# Relative velocity at a junction

1. Mar 2, 2013

### Woolyabyss

1. The problem statement, all variables and given/known data
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 travelling 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

2. Relevant equations

Vab = Va - Vb

3. 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 im 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.

2. Mar 2, 2013

### TSny

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.

3. Mar 2, 2013

### Woolyabyss

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?

4. Mar 2, 2013

### Woolyabyss

Since displacement is a vector quantity would the negative sign just be telling you that its displacement is at the opposite side of O?

5. Mar 2, 2013

### TSny

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: Mar 2, 2013
6. Mar 2, 2013

### Woolyabyss

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.

7. Mar 2, 2013

### TSny

Yes. That looks good. I guess the book is ignoring that answer.

8. Mar 2, 2013

### Woolyabyss

Alright thank you.

9. Mar 2, 2013

### Woolyabyss

Alright, thank you.