Before tackling this exercise, I fervently recommend you to try and work out the solution to this same situation, but with a reference system set on the ground, and not on a moving body, here: c3.42

What is now proposed is to re-do that same exercise, but with a reference system that happens to be mounted on our white ram´s back and hanging on to his horns. This means that if he moves either here or there, his position will always be zero, always zero, is that clear enough? If our guy decides to take off at full speed and accelerate in order to crash against the black ram, he will always be fixed in the zero position of the reference system, with velocity=0 and acceleration=0… and the whole universe will be moving and accelerating, backwards… and don´t even get me started on what happens to the black ram!

Do not despair, there is always a way out. The mobile reference system that allows for movement within itself is the white ram´s imagination. It is the white ram. Therefore, we will work with: V_BW,  velocity of the black ram as seen by the white ram, V_WG,  velocity of the white ram, as seen from the ground, by a frightened observer. And at last, V_BG, the velocity of our black ram, as seen by the same observer, on the ground. As always, it is true that 

As seen from the ground, the accelerations are the ones described in the exercise 3.12, which are a_BG = – 1,4 m/s² and a_WG = 1,6 m/s². Clearing this up, we get:

a_BW = a_BG – a_WG

a_BW = – 3 m/s²

This means that, as seen from the white ram´s position, who now feels like the center of the universe, the black ram is running backwards towards the center, towards him, with an acceleration of 3 m/s². It was only logical! I could´ve predicted this without knowing a thing about physics!

Now, let´s build our black ram´s kinematic equations (since the white ram´s all equal zero, zero, zero).

x = 24 m – 1,5 m/s² . t²

v = – 3 m/s² . t

Asking these equations about the meeting point is the same as asking them to express the zero position, which is, of course, the white ram´s position, who is still insisting on being the center of the universe.

0 m = 24 m – 1,5 m/s² . t_e²

v_e = – 3 m/s² . t_e

I knew it… two equations with two unknown variables. This setup is way easier than the original one! (The original exercise ended with a 4x4 equation system). From the first equation:

t_e² = 24 m / 1,5 m/s²

Now, with this solution, we head to the second equation, and we are left with…

v_e = – 3 m/s² . 4 s

If you take a closer look, there´s a lot to take in, areas that are the same, differences between magnitudes that are actually the same, etc, etc. Get inspired.