## Statement

difficulty Two marble players face each other with their marbles in hand. The game consists of throwing the marbles at the same time in a straight line so they hit each other. The players are located 36 meters from each other and player A launches its marble at 2 m/s and player B at 4 m/s, in a uniform rectilinear motion. Calculate that distance from player B at which the marbles will collide.

## Solution

Data

Considering that player A’s marble is at the origin of coordinates:

Marble A
X0=0 m
VA=2 m/s

Marble B
X0=36 m
VB=-4 m/s (moves toward the origin of the reference system)

Resolution

Considering the coordinate system described in the data, we will study the equation of position of each marble separately.

In u.r.m. the position of a body in motion is given by the following equation:

$x={x}_{0}+v·t$

Marble player A.

Substituting the values for this player in the equation of the u.r.m. we get that:

Marble player B.

Substituting back in the equation, but with the data of player B, we get:

Notice that since the motion is towards the origin of our system of reference the speed is negative.

Both marbles will impact when their positions are the same, i.e. XA=XB, therefore:

That is, they will collide after 6 s, but where? As we know when the impact occurs, simply replace that time into the equation of the position of any of the 2 marbles.

Therefore, the collision will happen at 12 meters of player A and at 24 m (36-12) of player B.

## Formulas worksheet

These are the main formulas that you must know to solve this exercise. If you are not clear about their meaning, we recommend you to check the theory in the corresponding sections. Furthermore, you will find in them, under the Formulas tab, the codes that will allow you to integrate these equations in external programs like Word or Mathematica.

Formulas
Related sections
$x={x}_{0}+v\cdot t$