## Statement

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*

X_{0}=0 m

V_{A}=2 m/s

*Marble B*

X_{0}=36 m

V_{B}=-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:

$$x={x}_{0}+v\xb7t$$

*Marble player A.*

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

$${x}_{A}=0+2\xb7tm\Rightarrow \phantom{\rule{0ex}{0ex}}\overline{){x}_{A}=2\xb7tm}$$

*Marble player B.*

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

$$\overline{){x}_{B}=36-4\xb7tm}$$

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. X_{A}=X_{B}, therefore:

$$XA=XB\Rightarrow \phantom{\rule{0ex}{0ex}}2\xb7t=36-4\xb7t\Rightarrow \phantom{\rule{0ex}{0ex}}t=\frac{36}{6}\Rightarrow \phantom{\rule{0ex}{0ex}}\overline{)t=6sg}\phantom{\rule{0ex}{0ex}}$$

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.

$${X}_{A}=2\xb7t\Rightarrow \phantom{\rule{0ex}{0ex}}{X}_{A}=2\xb76\Rightarrow \phantom{\rule{0ex}{0ex}}\overline{){X}_{A}=12m}$$

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