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Resultant Velocity

By Abigail Rogers
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A flat rectangular barge, $48m$ long and $20m$ wide, is headed directly across a stream at $4.5km/hr$. The stream flows at $3.8km/hr$. What is the velocity, relative to the river bed, of a person walking diagonally across the barge at $5km/hr$ while facing the opposite upstream bank?

OK - so doing the math with the angles I now get the speed of the man relative to the river bed to be 2.6637 - does this sound right? Still not sure I'm adding the vectors correctly.

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2 Answers

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Assuming, stream is along positive x direction,

  1. Barge velocity, $V_b = 4.5 \hat{j}$
  2. Stream velocity, $V_s = 3.8 \hat{i}$
  3. person velocity, $V_p = 5(48\hat{i}-20\hat{j})/sqrt(48^2+20^2)$

Net velocity of person = $V = V_b+V_s+V_p$

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Do it stepwise. His speed relative to the barge is 5km/hr ($V \_MB$) and you know the velocity of the barge relative to the riverbed ($V \_BR $).

If the velocity of the man relative to the riverbed is ($V \_MR$)

$$V \_MR = V \_MB + V \_BR $$ Make sure to consider only the components of the velocity in the relevant directions.

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