+\dfrac{1}{2}(2a)t^2}$

Both the objects start at the same point, but with different velocities and accelerations. Object 1 is initially moving faster and is ahead of Object 2, however, Object 2 has a larger acceleration and hence it soon catches up.

Now, When Object 2 overtakes Object 1, the distance covered by both of them would be the same. So we can write:

$\sf s_1=s_2 \\\\2ut+\frac{1}{2}at^2=ut+\frac{1}{2}(2a)t^2\\\\2ut-ut=at^2-\frac{1}{2}at^2\\\\ut=\frac{1}{2}at^2\\\\at^2-2ut=0\\\\t(at-2u)=0\\\\t=0 \quad OR \quad t=\dfrac{2u}{a}$

Here t=0 corresponds to the starting point A. So the second solution corresponds to the time t when Object 2 overtakes Object 1.

We can put this $\sf t=\dfrac{2u}{a}$ in either $\sf s_1$ or $\sf s_2$ to get the distance [Note that $\sf s_1=s_2$. So we are anyways going to get the same distance]

Let's put it in $\sf s_1$

$\sf s = s_1 = 2ut+\dfrac{1}{2}at^2 \\ \\ Putting \: t=\dfrac{2u}{a}\\\\\implies s=2u\times \dfrac{2u}{a} +\dfrac{1}{2}a\times \left(\dfrac{2u}{a}\right)^2 \\\\\implies s=\dfrac{4u^2}{a}+\dfrac{2u^2}{a} \\ \\\\\implies \boxed{\sf s=\dfrac{\bold{6}u^2}{a}}$

Thus, The distance moved by the object with respect to point A when one object overtakes the other is $\sf \dfrac{6u^2}{a}$.

Comparing with $\sf \dfrac{\alpha u^2}{a}$, we have:

$\huge{\boxed{\boldsymbol{\alpha=6}}}$

A Graph is also attached for conceptual understanding. For both objects, the distance-time graph is given. Object 1 is the green curve, while Object 2 is the blue curve.

We see that Object 1 is initially farther than Object 2, but the blue curve (Object 2) catches up soon and overtakes. The intersection of the curves represents the point when Object 2 overtakes Object 1.