Doubt regarding relative acceleration

[Dark85]

TL;DR

If two objects are separated at a distance and start from rest with a uniform acceleration, when will the object behind catch up?

Hey everyone,
I would just like to confirm if the following is correct:
If there are 2 bodies A and B and they both start from rest such that A is "M" meters behind B and once they start, they undergo a uniform acceleration of "x" m/s^2 for A and "y" m/s^2 for B, such that x>y, then A would catch up with B after:

 

[Baluncore]

You are correct.

 

[Dark85]

You are correct.

Alright, thank you!

 

[Ibix]

If you are going to post maths here, you probably want to read the LaTeX Guide, linked below the reply box. This:$$\sqrt{\frac{2M}{x-y}}$$is a lot more readable than your image.

Note that the first time LaTeX is added to a page the parser doesn't always kick in. If you see letters, back slashes and curly brackets instead of maths, refresh the page.

 

[Dark85]

Sure.. will do so next time

 

[PeroK]

TL;DR Summary: If two objects are separated at a distance and start from rest with a uniform acceleration, when will the object behind catch up?

Hey everyone,
I would just like to confirm if the following is correct:
If there are 2 bodies A and B and they both start from rest such that A is "M" meters behind B and once they start, they undergo a uniform acceleration of "x" m/s^2 for A and "y" m/s^2 for B, such that x>y, then A would catch up with B after:

View attachment 360146

Although you can use whatever notation you like, in kinematics its conventional to use $M$ for mass rather than distance, $x, y$ for the x and y coordinates and something like $a_1, a_2$ for two accelerations.

More generally, if two objects have initial displacements $x_1(0)$ and $x_2(0)$, initial velocities $v_1(0)$ and $v_2(0)$ and constant accelerations $a_1$ and $a_2$, then the displacements at time $t$ are the usual:
$$x_1(t) = x_1(0) + v_1(0)t + \frac 1 2 a_1t^2, \ \ x_2(t) = x_2(0) + v_2(0)t + \frac 1 2 a_2t^2$$And we can define the displacement of object 2 relative to object 1 as:
$$x_{21}(t) \equiv x_2(t) - x_1(t)$$And we have:$$x_{21}(t) = \big ( x_2(0) - x_1(0) \big ) + \big (v_2(0) - v_1(0) \big ) t + \frac 1 2 \big (a_2 - a_1 \big )t^2$$The important point is that this is equivalent to the displacement of an object with initial position $x(0) = x_2(0) - x_1(0)$, initial velocity $v(0) = v_2(0) - v_1(0)$ and constant acceleration $a = a_2 - a_1$.

This means that we can look at the scenario as though object 1 were at rest and only object 2 were moving. Or, vice versa.

Note finally that the concept of relative motion extents to the case of two and three dimensional motion.

 

[Dark85]

I did not understand how you got:
$x_1(t)=x_1(0)+v_1(0)t+\frac 1 2a_1t^2$
i.e. why the term $v_1(0)t$ must be present. Won't $\frac 1 2a_1t^2$ be enough to calculate for displacement since it's an accelerated motion? The bodies start accelerating right from the time when they start moving.

 

[PeroK]

I did not understand how you got:
$x_1(t)=x_1(0)+v_1(0)t+\frac 1 2a_1t^2$
i.e. why the term $v_1(0)t$ must be present. Won't $\frac 1 2a_1t^2$ be enough to calculate for displacement since it's an accelerated motion? The bodies start accelerating right from the time when they start moving.

The general case has an initial velocity. If you are driving at constant speed, then you can start to accelerate or decelerate.

 

[Dark85]

The general case has an initial velocity. If you are driving at constant speed, then you can start to accelerate or decelerate.

Right my bad, I thought you were referring solely to the case I mentioned above.