# Ferry Boat HW Problem ( HELP)

1. Jan 11, 2006

### flyingdonkey21

Two ferryboat each traveling at different constant speed start at the same instant from opposite sides of the Hudson River, one going from New York City to Jersey City and the other from Jersey City to New York City. They pass one another at a point 720 yard from New York City shore. After arriving at their respective destinations, each boat spends precisely 10 minutes at the opposite shore to change passengers before switching directions. On the return trip, the two boats meet at a point 400 yards from the Jersey shore.
What is the width of the river?
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I look at this and my intuitiom says to use two linear motion w/ constant accelaration equations and set them equal to each other. However, I look at them all and I don't see how I can do this problem without the velocities of the two boats given. If not providing me with a solution to this, can someone at least point me in the right direction? Thank you.
Boat 1
a=0
v=CONSTANT
v=0 from t1 to t1+10
Boat 2
a=0
v=CONSTANT
v=0 from t2 to t2+10
pass @ x=720 and x=(total distance-400)

flyingdonkey21

2. Jan 11, 2006

### Hootenanny

Staff Emeritus
At the point of crossing they will have both travelled the same distance and been travelling for the same period of time. (S1=S2 and T1=T2).

3. Jan 11, 2006

### flyingdonkey21

?????

How will they have traveled the same distance and the same time if they are traveling at two different velocities? After t seconds wouldn't the boat with the higher velocity have covered more distance?

4. Jan 12, 2006

### flyingdonkey21

anymore hints????

i still don't comprehend how to solve this problem?.....anyone able to give a little more insight?....thanks

5. Jan 12, 2006

### flyingdonkey21

solvable????

is there even enough information in this problem to solve it?

6. Jan 13, 2006

### flyingdonkey21

aha!!!

i got it finally....thanks whoever gave it a shot

7. Jan 13, 2006

### andrevdh

Maybe you should enligten us on how you solved it? Seeing that nobody else seemed able to do it (it got quite a lot of views!). I thought about it and read up on relevant theory, but also had the opinion that the amount of given information seemed hopelessly insufficient.

8. Jan 13, 2006

### flyingdonkey21

Sorry, I didn't have time to post the solution last night, but here it goes.
There are two different ways (maybe more) to solve this problem. One requires a lot of algebra and I do not have the patience to write it all out, but it basically comes down to having two equations with arbitrary V1 and V2 and the velocities drop out and you are left with one equation with the width of the river the only unknown. Method number two is much simpler. If you are trying to follow along I'd sketch a diagram of the problem so that my explanation will be a bit clearer. Obviously one boat is a bit slower than the other boat. The slower boat travels 720 yards in after t1 (time 1). The faster boat travels the lenght of the river minus the 720 yards (w-720) after t1. At this point the two boats have traveled a combined distance of w. After both boats make it all the way across the river they have traveled a total of 2(w) after t2. Since both boats rest at the opposite shores for the same amount of time, we do not need to take the rest into account. Now, when they meet the second time they are 400 yards away from the Jersey shore. And the total combined distance traveled by the two boats is now 3w after 3t. Since the boats have constant speed, the slower boat has now traveled 3(720 yards) which is 2160 yards. Knowing that at their second meeting they met 400 yards from the shore, we can safely assume that the slow boat has traveled w+400 yards. Now set the this equal to 2160 yards. 2160=w+400 ...... w=1760 yards ....I know this my not intuitively make sense at first but my answer was confirmed by my professor today as well as the alternate solution.

9. Jan 15, 2006

### andrevdh

The difficult part to understand is your statement that "Since the boats have constant speed, the slower boat has now traveled 3(720 yards) which is 2160 yards." Which could be mathematically proven as follows:
Assume the elapsed time when they first meet is $t_1$ and when they meet again is $t_3$, then we have that the total distances traveled by these two boats at these two particular times are given by
$$t_1\ v_1\ +\ t_1\ v_2\ =\ w$$ ...(1)
and
$$t_3\ v_1\ +\ t_3\ v_2\ =\ 3w$$ ...(2)
substituting w in (1) for w in (2) gives $t_3\ =\ 3t_1$. I assumed that the water speed of the boats were constant, and therefore that their speeds relative to the embankment would be altered by the speed of the river. This way one ends up with 8 or 9 unknowns! Thank you for your response.

Last edited: Jan 15, 2006