Question

A protester carries his sign of protest, starting from the origin of an${}^{\mathbf{x}\mathbf{y}\mathbf{z}}$coordinate system, with the${}^{\mathbf{x}\mathbf{y}}$plane horizontal. He moves${}^{\mathbf{40}\mathbf{}\mathbf{m}}$in the negative direction of the${}^{\mathbf{x}}$axis, then${}^{\mathbf{20}\mathbf{m}}$along a perpendicular path to his left, and then${}^{\mathbf{25}\mathbf{m}}$up a water tower. (a) In unit-vector notation, what is the displacement of the sign from start to end? (b) The sign then falls to the foot of the tower. What is the magnitude of the displacement of the sign from start to this new end?

Short answer

(a) The displacement of the sign from start to end is ${}^{-40\hat{\mathrm{i}}-20\hat{\mathrm{j}}+25\hat{\mathrm{k}}}$.

(b) The magnitude of the displacement of the sign from start to new end (foot of the tower) ${}^{45\mathrm{m}}$.

Step-by-step solution

Given data

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.

Understanding the displacement

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.

(a) Determination of the displacement

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, $d=\left(-40\hat{\mathrm{i}}-20\hat{\mathrm{j}}+25\hat{\mathrm{k}}\right)\mathrm{m}$

(b) Determination of the magnitude of displacement when the sign falls to the foot of the tower

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,

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

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

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

Thus, the magnitude of displacement is 45 m.