Which Jet Ski Crosses the River Faster?

[wadini]

Two people take identical Jet Skis across a river, traveling at the same speed relative to the water. Jet Ski A heads directly across the river and is carried downstream by the current before reaching the opposite shore. Jet Ski B tavels in a direction that is 35 degrees upstream and arrives at the opposite shore directly across from the starting point. a) Which Jet Ski reaches the opposite shore in the least amount of time? b.) Confirm your answer to part a by finding the ratio of the time it takes for the two Jet Skis to cross the river.

 

[LowlyPion]

What have you tried?

 

[belliott4488]

Interesting problem. Is there anything in particular you wanted to discuss about it?

 

[wadini]

OKay so I figured out that Jet Ski gets there faster and now I have to find the ratio of the time...and I think I do that by doing something like showing how Vsw + Vwg with sw meaning skier on the water and gw meaning the ground water...I was thinking of assigning random numbers for A and B and then showing how from there but the thing is I don't really know how to prove it...I just know because of common sense that Skier A will reach there faster... what do you guys think?

 

[wadini]

sorry Jet Ski A ** gets there faster

 

[tiny-tim]

Two people take identical Jet Skis across a river, traveling at the same speed relative to the water. Jet Ski A heads directly across the river and is carried downstream by the current before reaching the opposite shore. Jet Ski B tavels in a direction that is 35 degrees upstream and arrives at the opposite shore directly across from the starting point. a) Which Jet Ski reaches the opposite shore in the least amount of time? b.) Confirm your answer to part a by finding the ratio of the time it takes for the two Jet Skis to cross the river.

Hi wadini!

Use a vector triangle for each Jet Ski …

what do you get?

 

[wadini]

A is like x
and B is like y

should I do something like cos(35) to get x and then do something from there?

 

[wadini]

actually I have no idea how to go about this truthfully.

 

[tiny-tim]

vector triangle

Do you know what a vector triangle is?

Have you been shown how to draw one?

 

[wadini]

No, I have no idea what a vector triangle is.

 

[wadini]

I will google it.

 

[wadini]

okay so it is something about all the points of the triangle meeting at the same point which works because the skiers do eventually meet at the same point but one just gets there faster than the other.

 

[tiny-tim]

ok … velocities are vectors, and so they obey the vector law of addition

in other words, you can add velocities the same way you add vectors.

For the second Jet Ski, draw arrows representing the three velocities …

that's V_BR, the velocity of B relative to the river,

V_BG, the velocity of B relative to the ground,

and V_RG, the velocity of the river relative to the ground.

Those three vectors should make a closed-up triangle.

Then do a triangle for A also, and then put the two triangles next to each other

 

[wadini]

so if I drew this correctly the first and second velocities relative to the water is the same...but I don't really know. I don't understand what I am doing. Why would the speed relative to the water be the same...shouldn't they be different because skier number 2 is going at an angle and gets to the shore after skier number 1 ? I am unbelievably confused!

 

[wadini]

and how am I supposed to get the time ratio from all this?

 

[LowlyPion]

and how am I supposed to get the time ratio from all this?

Draw the vectors out.

In both cases you have a right triangle don't you?

Except in the one case you have traveled the length of the hypotenuse.
In the other you traveled one leg.

What is the ratio of that leg to the hypotenuse? Anything pop to mind?

 

[LowlyPion]

Why would the speed relative to the water be the same...

Because that's the only speed the jet ski can go. It's going constant velocity in the water. The vector of the current moves it, but in the water frame of reference it is still going the same speed whether up stream or down.