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General Physics I:

Subsection 3.1.3 Relative velocity

(a)
(b)
(c)
Figure 3.1.9.
It is the velocity that an object \(A\) would appear (to be moving or at rest) to an observer on the object \(B\) and vice-versa. If \(v_{AB}\) is a velocity vector of an object \(A\) relative to another object \(B\text{,}\) then
\begin{equation*} \vec{v}_{AB} = - \vec{v}_{BA} \end{equation*}
and the basic equation for relative velocity becomes:
\begin{equation*} \vec{v}_{az} = \vec{v}_{ab} + \vec{v}_{bc} + \cdots + \vec{v}_{yz} \end{equation*}
From Figure 3.1.9.(a) we have
\begin{equation*} \vec{v}_{A}+\vec{v}_{BA} = \vec{v}_{B} \end{equation*}
\begin{equation*} -\vec{v}_{A}+\vec{v}_{B} = \vec{v}_{BA}; \end{equation*}
\begin{equation*} \therefore \vec{v}_{BA} =\vec{v}_{B}-\vec{v}_{A} \end{equation*}
where \(\vec{v}_{A} \) and \(\vec{v}_{B}\) are called absolute velocities of objects A and B, respectively. Consider an airplane \(A\) is coming with velocity \(\vec{v}_{A} =600 \quad km/h\) due \(\theta=35^{o}\) south of east and another one \(B\) is coming with velocity \(\vec{v}_{A} =700 \quad km/h\) due west as shown in Figure 3.1.9.(c). Find the relative velocity of \(B\) with respect to \(A\text{.}\) To solve such problem assume that object \(A\) is at rest. The object \(A\) must be the object with respect to which we are finding the velocity of object \(B\text{.}\)
To assume that the object \(A\) is at rest take its velocity in negative direction to draw a parallelogram of adjacent vectors \(\vec{v}_{B}\) and \(-\vec{v}_{A}\) so that their resultant velocity is \(\vec{v}_{BA} \text{.}\) From Figure 3.1.9.(c), we have -
\begin{equation*} \vec{v}_{BA} = -\vec{v}_{A}+\vec{v}_{B}; \end{equation*}
\begin{equation*} v_{Bx}=v_{B}\cos\theta = 491.5 \quad km/h, \end{equation*}
\begin{equation*} v_{By}=v_{B}\sin\theta = 344 \quad km/h, \end{equation*}
\begin{equation*} \text{and} v_{A}=700 \quad km/h \end{equation*}
\begin{equation*} v_{Rx}=v_{A}+v_{Bx}=1191.5 \quad km/h; \end{equation*}
\begin{equation*} v_{Ry}=v_{By}=344 \quad km/h \end{equation*}
\begin{equation*} \therefore v_{BA} = \sqrt{v^{2}_{Rx}+ v^{2}_{Ry}} = 1240.16 km/h \end{equation*}
where the direction of relative velocity is given by
\begin{equation*} \phi = \tan^{-1}\left(\frac{v_{Ry}}{v_{Rx}}\right) = 16^{o} \quad \text{S of E}. \end{equation*}