| Both sides previous revision Previous revision Next revision | Previous revision |
| course_planning:183_projects:s23_week_1_problem_toy_boat_problem_solution [2022/12/20 21:53] – [Project 1: Part B: Zevo toy boats] hallstein | course_planning:183_projects:s23_week_1_problem_toy_boat_problem_solution [2023/01/11 15:56] (current) – hallstein |
|---|
| To figure out the speed of the boat with respect to the water ($v_{\rm b/w}$), you must first recognize that vectors can be broken down into components. Since the water has no velocity perpendicular to the shore, if we send the boat directly across the river it will traverse the width $W$ in a time $t_{\rm across}$ having velocity $v_{\rm b/w}$. So, we send it across and measure the time it takes $t_{\rm across}$. Since this is constant velocity motion, we have $$x(t)=v\,t\quad\longrightarrow\quad W=v_{\rm b/w}t_{\rm across},$$ which we can simply solve for the desired: $$v_{\rm b/w}=\frac{W}{t_{\rm across}}.$$ <wrap caution>$t_{\rm across}=2\,{\rm s}$</wrap> $\quad\longrightarrow \quad v_{\rm b/w}=10\,{\rm m/s}.$ | To figure out the speed of the boat with respect to the water ($v_{\rm b/w}$), you must first recognize that vectors can be broken down into components. Since the water has no velocity perpendicular to the shore, if we send the boat directly across the river it will traverse the width $W$ in a time $t_{\rm across}$ having velocity $v_{\rm b/w}$. So, we send it across and measure the time it takes $t_{\rm across}$. Since this is constant velocity motion, we have $$x(t)=v\,t\quad\longrightarrow\quad W=v_{\rm b/w}t_{\rm across},$$ which we can simply solve for the desired: $$v_{\rm b/w}=\frac{W}{t_{\rm across}}.$$ <wrap caution>$t_{\rm across}=2\,{\rm s}$</wrap> $\quad\longrightarrow \quad v_{\rm b/w}=10\,{\rm m/s}.$ |
| |
| To figure out the speed of the river, you must recognize that you can add relative velocity vectors, and you are taking time measurements with respect to the shore: $$\vec{v}_{\rm b/s}=\vec{v}_{\rm b/w}+\vec{v}_{\rm w/s}.$$ Now, since we don't have a ruler, we have to eliminate distance in all measurements. This is done by sending the boat up and down the river along the same distance (mind you the times will be different). As we send the boat downstream, we have a speed with respect to the shore $$v_{\rm b/s}^{\rm down}=v_{\rm b/w}+v_{\rm w/s}.$$ As we send the boat upstream, we have a speed with respect to the shore $$v_{\rm b/s}^{\rm up}=v_{\rm b/w}-v_{\rm w/s}.$$ Again, since this is constant velocity, we have $$D=v_{\rm b/s}^{\rm down}t_{\rm down}\qquad\mbox{and}\qquad D=v_{\rm b/s}^{\rm up}t_{\rm up}.$$ Making the requisite substitutions and solving for the desired, we come to find $$v_{\rm w/s}=v_{\rm b/w}\bigg(\frac{t_{\rm up}-t_{\rm down}}{t_{\rm up}+t_{\rm down}}\bigg).$$ <wrap caution>$t_{\rm down}=1\,{\rm s},\,\,\,t_{\rm up}=3\,{\rm s}$</wrap> $\quad\longrightarrow\quad v_{\rm w/s}=5\,{\rm m/s}.$ | To figure out the speed of the river, you must recognize that you can add relative velocity vectors, and you are taking time measurements with respect to the shore: $$\vec{v}_{\rm b/s}=\vec{v}_{\rm b/w}+\vec{v}_{\rm w/s}.$$ Now, since we don't have a ruler, we have to eliminate distance in all measurements. This is done by sending the boat up and down the river along the same distance (mind you the times will be different). As we send the boat downstream, we have a speed with respect to the shore $$v_{\rm b/s}^{\rm down}=v_{\rm b/w}+v_{\rm w/s}.$$ As we send the boat upstream, we have a speed with respect to the shore $$v_{\rm b/s}^{\rm up}=v_{\rm b/w}-v_{\rm w/s}.$$ |
| | <WRAP tip> |
| | == Tutor - Leading Questions == |
| | If students are having a difficult time with this, you can ask the following leading questions: |
| | * **Question**: When do you expect the boat to move faster relative to the shore: when it is pointed directly upstream/against the current or downstream with the current? |
| | * **Expected Answer**: ...we expect it to move faster relative to shore when it is pointed with the current... |
| | * **Question**: Algebraically, what is different? |
| | * **Expected Answer**: ...when upstream the two velocities are opposite each other and we need to subtract one from the other, but when it is pointed downstream they are in the same direction and add... |
| | |
| | </WRAP> |
| | |
| | Again, since this is constant velocity, we have $$D=v_{\rm b/s}^{\rm down}t_{\rm down}\qquad\mbox{and}\qquad D=v_{\rm b/s}^{\rm up}t_{\rm up}.$$ Making the requisite substitutions and solving for the desired, we come to find $$v_{\rm w/s}=v_{\rm b/w}\bigg(\frac{t_{\rm up}-t_{\rm down}}{t_{\rm up}+t_{\rm down}}\bigg).$$ <wrap caution>$t_{\rm down}=1\,{\rm s},\,\,\,t_{\rm up}=3\,{\rm s}$</wrap> $\quad\longrightarrow\quad v_{\rm w/s}=5\,{\rm m/s}.$ |
| |
| <WRAP tip> | <WRAP tip> |
| </WRAP> | </WRAP> |
| |
| <code Python boat_sol.py> | <WRAP download 60%>Solution code for Project 1: part B\\ https://www.glowscript.org/#/user/pcubed/folder/solutions/program/RiverCrossingSolution/edit</WRAP> |
| | |
| | |
| | Solution Code: |
| | |
| | <code> |
| GlowScript 2.9 VPython | GlowScript 2.9 VPython |
| | |
| t = t + dt | t = t + dt |
| </code> | </code> |
| | |
| | Changes to given code: |
| | {{course_planning:project_solutions:boat-codesoln.jpg}} |