Methods to use in problems with equations of motion, where is required to determine a particular acceleration

Question

I've a doubt on problems with manipulation of equations of motion. I'll make an example of a situation I'm confused about.

Let $A$ and $B$ be two points that move on a line with constant speed, $A$ is behind $B$ but it can reach $B$ since $v_A>v_B$. At $t=0$, when the distance between the two points is $d$, $A$ is slowed by $-a$ deceleration. Determine the relation between $v_A,v_B,d,a$ such that $A$ and $B$ do not collide (i.e. $A$ get $B$ with speed equal to $v_B$ and not greater).

My question is on problem like this one, when it is required, among the others, to find an acceleration (deceleration) that gives to a point a particular speed at a given time/position.

In theory what I would do is to write down the two equations of motion of $x_A(t)$ and $x_B(t)$ and try to solve for $t$ the equation $x_A(t)=x_B(t)$, thus I find a time, say $t^*$ at which the position of the points are equal. In this way I can easily impose $v_A(t^*)=v_B$ and get the result. Nevertheless it is quite hard to solve $x_A(t)=x_B(t)$ in the sense that I don't get a easy result.

In fact my book uses another strategy: firstly it finds the time $t^*$ at which $v_A(t^*)=v_B$, then it writes down the equation $x_A(t)=x_B(t)$ and it substitutes the $t^*$ previously found, such that $x_A(t^*)=x_B(t^*)$ and it gets the relation required from this last equation.

I don't understand why this method is correct, and if it is equivalent to the one I proposed. In other words, is it the same to firstly get the time in which $A$ and $B$ have the same position and then impose that in that moment $v_A(t)=v_B$, or firstly find the moment in which $v_A(t)=v_B$ and then impose that in that moment $A$ and $B$ have the same position? Does something guarantees that $A$ reaches the speed $v_B$ only when it is in the same position of $B$?

Similar things happen in the kind of problems I said, to solve the quadratic equation $x_A(t)=x_B(t)$ for $t$ is not the best way, but I don't understand why the other methods (like the one of my textbook) are equivalent.

Thanks a lot for your help

Answer 1

There are often lots of ways to solve problems and finding the simplest way is something that comes with experience. The reason your teacher keeps setting you so many problems is to give you that experience.

One of the common tricks is choosing a suitable reference frame to do the calculation. As watched by a stationary observer it's a somewhat complicated problem, but suppose you are travelling alongside $B$. In that case you see $B$ to be stationary and $A$ is approaching from a distance $d$ at a relative speed of $v = v_A - v_B$. Now the calculation is really simple because you can use the SUVAT equation:

$$ v^2 = u^2 + 2as $$

with $u = v_A - v_B$, $s = d$ and $v = 0$, and the answer drops straight out.

If I understand your question correctly the book is using another strategy. It works backwards by starting with $v_A = v_B$ when the distance between $A$ and $B$ is zero then works backwards to find the relative velocity when the distance between $A$ and $B$ is $d$. Fair enough, though I think my method is simpler.