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.
2 Answers
Assuming, stream is along positive x direction,
- Barge velocity, $V_b = 4.5 \hat{j}$
- Stream velocity, $V_s = 3.8 \hat{i}$
- person velocity, $V_p = 5(48\hat{i}-20\hat{j})/sqrt(48^2+20^2)$
Net velocity of person = $V = V_b+V_s+V_p$
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.
