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PLZ HELP!!!!!! 6. A boat is traveling west across a river. The river is 400.0m wide and the speedometer reading on the boat is 30.0 km/hr. The current in the river is flowing south at 5km/hr. The boat driver is not compensating the current. c) How far downstream did the boat travel? d) What was the total distance traveled by the boat?

Physics
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You didn't state what you need to find, but it looks like you need to find the component of the velocity along the width of the river. If the width of the river is in the x direction flow of the river is in the y direction, the resultant velocity will be \[v ^{2}=v _{x}^{2}+v _{y}^{2}\]where v is the resultant velocity; vx is the velocity in the x direction; and vy is the velocity in the y direction. You should be able to use that to find vx.
@PsiSquared Since the speedometer is measured relative to the water, and the boat is only moving west relative to the water it's in (ideally), wouldn't the velocity across the river be \(30\ \rm km/hr\)? In your scenario, \(v_x=30\ \rm km/hr\) and is thus known? The downstream velocity would be \(v_y=5\ \rm km/hr\), as given. Then \(\vec v=(v_x,~v_y)\) would be unknown. But the question might have changed in that content anyway, since the questions were not posted at the time you responded. @aminehr I'd look into how long it'd take to cross the river, basic \(v=\dfrac{\Delta d}{\Delta t}\). Then see how far downstream you'd go in that time. Same equation! To find the total distance, you'll want to use the Pythagorean theorem using the river width and downstream travel distance. Here's why:|dw:1398128439699:dw|
okay but when you use v=d/t to find the time and plug in the value back into that equation to find the distance, wont the answer be the 400 m that i used for the distance?

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Well, you find the time across the river using \(v=30\ \rm km/hr\) and \(d=400.0m\) But convert the units to be the same!! I suggest that you use kilometers. When you try to find the downstream distance, you'll use \(v=5\rm\ km/hr\) and \(t=\rm what\ you\ got\ above\uparrow\)
oh okay i get it thx. i forgot to convert the 400m to km.
Okay!
@theEric, @aminehr: the boat's speedometer only measures speed with respect to the relative flow of water. The relative flow of water is the sum of two vectors: the vector defying westward travel and the vector defining downstream travel. The speedometer can't tell you which direction the river is flowing. It's much the same as the airspeed indicator on an airplane. It only tells you the speed of air flowing over the plane, and the speed of that air is the sum of two velocity vectors: the velocity of the plane and the velocity of the wind (not the relative wind). So to solve this problem we have to find the velocity components: the velocity west across the river and the velocity downstream. The downstream velocity is given. Once we have the westward velocity, we can calculate how long it takes to cross the river, and from that we can then calculate how far down the river the boat travelled.
@PsiSquared I think we're perceiving the problem differently. To both of you, here's what I think. See what you think of what I think! First I'll look at it from an informal point of view, then I'll use vector reasoning as @PsiSquared did. So, the boat is only propelling itself westward. But it is still going downstream. The boat propels itself westward, and so it moves west relative to the water (the body it's propelling itself within). That, what is relative to the water, is what the speedometer is reading (\(30\rm\ km/hr\ west\))*. It is moving downstream, but \(\rm with\) the water. So the speedometer does not pick up on that. It's downstream velocity, since the boat does not propel itself downstream or upstream, is just going to be the velocity of the water (\(5\rm\ km/hr\ south\))**. Thus, its total velocity*** can be found by vector-adding the two velocities. Since south is perpendicular to west, the velocities are orthogonal, and so the magnitude (like length) of their vector sum can be found by the Pythagorean theorem. * With respect to water and ground ** With respect to ground. (It's zero, or still, with respect to water) *** With respect to ground I'll post this so that I don't lose it! If there are errors in it, please let me know! I will proceed to make a visual for the vector additions for this problem and the plane scenario that @PsiSquared introduced!
|dw:1398207973870:dw| |dw:1398207568318:dw|
Here I tried to make the vector lengths be more different, so that you can tell what's going on better.|dw:1398208368561:dw|

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