# Math Help - vectors

1. ## vectors

a carrom board (4ft*4ft sq) has the queen at the centre.the queen ,hit by thje striker moves to the front edge ,rebounds
and goes in the hole behind the striking line.find themagnitud eof displacementof the queen
(A)from to the centre to the front edge
(B)from the front to the hole
(c)from the centre to the hole

2. Originally Posted by vritikagupta
a carrom board (4ft*4ft sq) has the queen at the centre.the queen ,hit by thje striker moves to the front edge ,rebounds
and goes in the hole behind the striking line.find themagnitud eof displacementof the queen
(A)from to the centre to the front edge
(B)from the front to the hole
(c)from the centre to the hole
(I really need to learn how to do pictures! Sorry about that. It's going to be a bit hairy trying to explain this.)

Since we don't have a specified coordinate system I'm assuming you want the magnitude of the displacements?

The queen moves from the center of the board to one corner. So the magnitude of the displacement is just half the diagonal of the board: 1/2*4*sqrt(2) ft = 2*sqrt(2) ft.

For reference purposes I will call the direction from the queen's initial position directly to the striking rail the "vertical" direction and the direction at right angles to this "horizontal."

For a) and b) note that the angle the queen strikes the rail at is going to be the same as the angle that the queen rebounds from the rail at. (Angle of incidence is equal to the angle of reflection.) If you look at the situation carefully enough you will see that the magnitude of the horizontal displacement component of the queen from the center to the rail is 1/3 of 1/2 the length of the rail. (It takes the queen twice the time to get from the rail to the pocket as it does for the queen to get from the center to the rail. The overall magnitude of the horizontal displacement component is equal to 1/2 the rail.)

So for a) the queen's vertical displacement is 2 ft and the horizontal displacement is 1/3*1/2*4 ft = 2/3 ft. Thus the magnitude of the displacement will be sqrt(2^2 + (2/3)^2) = sqrt(40/9) = 2/3*sqrt(10) ft.

For b) the queen's vertical displacement is 4 ft. (technically -4 ft depending on which direction you choose to be positive) and the horizontal displacement will be 2*(2/3) ft. (Twice the amount as in a)). So the magnitude of the displacement will be sqrt(4^2 + (4/3)^2) = 4/3*sqrt(10) ft. (We should expect this to be twice the displacement as in a), so this checks.)

-Dan

3. Hello, vritikagupta!

This is topsquark's solution . . . with a diagram.

A carom board (4ft*4ft sq) has the queen at the centre.
The queen, hit by the striker, moves to the front edge, rebounds,
and goes in the hole behind the striking line.
Find the magnitude of displacement of the queen
(A) from to the centre to the front edge
(B) from the front to the hole
(c) from the centre to the hole.

Make a sketch and you can almost "eyeball" the problem!
Code:
               2        B     C           D
* - - - - - - - - * - - * - - - - - *
|                 :   θ/ \θ         |
|                2:   /   \         |
|                 :  /     \        |
|                 : /       \       |
|                 :/         \      |
|                A*           \     |4
|                              \    |
|                               \   |
|                                \  |
|                                 \ |
|                                  \|
* - - - - - - - - - - - - - - - - - *E

Since the angle of incidence (θ) equals the angle of reflection,
. . ∆ABC ~ ∆CDE

Then .BC:CD .= .BA:DE .= .2:4 .= .1:2

Since BD = 2, we have: .BC = 2/3, .CD = 4/3

Then: AC² .= .AB² + BC² .= .2² + (2/3)² .= .40/9
. . . . . . . . . . . . . . .__
. . Hence: .AC .= .2√10/3 . (A)

. . . . . . . . . . . . . . . . . __
And: .CE .= .2·AC .= .4√10/3 . (B)
. . . . . . . . . . . . . . .__
And: .AC + CE .= .6√10/3 . (C)

4. Originally Posted by Soroban
Hello, vritikagupta!

This is topsquark's solution . . . with a diagram.

Make a sketch and you can almost "eyeball" the problem!
Code:
               2        B     C           D
* - - - - - - - - * - - * - - - - - *
|                 :   θ/ \θ         |
|                2:   /   \         |
|                 :  /     \        |
|                 : /       \       |
|                 :/         \      |
|                A*           \     |4
|                              \    |
|                               \   |
|                                \  |
|                                 \ |
|                                  \|
* - - - - - - - - - - - - - - - - - *E
Since the angle of incidence (θ) equals the angle of reflection,
. . ∆ABC ~ ∆CDE

Then .BC:CD .= .BAE .= .2:4 .= .1:2

Since BD = 2, we have: .BC = 2/3, .CD = 4/3

Then: AC² .= .AB² + BC² .= .2² + (2/3)² .= .40/9
. . . . . . . . . . . . . . .__
. . Hence: .AC .= .2√10/3 . (A)

. . . . . . . . . . . . . . . . . __
And: .CE .= .2·AC .= .4√10/3 . (B)
. . . . . . . . . . . . . . .__
And: .AC + CE .= .6√10/3 . (C)
Thank you for posting the diagram!

I only have one comment to make... About part c), the question is looking at the displacement of the queen from the center to the pocket. This would be on the direct line from the center to the pocket, not along the path taken. (If they were asking for the distance your answer would be correct.)

-Dan

5. Hello, Dan!

I considered that interpretation of part (c), but rejected it.

It would be a rather silly question.
We could have answered it first, with no reference to the queen.
["Find the distance from the center of a square to one of its vertices.")

Yet the phrase "displacement of the queen" indicates that you are correct.

6. Originally Posted by Soroban
Hello, Dan!

I considered that interpretation of part (c), but rejected it.

It would be a rather silly question.
We could have answered it first, with no reference to the queen.
["Find the distance from the center of a square to one of its vertices.")

Yet the phrase "displacement of the queen" indicates that you are correct.
Yes, the distinction between distance and displacement causes all sorts of difficulties for my Physics students.

-Dan