Relative velocity at a junction

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Homework Help Overview

The problem involves two cars approaching a junction from intersecting roads at an angle, with specific speeds and distances from the junction. The original poster seeks to determine various aspects of their motion, including the time they reach the junction, their relative distances, and their velocities with respect to each other.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore equations for the distances of the cars from the junction over time and question the interpretation of negative distances. They discuss the conditions under which the cars are equidistant from the junction and consider multiple scenarios for their positions.

Discussion Status

Some participants have provided guidance on setting up equations for the distances of the cars and have noted the ambiguity in the interpretation of "distance." There is ongoing exploration of different solutions, with no explicit consensus reached on the correct interpretation or method.

Contextual Notes

Participants note that the problem's wording may lead to different interpretations regarding the distances and positions of the cars relative to the junction, which affects the solutions being discussed.

Woolyabyss
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Homework Statement


Two straight roads intersect at point O and cross at an angle Θ to one another, such that tan Θ = 3/4.Two cars, A and B are traveling towards O on these roads,A at 5 m/s and B at 8 m/s

At a certain moment.A is 100 m from the junction and B, on the other road, is 200 m from the junction. Find
(i)the time at which A reaches O
(ii) The distance between A and B at this time.
(iii) The magnitude and direction of the velocity of A with respect to B
(iv) The shortest distance between them.
(v)the time at which they are nearest to one another
(vi)the time when they are equidistant from O


Homework Equations



Vab = Va - Vb

The Attempt at a Solution



I got every part except (vi)

(i) 100/5 = 20 seconds

(ii) 200 - 8(20) = 40m

(iii) Vab = ( -5(4/5)i -5(3/5)j ) -(-8i) = 4i -3j m/s

|Vab| = 5m/s
inverse tan 3/4 = 36.87 degrees south of east

(iv)
40(sinΘ) = 40(3/5) = 24 m = shortest distance

(v)
Pythagoras theorem (40^2 -24^2)^(1/2) = 32m

32/5 =6.4 seconds

20 + 6.4 = 26.4 seconds
(vi)
This is the part I am stuck on.I know I need the distance from O when they are equidistant but I just can't seem to think of any way that I could get none of the examples in my book
covered this any help would be appreciated.
 
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Woolyabyss said:
This is the part I am stuck on.I know I need the distance from O when they are equidistant but I just can't seem to think of any way that I could get none of the examples in my book
covered this any help would be appreciated.

See if you can write an equation for how far A is from O for any time t. Similarly for how far B is from O at any time t.
 
Alright thanks, I just jotted down this down randomly
5t - 100 = 8t - 200... 3t = 100

t = 33.33 seconds

This is the right answer according to the back of my book but for some reason I feel as though the method I just used isn't the right one since if I picked any time before A passed O I would get a negative distance.
Any thoughts?
 
TSny said:
See if you can write an equation for how far A is from O for any time t. Similarly for how far B is from O at any time t.

Since displacement is a vector quantity would the negative sign just be telling you that its displacement is at the opposite side of O?
 
OK, you set up equations such that positive positions are interpreted as positions beyond O while negative positions are before reaching O. That's fine.

The question is a little ambiguous. Usually "distance" is interpreted as positive independent of which side of O you're on. You have found a solution for when both cars are past O. Can you find another solution where one car is beyond O while the other has not yet reached O (but they are the same distance from O)?
 
Last edited:
TSny said:
OK, you set up equations such that positive positions are interpreted as positions beyond O while negative positions are before reaching O. That's fine.

The question is a little ambiguous. Usually "distance" is interpreted as positive independent of which side of O you're on. You have found a solution for when both cars are past O. Can you find another solution where one car is beyond O while the other had not yet reached O (but they are the same distance from O)?

Would it be 5t -100 = 200 - 8t

t=23.077 seconds

bearing in mind the previous answer I got is the only one that is given as a solution at the back of my book.
 
Yes. That looks good. I guess the book is ignoring that answer.
 
TSny said:
Yes. That looks good. I guess the book is ignoring that answer.

Alright thank you.
 
TSny said:
Yes. That looks good. I guess the book is ignoring that answer.

Alright, thank you.
 

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