Solved Problem on Relative Motion
ES
FR
PT-BR
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A boat has a constant speed, its speed relative to the water has a magnitude equal to 5 m/s. The river current has a constant speed relative to the bank equal to 3 m/s. Determine the magnitude of the velocity of the boat relative to the river banks in the following situations:
a) The boat navigates in the direction of the current (downstream);
b) The boat navigates in the opposite direction to the current (upstream);
c) The boat navigates in the direction perpendicular to the current.
Problem data:
- Speed of the boat relative to water: vb/a = 5 m/s;
- Speed of the river current relative to the banks: va = 3 m/s.
a) The resultant vector
\( {\vec v}_b \)
(velocity of the boat relative to the banks) has a magnitude equal to the sum of the magnitudes of
vectors
\( {\vec v}_a \)
and
\( {\vec v}_{b/a} \),
the boat and the stream have the same direction (Figure 1)
\[
\begin{gather}
\bbox[#99CCFF,10px]
{{\vec v}_b={\vec v}_{b/a}+{\vec v}_a} \tag{I}
\end{gather}
\]
in magnitude
\[
\begin{gather}
v_b=v_{b/a}+v_{a}\\[5pt]
v_b=5\;\mathrm{m/s}+3\;\mathrm{m/s}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{v_{b}=8\;\mathrm{m/s}}
\end{gather}
\]
b) The resultant vector
\( {\vec v}_b \)
has a magnitude equal to the difference of the magnitudes of vectors
\( {\vec v}_a \)
and
\( {\vec v}_{b/a} \),
the boat and the stream have opposite directions, applying the equation (I), in magnitude (Figure 2)
\[
\begin{gather}
v_b=v_{b/a}+(-v_a)\\[5pt]
v_b=5\;\mathrm{m/s}-3\;\mathrm{m/s}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{v_b=2\;\mathrm{m/s}}
\end{gather}
\]
c) The boat arrives at a downstream point relative to the starting point (Figure 3-A), and the resultant
speed will be given by the Pythagorean Theorem, where the resultant vector
\( {\vec v}_b \)
represents the hypotenuse and the vectors
\( {\vec v}_a \)
and
\( {\vec v}_{b/a} \)
represent the legs (Figure 3-B), applying the equation (I) in magnitude
\[
\begin{gather}
v_b^2=v_{b/a}^2+v_a^2\\[5pt]
v_b^2=(5\;\mathrm{m/s})^2+(3\;\mathrm{m/s})^2\\[5pt]
v_b=\sqrt{25\;\mathrm{m^2/s^2}+9\;\mathrm{m^2/s^2}\;}\\[5pt]
v_b=\sqrt{34\;\mathrm{m^2/s^2}}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{v_b=5.8\;\mathrm{m/s}}
\end{gather}
\]
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .