# Relative velocity and vectors

1. May 15, 2010

### thereddevils

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

A car A and a bicycle B are moving along two straight roads which cross at O at 90 degree . A moves to the east at a constant speed of 48km/h , whereas B moves towards the north at a speed of 14 km/h . At the time when B passes O , A is 400 m away from O and has yet to pass through O . Calculate the distance of the car and the bicycle from O when they are at the closest distance from each other .

2. Relevant equations

3. The attempt at a solution

Well , i have calculated the shortest distance between the car and the bicycle which is 112 m and its correct .However , i have a hardtime calculating their distances from O respectively .

2. May 15, 2010

### tiny-tim

Hi thereddevils!
I assume you used relative velocity for the first part?

Now compare that with the relative displacement to find the time.

3. May 15, 2010

### thereddevils

Re: vectors

ok , i found that the time is 8 s . Do i multiply this time with the respective velocities of A and B to get their displacement at this time ?

4. May 15, 2010

### tiny-tim

Is that 8 s from when B passes O, and A is 400 m west of O?

In that case, yes, add time times velocity to those positions.

5. May 15, 2010

### D H

Staff Emeritus
Re: vectors

How did you get that result? With respect to what event are you measuring time?

If you show how you arrived at that result we can help you take the next step.

6. May 15, 2010

### thereddevils

Re: vectors

sorry , i still don get it .

I drew a triangle , with A moving in the path of AA' with B assumed to be stationary , A moves till a point when its perpendicular to B , and the relative distance travelled by A is
400 cos 16.3 and the relative velocity is 50 so the time taken is 8 s .

i drew another triangle too before that , AOB , where A and B (right angle to each other)moving towards O . I am not sure how to insert the 8 s here , or the multiply 8 by velocity of A ? 8*48=384 but this is not measured from O , i am not told the distance of OA .

7. May 15, 2010

### tiny-tim

Hi thereddevils!
erm … it's 50 km/hr, not m/s.
Why are you drawing triangles?

Once you find the time taken, that's from when B is at O, and A is 400 m west of O, so just go the appropriate distances north and east respectively.

8. May 15, 2010

### thereddevils

Re: vectors

thanks Tiny , but the time i found is when A and B are closest to each other , are they closest when B is at O and A is 400 m west of B ?

9. May 16, 2010

### tiny-tim

hmm … let's review what you've found so far …

You found the relative velocity of B from A was 50 km/hr at tan-114/48 south of east.

You then found (still in your first post) that the initial relative displacement AB was 400, so the closest relative displacement AB would be 400 times the cos of that angle, ie 400 times 14/50, or 112m.

Now you need to find the time, so you divide the distance A travels (still in the relative frame) by the speed, and the distance will be 400 times the sin of that angle, ie 400 times 48/50, or 396m divided by 50 km/hr.

Up till now, you've been using relative velocity and displacement, but now you've found t, you can go back to the stationary frame to get the distance along the roads, using that t.

And those distances along the roads will be from the initial positions, ie B at O and A 400 m west of O.

10. May 16, 2010

### thereddevils

Re: vectors

thanks a lot Tiny for taking this trouble ,

14/40=d/400 and d=384 and this is the relative distance travelled by A

and relative velocity of A is 50 , and 384=50t , t is 7.68 s .

then i convert 48km/h to 13.33 m/s and 14km/h to 3.89 m/s

relative Distance travelled by B = 3.89 x 7.68 = 29.9 m

relative distance travelled by A = 400-(13.33 x 7.68) = 297.6 m

but the answer given is 31.4 m and 292.5 m

Or maybe they rounded off the time to 8 ?

Last edited: May 16, 2010
11. May 16, 2010

### tiny-tim

Hi thereddevils!

I make it 31.36 m for A and 170.52 m for B … I don't know where 292.5 m comes from, but I'm pretty sure it's wrong (31.362 + 107.523 = 1122, and 31.36/107.52/112 = 14/48/50).

oh, btw, that's a much better way of getting the final result, since it doesn't involve calculating the time … we know the distance is 112, and it must be at right-angles to the relative velocity, so again it'll be a 14/48/50 triangle.

(I can't spot the error in your calculations, but I would have recommended not converting to seconds … if you'd kept the time in hours, it would have simplified the arithmetic. )

12. May 17, 2010

### thereddevils

Re: vectors

ok , i will give it another shot . Btw , for that 292.5 , we measure the distance from its initial displacement which is B at O (moving towards the north) and A 400 m west of B (moving to east) , so after certain time , t , it has travelled a distance of d so its distance from O would be 400 - d ?

Last edited: May 17, 2010
13. May 17, 2010

### tiny-tim

I have it the other way round … my 31.36 is the 400 - d.