Q2.12 Relative velocities 2 (S)

A man rows a dinghy across a river of width 1 km, in which the current has a uniform speed of 3 kmhr-1. The man rows at a steady speed of 4 kmhr-1 with respect to the water.
  1. Suppose first that the man maintains a bearing perpendicular to the current flow.
    • [(a)] Construct a vector triangle relating the velocities of the man assigned in the reference frame of the river, and the reference frame of the bank.
    • [(b)] Deduce the velocity of the man relative to the bank, and the time for the crossing.
  2. Now suppose that the bearing is adjusted so that, as seen from the bank, the dinghy moves perpendicular to the bank.
    • [(a)] Construct the new velocity vector triangle.
    • [(b)] Recalculate the velocity of the man relative to the bank, and the time for the crossing.
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Hint

You will need Key Point 1.9. The vectors you need to relate are $\vec{v}_{DB}$, $\vec{v}_{DW}$ and $\vec{v}_{WB}$ where D B and W stand for dinghy, bank and water.
 

Solution

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Solution

Let subscripts D B and W label the dinghy, the bank and the water respectively.

According to Key Point 1.9 

\[ \vec{v}_{DB}= \vec{v}_{DW} +\vec{v}_{WB} \]
where

  • $\vec{v}_{DB}$ means the velocity of the dinghy relative to the bank
  • $\vec{v}_{DW}$ means the velocity of the dinghy relative to the water
  • $\vec{v}_{WB}$ means the velocity of the water relative to the bank.
  1. In the first case (where an observer floating downstream with the current sees the dinghy moving perpendicular to the current flow) we know that $\vec{v}_{DW}$ is perpendicular to $\vec{v}_{WB}$

    Thus the vector triangle must be of the form shown.

    According to a bank observer the dinghy is carried downstream by the current.

    Using vDW=4kmh-1 and vWB=3kmh-1 we find that the magnitude of $\vec{v}_{DB}$ is

    vDB=5 kmhr-1

    and its direction is at an angle with respect to the bank of

    \[ \phi= \tan^{-1}(4/3) = 53 ^{\circ} \hspace{0.5cm}\mbox{\rm (2sf)}\hspace{0.5cm} \]

    Graphic - No title - boatbearinga

    The time for the crossing is (with w the river width)

    \[ t_{\mbox{\rm cross}} = \frac{w}{v_{DW}} = \frac{1}{4} hr \]

  2. In the second case (where an observer on the bank sees the dinghy moving perpendicular to the bank) we know that $\vec{v}_{DB}$ is perpendicular to $\vec{v}_{WB}$. Thus the vector triangle must be of the form shown.

    The bearing is now such that the upstream component of the dinghy’s velocity relative to the water just cancels the effect of the current.

    From the revised vector triangle we read off

    \[ v_{DB} = \sqrt{4^2 -3 ^2} =\sqrt{7} =2.6 \,km \cdot hr^{-1}\hspace{0.5cm}\mbox{\rm (2sf)}\hspace{0.5cm} \]
    perpendicular to the bank.

    Graphic - No title - boatbearingb

    The time for the crossing is now

    \[ t_{\mbox{\rm cross}} = \frac{w}{v_{DB}} = 0.38 hr \hspace{0.5cm}\mbox{\rm (2sf)}\hspace{0.5cm} \]