(3) \( 8 \mathrm{~km} / \mathrm{h} \) In still water a ship can produce a maximum velocity of \( 5 \mathrm{~m} / \mathrm{s} \) in any direction. This same ship is placed in a river with a current of \( 5 \mathrm{~m} / \mathrm{s} \) eastward. With what velocity should the boat be operated with in order to reach point \( Y \) from point \( \mathrm{X} \) ? (Note: Point \( \mathrm{Y} \) is due north of point \( \mathrm{X} \) ) (1) \( 5 \mathrm{~m} / \mathrm{s} \) due west until the ship is far enough out, then \( 5 \mathrm{~m} / \mathrm{s} \) due north. (2) Simultaneously, \( 2.5 \mathrm{~m} / \mathrm{s} \) Due West and \( 2.5 \) \( \mathrm{m} / \mathrm{s} \) Due North. (3) \( 5 \mathrm{~m} / \mathrm{s} \) Due North. (4) \( 5 \mathrm{~m} / \mathrm{s} \) Due North until the ship has reached the northern shore. Then \( 5 \mathrm{~m} / \mathrm{s} \) Due West until the ship reaches point \( \mathrm{Y} \). (5) The ship's engine is not powerful enough to reach point \( Y \). 1. Base your answer to the following question on the diagram below which shows an object of weight \( W \) is suspended from two massless
Added by Alston Z.
Close
Step 1
The object is suspended from two massless objects, so its net weight is zero. Show more…
Show all steps
Your feedback will help us improve your experience
Khoobchandra Agrawal and 76 other Physics 101 Mechanics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
A 400.0 -m-wide river flows from west to east at 30.0 $\mathrm{m} / \mathrm{min}$ . Your boat moves at 100.0 $\mathrm{m} / \mathrm{min}$ relative to the water no matter which direction you point it. To cross this river, you start from a dock at point $A$ on the south bank. There is a boat landing directly opposite at point $B$ on the north bank, and also one at point $C, 75.0 \mathrm{m}$ downstream from $B(\text { Fig. } 3.53)$ . (a) Where on the north shore will you land if you point your boat perpendicular to the water current, and what distance will you have traveled? (b) If you initially aim your boat directly toward point $C$ and do not change that bearing relative to the shore, where on the north shore will you land? (c) To reach point $C :(i)$ at what bearing must you aim your boat, (ii) how long will it take to cross the river, (iii) what distance do you travel, and (iv) and what is the speed of your boat as measured by an observer standing on the river bank?
Ships $\mathrm{A}$ and $\mathrm{B}$ are in the high seas. Ship $\mathrm{A}$ is sailing from South towards North at $5 \mathrm{~m} \mathrm{~s}^{-1}$. Ship B is sailing at $5 \sqrt{10} \mathrm{~m} \mathrm{~s}^{-1}$ in a direction making an angle $\theta$ towards West of South such that $\tan \theta=\sqrt{1.5}$. At $\mathrm{t}=0$, ship $\mathrm{B}$ was at $2000 \mathrm{~m}$ directly North of A. Determine the time and distance of their closest approach. Treat high seas as a flat, plane surface.
A boat can travel at a speed of $8 \mathrm{~km} / \mathrm{h}$ in still water on a lake. In the flowing water of a stream, it can move at $8 \mathrm{~km} / \mathrm{h}$ relative to the water in the stream. If the stream speed is $3 \mathrm{~km} / \mathrm{h}$, how fast can the boat move past a tree on the shore when it is traveling $(a)$ upstream and $(b)$ downstream? (a) If the water was standing still, the boat's speed past the tree would be $8 \mathrm{~km} / \mathrm{h}$. But the stream is carrying it in the opposite direction at $3 \mathrm{~km} / \mathrm{h}$. Therefore, the boat's speed relative to the tree is $8 \mathrm{~km} / \mathrm{h}-3 \mathrm{~km} / \mathrm{h}=5 \mathrm{~km} / \mathrm{h}$. (b) In this case, the stream is carrying the boat in the same direction the boat is trying to move. Hence, its speed past the tree is $8 \mathrm{~km} / \mathrm{h}+3 \mathrm{~km} / \mathrm{h}=11 \mathrm{~km} / \mathrm{h}$.
Recommended Textbooks
University Physics with Modern Physics
Physics: Principles with Applications
Fundamentals of Physics
Watch the video solution with this free unlock.
EMAIL
PASSWORD