Question

A man goes for a walk, starting from the origin of an xyz coordinate system, with the xy plane horizontal and the x axis eastward. Carrying a bad penny, he walks$\mathbf{1300}\mathbf{}\mathbf{m}$east,$\mathbf{2200}\mathbf{}\mathbf{m}$north, and then drops the penny from a cliff $\mathbf{410}\mathbf{}\mathbf{m}$high. (a) In unit-vector notation, what is the displacement of the penny from start to its landing point? (b) When the man returns to the origin, what is the magnitude of his displacement for the return trip?

Short answer

a) displacement of the penny from start to its landing point is$\left(1300\mathrm{m}\right)\hat{\mathrm{i}}+\left(2200\mathrm{m}\right)\hat{\mathrm{j}}=\left(-410\mathrm{m}\right)\hat{\mathrm{k}}$

b) magnitude of displacement for the return trip is$2.56\times {10}^3\mathrm{m}/\mathrm{s}$

Step-by-step solution

To understand the concept of the problem

Using the vector algebra and simple mathematical operations, the unknown quantities can be found. Here the vector addition is to be used to find the magnitude of vector.

In this problem, the displacement vector $\stackrel{\rightharpoonup }{d}$can ne written as,

$$\begin{gathered} \overrightarrow{d}=\overrightarrow{x}\hat{\mathrm{i}}+\overrightarrow{y}\hat{\mathrm{j}}+\overrightarrow{z}\hat{\mathrm{k}} \\ \left(\mathrm{i}\right) \end{gathered}$$

With magnitude,

$$\begin{gathered} \left|\overrightarrow{d}\right|=\sqrt{x^2+y^2+z^2} \\ \left(\mathrm{ii}\right) \end{gathered}$$

Given are,

role="math" localid="1656308404966" $$\begin{gathered} x=1300\mathrm{m} \\ \mathrm{y}=2200\mathrm{m} \\ \mathrm{z}=410\mathrm{m} \end{gathered}$$

To find the displacement of the penny from start to its landing point

Substituting the values of x, y, and z, in equation (i) the displacement from start to landing point is given by,

$\overrightarrow{d}=\left(1300\mathrm{m}\right)\hat{\mathrm{i}}+\left(2200\mathrm{m}\right)\hat{\mathrm{j}}+\left(-410\mathrm{m}\right)\hat{\mathrm{k}}$

To find the magnitude of the displacement for the return trip

In this case the final position and the initial position of the man is same. Thus, the net displacement is zero. Therefore, the vector ${\hat{d}}^{\text{'}}$for the return trip which can be written as,

$\overrightarrow{d^{\text{'}}}=\left(-1300\mathrm{m}\right)\hat{\mathrm{i}}+\left(-2200\mathrm{m}\right)\hat{\mathrm{j}}$

Thus the magnitude ofis given by,

$$\begin{gathered} d^{,}=\sqrt{{\left(-1300\mathrm{m}\right)}^2+{\left(-2200\mathrm{m}\right)}^2} \\ {\overrightarrow{d}}^{\text{'}}=2.56\times {10}^3\mathrm{m}/\mathrm{s} \end{gathered}$$