What is the width of the Hudson River?

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Homework Help Overview

The problem involves two ferryboats traveling across the Hudson River at constant speeds, with the goal of determining the width of the river based on their meeting points. The scenario includes specific distances from the shores where the boats meet and the time spent at each shore.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to set up equations based on the distances traveled by each boat and their speeds. They express uncertainty about how to relate these equations to find the width of the river. Other participants raise questions about unit conversions and the accuracy of measurements.

Discussion Status

The discussion is ongoing, with participants providing hints and questioning assumptions about unit conversions. There is no explicit consensus on the approach to take, and multiple interpretations of the problem are being explored.

Contextual Notes

There are discussions about the accuracy of unit conversions between yards and feet, with some participants expressing confusion about the definitions of these units. The original poster's equations are based on the assumption of constant speed and no acceleration.

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Physics Problem using constant speed.

hey guys, here the problem.

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?


I drew out diagrams of the river and added in the all the nessasary details. I'm using X=Xi+Vit+1/2at^2. For boat A, solved and got 720=Viat (Via is the initial speed of boat A) and for boat B I solved and got x-720=Vibt (Vib is the intial speed of boat B) (I'm using x as the total width of the river). There is no acceleration, so the 1/2at^2 will cancel out.

For the second time they pass each other i got the equations
Boat A=400=Viat
Boat B=x-400=Vibt

I was thinking that you have to set these equations equal to one another and some how get a ratio, but, these equations are getting me nowhere. Any tips to get me started in the right direction from here will be greatly appreciated :smile:
 
Last edited:
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1 yd=30'=0.9144 m.

Daniel.
 
Last edited:
dextercioby,

"Nothing is more practical than a good theory"

Perhaps, but the best theory in the world won't do you much good if you think 1 yd = 30' :wink:
 
hmm, thanks for the hints, do you guys think that it'll be easier in yards?
 
what is the numerical solution yoyoma ?

marlon
 
i don't know the solution to it marlon.
 
"feet" i meant...' is for inch." is for feet.I reversed them.It's natural,i'm not an American.

Daniel.
 
dextercioby,

""feet" i meant...' is for inch." is for feet.I reversed them.It's natural,i'm not an American."

I guess we Americans are also the only ones who know that a yard is 36 inches, not 30. We are sooooooo smart! :wink:
 
What? A yard is 30 feet,i.e.360 inches...

Daniel.
 

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