## Homework Statement

A person in a rowboat crosses a river which flows with 4.043 km/hr. The person rows with maximum velocity of 3.138 km/hr relative to the water (measured while rowing on a still lake) and heads the boat straight across the river - along a line at 90 degrees with respect to the river shore pointing to a point on the opposite shore. How far downstream from this point in km does the person reach the opposite shore if the river is 0.2773 km wide.

## Homework Equations

VAC=VAB+VBC, then
Vbg=Vbw+Vwg

## The Attempt at a Solution

I tried tan-1 (4.043/3.138)= 52.2 as theta
3.138(cos)(52.2)=1.92
then i thought i have to divide the width. (.2773/1.92)=.1444
but thats wrong, i tried this problem over and over again and i just dont know what im doing wrong.

vela
Staff Emeritus
Homework Helper

## Homework Statement

A person in a rowboat crosses a river which flows with 4.043 km/hr. The person rows with maximum velocity of 3.138 km/hr relative to the water (measured while rowing on a still lake) and heads the boat straight across the river - along a line at 90 degrees with respect to the river shore pointing to a point on the opposite shore. How far downstream from this point in km does the person reach the opposite shore if the river is 0.2773 km wide.

## Homework Equations

VAC=VAB+VBC, then
Vbg=Vbw+Vwg

## The Attempt at a Solution

I tried tan-1 (4.043/3.138)= 52.2 as theta
3.138(cos)(52.2)=1.92
then i thought i have to divide the width. (.2773/1.92)=.1444
but thats wrong, i tried this problem over and over again and i just dont know what im doing wrong.
Can you explain why you did what you did in your solution? You should have a good reason for each step.

Here are some questions to consider:

What are the units of 1.92? Do they match what you think it's supposed to represent?

What are the units of your final answer? Are they consistent with what you're trying to find?

Would you mind posting a diagram of what you did. The diagram is very important

Im trying to follow the notes from my professsor that i have for a similar problem. I set up a triangle and got opposite over adjacent thats why i used tan-1.
1.92km/h , after i divided out the width I multiplied .1444 by 60 because its km/hour and got 8.7 km/h. That answer is still wrong. Im not sure what else to try.

I dont know how to post a diagram on this. I have a right triangle. with bw=3.138 on the y axis, and wg= 4.043 on the x axis, with bg that im trying to find on the hypotenuse.

Don't try to work it out using your diagram. I find that for this type of Relative Velocity question you just use it as a visualisation aid. It sets up the maths, and the solving tends to be 100% maths.

Try this, I hope you're familiar with i and j notation.

River Velocity... Vr = 4.043i + 0j
ACTUAL Boat Velocity... Vb = XcosA i + XsinA j
Boat relative to river... Vbr = 0i + 3.138j

Do you understand this? Can you figure out the rest?

Im sorry, im not familiar with the i and j notation.

i axis = x axis
j axis = y axis

Basically the same co-ordinate system just different letters

vela
Staff Emeritus