91Ó°ÊÓ

Displacement in the negative x direction is $40\mathrm{m}$

Displacement in perpendicular direction is 20 m.

Displacement up a water tower is 25 m.

Displacement may be defined as the shortest distance between two points.

The magnitude of the displacement ${}^d$can be expressed in three dimension as:

$d=\sqrt{{\left(∆x\right)}^2{\left(∆y\right)}^2{\left(∆z\right)}^2}$ … (i)

Here, ${}^{∆x}$, ${}^{∆y}$, and ${}^{∆z}$can be consider as width, length, and height respectively.

The three displacements of the sign are given as:

$$\begin{gathered} 40\mathrm{m}\mathrm{along}-\mathrm{x}\mathrm{axis},\mathrm{so}\mathrm{it}\mathrm{is}-40\hat{\mathrm{i}} \\ 20\mathrm{m}\mathrm{along}-\mathrm{y}\mathrm{axis},\mathrm{so}\mathrm{it}\mathrm{is}-20\hat{\mathrm{j}}\mathrm{and} \\ 25\mathrm{m}\mathrm{along}\mathrm{the}\mathrm{z}\mathrm{axis},\mathrm{so}\mathrm{it}\mathrm{is}25\hat{\mathrm{k}} \end{gathered}$$

The displacement of the point start to end is, $d=\left(-40\hat{\mathrm{i}}-20\hat{\mathrm{j}}+25\hat{\mathrm{k}}\right)\mathrm{m}$

The sign falls to the foot of the tower, so there is no displacement of sign along the z axis.

So the displacement vector is,

$\stackrel{→}{d}=\left(-40\hat{\mathrm{i}}-20\hat{\mathrm{j}}\right)\mathrm{m}$

The magnitude of the displacement of the sign is calculated as follows:

$$\begin{gathered} \left|\stackrel{→}{d}\right|=\sqrt{{\left(-40\mathrm{m}\right)}^2+{\left(-20\mathrm{m}\right)}^2} \\ =44.7\mathrm{m} \\ ≈45\mathrm{m} \end{gathered}$$

Thus, the magnitude of displacement is 45 m.