Boats A & B: Speed & Separation

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Homework Help Overview

The problem involves two boats departing from the same shore and traveling at different speeds and directions. The objective is to determine the speed of boat A relative to boat B and the time it takes for the boats to be 800 feet apart.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the relative velocity of the two boats using vector components but struggles with determining the time when they are 800 feet apart.
  • Some participants question the method of calculating time based on the relative velocity found.
  • Others suggest that the relationship between distance, speed, and time should be applied to find the time taken for the specified separation.

Discussion Status

Participants are exploring the calculations related to relative velocity and the time it takes for the boats to reach a certain distance apart. Some guidance has been provided regarding the use of the relative velocity to determine the time, but there is no explicit consensus on the approach being taken.

Contextual Notes

There appears to be confusion regarding the application of the relative velocity in the context of the problem, as well as the correct interpretation of the time calculation based on the distance and speed. The original poster expresses difficulty in completing the necessary calculations.

VinnyCee
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Two boats leave the shore at the same time and travel in the directions shown. If [itex]v_A\,=\,20\,\frac{ft}{s}[/itex] and [itex]v_B\,=\,15\,\frac{ft}{s}[/itex], determine the speed of boat A with respect to boat B. How long after leaving the shore will the boats be 800 ft apart?

http://img366.imageshack.us/img366/8164/engr204problem121995cj.jpg

I figured the relative velocity:

[tex]\overrightarrow{v_A}\,=\,{\left(-20\,sin\,30\right)\,\widehat{i}\,+\,\left(20\,cos\,30\right)\,\widehat{j}}\,\frac{ft}{s}[/tex]

[tex]\overrightarrow{v_B}\,=\,{\left(15\,cos\,45\right)\,\widehat{i}\,+\,\left(15\,sin\,45\right)\,\widehat{j}}\,\frac{ft}{s}[/tex]

[tex]\overrightarrow{v_{A/B}}\,=\,\overrightarrow{v_B}\,-\,\overrightarrow{v_A}[/tex]

[tex]\overrightarrow{v_{A/B}}\,=\,{(20.6)\,\widehat{i}\,+\,(-6.71)\,\widehat{j}}\,\frac{ft}{s}[/tex]

[tex]v_{A/B}\,=\,\sqrt{20.6^2\,+\,(-6.71)^2}\,=\,21.7\,\frac{ft}{s}[/tex]

[tex]\theta\,=\,tan^{-1}\,\right(\frac{6.71}{20.6}\right)\,=\,18[/tex]

Here is where I can't figure out how to get the time t for when the boats are 800 ft apart. I try to complete the triangles above and solve for the length of one of the sides and get a wrong answer when I divide by the velocity to get the time since the velocity is constant. Please help!
 
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You have found the relative velocity, i.e. how fast they are moving apart. How long will it take them to become 800m apart?
 
Last edited:
Thanks!

It seems the simpler the solution, the tougher it is to find, for me anyways!

The right answer:

[tex]t\,=\,\frac{800\,ft}{21.7\frac{ft}{s}}\,=\,36.9\,s[/tex]
 
VinnyCee said:
Thanks!

It seems the simpler the solution, the tougher it is to find, for me anyways!

The right answer:

[tex]t\,=\,\frac{800\,ft}{21.7\frac{ft}{s}}\,=\,36.9\,s[/tex]

Looks good to me :smile:
 
The velocity of A relative to B is the apparent velocity A will have for an observer on B. It tells us how object A moves with respect to object B. That is as if object B were stationary how would object A move (with respect to it). So [tex]V_A - V_B[/tex] removes from the velocity of A the part similar to the velocity of B (their relative motion would be zero if they had similar velocities). What is left is the remaining motion of object A w.r.t. object B. So you end up with the velocity of object A with respect to object B, usually indicated as [tex]V_{AB}[/tex].
 
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