Physics Explained Through a Video Game/Calculating Displacement

Topic 1.2 - Calculating Displacement

We can calculate for the exact displacement of an object in a particular direction by considering the final and initial positions of an object in that direction. Then, we subtract the initial position value from the final position value to get the change in position, which is displacement.

This allows for us to quantitatively look at the change in an object's position, which will later make it possible to make clearer predictions about objects in motion.

Formula for displacement in the X (horizontal) direction: ${\vec {d}}_{x}=x_{\text{after}}-x_{\text{before}}$

Formula for displacement in the Y (vertical) direction: ${\vec {d}}_{y}=y_{\text{after}}-y_{\text{before}}$

Players on "Royal Curse" by GayfishDeluxe with an overlay of a horizontal and a vertical displacement component being added. The task is to calculate the net displacement.

If we have a displacement that has both a vertical and a horizontal component, we can add these displacement vectors together to get the net displacement vector.

Question 1: Suppose that there's an in-game situation where the player Hello (colored white) wants to knock Jensen (colored grey) off the board. At a particular time in the game, Hello needs to have a ${\vec {d}}_{x}$ of ${\vec {d}}_{horizontal}=6\;{\textsf {meters}}$ and a ${\vec {d}}_{y}$ of ${\vec {d}}_{vertical}=8\;{\textsf {meters}}$ in order to reach Jensen. How far does Hello need to travel in order to reach Jensen, assuming that Jensen is at the same spot when Hello arrives?

Solution:

In order to calculate how far Hello needs to travel, we need to find ${\vec {d}}_{net}$. Using the image on the right, we can see that if we were to draw ${\vec {d}}_{vertical}$ from where ${\vec {d}}_{horizontal}$ ends, we form a right angle between these two displacement components. Also, we know that because the net displacement is the shortest path between two points, ${\vec {d}}_{net}$ must be a line directly connecting Hello to Jensen.

Thus, we can form a right angle triangle that is outlined by ${\vec {d}}_{net}$ and its components. Because of the Pythagorean Theorem where $a^{2}+b^{2}=c^{2}$ where a and b are components and c is ${\vec {d}}_{net}$, we can show that:

$a^{2}+b^{2}=c^{2}\implies ({\vec {d}}_{x})^{2}+({\vec {d}}_{y})^{2}=({\vec {d}}_{net})^{2}$

$\implies ({\vec {d}}_{horizontal})^{2}+({\vec {d}}_{vertical})^{2}=({\vec {d}}_{net})^{2}$

$\implies {\vec {d}}_{net}=\pm {\sqrt {({\vec {d}}_{horizontal})^{2}+({\vec {d}}_{vertical})^{2}}}$

$\implies {\vec {d}}_{net}=\pm {\sqrt {(6{\textsf {meters}})^{2}+(8\;{\textsf {meters}})^{2}}}=\pm {\sqrt {100\;{\text{meters}}}}=\pm 10\;{\text{meters}}$

Note that any displacement can either be positive or negative, depending on what is convenient for the situation. In this case, because we are looking for the magnitude of distance between Hello and Jensen, we can allow for ${\vec {d}}_{net}$ to be positive such that: ${\vec {d}}_{net}=10\;{\text{meters}}$ (using 1 significant figure).