91Ó°ÊÓ

You can find the relative velocity of an object with respect to a frame of reference if its velocity with respect to another frame of reference and the relative velocity of the two reference frames are known. Take, for example, the following.

If an object A moves with velocity${\stackrel{→}{v}}_{\mathrm{AB}}$with respect to a frame of reference B and object B moves with velocity${\stackrel{→}{v}}_{\mathrm{BC}}$with respect to another frame of reference C, then the relative velocity of A with respect to C is given by:

${\stackrel{→}{v}}_{\mathrm{AC}}={\stackrel{→}{v}}_{\mathrm{AB}}+{\stackrel{→}{v}}_{\mathrm{BC}}$

The magnitude of the velocity of Huck Finn with respect to his raft is $v_{\mathrm{HR}}=0.70{\mathrm{ms}}^{-1}$.

The magnitude of the velocity of the raft with respect to the bank of the river is$v_{\mathrm{RB}}=1.50{\mathrm{ms}}^{-1}$.

Let the velocity of Huck Finn relative to the river bank be ${\stackrel{→}{v}}_{\mathrm{HB}}$.

The velocity of Huck Finn relative to the river bank is equal to the vector sum of the velocity of Huck Finn with respect to his raft and the velocity of the raft with respect to the bank of the river, i.e.,

${\stackrel{→}{v}}_{\mathrm{HB}}={\stackrel{→}{v}}_{\mathrm{HR}}+{\stackrel{→}{v}}_{\mathrm{RB}}$

Since the motion of Huck Finn is perpendicular to the raft motion with respect to the shore, his motion can be shown as:

Here,$\mathrm{θ}$ is the angle which the velocity vector of Huck Finn makes with respect to the river bank.

From the figure, the magnitude of the velocity of Huck Finn relative to the riverbank can be found using the Pythagoras theorem as:

$\begin{array}{c}{v}_{\mathrm{HB}}=\sqrt{v_{\mathrm{HR}}^2+v_{\mathrm{RB}}^2}\\ =\sqrt{{\left(0.70{\mathrm{ms}}^{-1}\right)}^2+{\left(1.50{\mathrm{ms}}^{-1}\right)}^2}\\ =\sqrt{2.74{\mathrm{ms}}^{-1}}\\ =1.65{\mathrm{ms}}^{-1}\end{array}$

From the figure, angle $\mathrm{θ}$can be found as follows:

$\begin{array}{c}\tan \mathrm{θ}=\frac{v_{\mathrm{HR}}}{v_{\mathrm{RB}}}\\ =\frac{\left(0.70{\mathrm{ms}}^{-1}\right)}{\left(1.50{\mathrm{ms}}^{-1}\right)}\\ =0.467\\ \mathrm{θ}={\tan }^{-1}\left(0.467\right)\\ =25°\end{array}$

Thus, the velocity of Huck Finn relative to the river bank is $1.65{\mathrm{ms}}^{-1}$in the direction that makes an angle of $25°$with the river bank.