- #1

DaalChawal

- 96

- 0

Question 3

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- MHB
- Thread starter DaalChawal
- Start date

- #1

DaalChawal

- 96

- 0

Question 3

- #2

skeeter

- 1,104

- 1

Let $v_1 > v_2$ be the respective speeds (rates) in feet per second of the two particles.

Moving in opposite directions, their rate of closure is $(v_1+v_2)$

Moving in the same direction, their rate of opening is $(v_1-v_2)$

Set up a system of two equations and solve for both speeds.

Moving in opposite directions, their rate of closure is $(v_1+v_2)$

Moving in the same direction, their rate of opening is $(v_1-v_2)$

Set up a system of two equations and solve for both speeds.

Last edited by a moderator:

- #3

DaalChawal

- 96

- 0

- #4

skeeter

- 1,104

- 1

If the two particles move in opposite directions, the

If the two particles move in the same direction, the faster particle will move ahead of the slower particle, hence the

Using this method, I get $v_1 = 18 \, ft/sec$ and $v_2 = 12 \, ft/sec$. You can check the results yourself.

- #5

HOI

- 923

- 2

This problem has nothing to do "velocity". The problem asks for their speeds, not their velocities.

Share:

- Last Post

- Replies
- 1

- Views
- 517

- Replies
- 13

- Views
- 569

- Replies
- 7

- Views
- 307

- Last Post

- Replies
- 19

- Views
- 304

- Last Post

- Replies
- 3

- Views
- 917

- Last Post

- Replies
- 3

- Views
- 280

- Replies
- 1

- Views
- 369

- Replies
- 13

- Views
- 686

- Replies
- 3

- Views
- 419

- Last Post

- Replies
- 2

- Views
- 718