202104150729 Homework 1 (Q1)

A man starts from the origin and walks $30\,\mathrm{m}$ due east in $25\,\mathrm{s}$ . Then, he walks $10\,\mathrm{m}$ due south in $10\,\mathrm{s}$ and $18\,\mathrm{m}$ due northwest in $15\,\mathrm{s}$ . Please take the paths due east and due north as the positive $x$ and $y$ directions respectively in the Cartesian plane.

(a) Sketch the man’s path on the Cartesian plane.

(b) What is the average velocity of the man?

(c) What is the average speed of the man?

Solution.

(The solution below is based on the manuscript of 2014-2015 PHYS1250 Fundamental Physics Homework Solutions.)

(a)

(b) The average velocity of the man is

$\begin{aligned} \mathbf{OC} & = \mathbf{OA} + \mathbf{AB} + \mathbf{BC} \\ & = 30\,\hat{\mathbf{i}} - 10\,\hat{\mathbf{j}} + 18\bigg( -\frac{1}{\sqrt{2}}\,\hat{\mathbf{i}} + \frac{1}{\sqrt{2}}\,\hat{\mathbf{j}} \bigg) \\ & = \bigg( 30-\frac{18}{\sqrt{2}} \bigg) \hat{\mathbf{i}} + \bigg( \frac{18}{\sqrt{2}}-10 \bigg) \,\hat{\mathbf{j}}\\ & = 17.27\,\hat{\mathbf{i}} + 2.73\,\hat{\mathbf{j}} \end{aligned}$

$\begin{aligned} \therefore \mathbf{v}_{\textrm{avg}}& =\frac{\mathbf{OC}}{\Delta t} \\ & = \frac{17.27\,\hat{\mathbf{i}} + 2.73\,\hat{\mathbf{j}}}{50} \\ & = 0.35\,\hat{\mathbf{i}} + 0.055\,\hat{\mathbf{j}} \\ \therefore\quad |\mathbf{v}_{\textrm{avg}}| & = \sqrt{0.35^2+0.055^2} = 0.35\,\mathrm{m\, s^{-1}} \end{aligned}$