Wednesday, December 14, 2011

Pedagogy Y Secondary Text 3


The Communicator explained how to call to other players on the field.  He did this by calling for the ball if the closest person controlled it.  Given the player’s x co-ordinate is x1, the x co-ordinate of the player to test is x2 and the current distance between the players he is testing is a, he did this by first testing whether x1 - a is less than or equal to x2 or x2 is less than or equal to x1 + a. Given the player’s y co-ordinate is y1, the y co-ordinate of the player to test is y2 and the current distance between the players he is testing is a, he also tested whether y1 - a is less than or equal to y2 or y2 is less than or equal to y1 + a.  If the closest player controlled the ball, then he called to him for it.  In this way, the Communicator explained how to call to other players on the field by calling for the ball if the closest person controlled it.
The Classical Studies and Archeology lecturer discussed how to test whether to agree with a colleague’s reason for catching a ball.  She did this by testing whether the ball had been hit in a certain direction.  If the player to hit the ball to was 1 unit in front, then the ball was hit forward.  If the player to hit the ball to was 1 unit to the left, then the ball was hit left.  If the player to hit the ball to was 1 unit to the right, then the ball was hit right.  In this way, the Classical Studies and Archeology lecturer discussed how to test whether to agree with a colleague’s reason for catching a ball by testing whether the ball had been hit in a certain direction.
The Communication Skills lecturer identified a pivotal moment.  He did this by calculating at what speed and in what direction the paddle should be moved to hit the ball that is travelling towards one end of the court.  He first calculated that the ball, that started 1 metre to the right and 1 metre to the south of the paddle travelled perpendicularly towards the north side of the court and where d = distance travelled (m), and s = speed (ms-1), in the time t = d / s = 1 m / 1 ms-1  = 1 s.  Given that d = distance travelled (m), and t = tine (s) he then calculated that in order to hit the ball, the paddle would need to travel at the speed s = d / t = 1 m / 1 s = 1 ms-1.  He then calculated that because the difference between the ball’s final x co-ordinate and the paddle’s initial x co-ordinate was greater than 0, the paddle would move right to hit the ball.  In this way, the Communication Skills lecturer identified a pivotal moment by calculating at what speed and in what direction the paddle should be moved to hit the ball that is travelling towards one end of the court.
The first Music Therapy lecturer computed how a tennis player should interact with his opponent.  He did this by computing how to shake one’s opponent’s hand after a tennis match.  He did this by computing that given y1 = player 1’s arm length = 0.8 m, y2 = player 2’s arm length = 0.7 m and y3 = hand overlap = 0.1 m, the total arm length = dy = y1 + y2 – y3 = 0.8 + 0.7 – 0.1 m = 1.4 m.  Then, he computed that player 1’s right heel that was under his shoulder was dy / 2 = 1.4 / 2 m = 0.7 m behind the net.  Also, he computed that if player 1’s gait = 0.4 m, and the x co-ordinate of his right heel was 4 m, then the x co-ordinate of his left heel was 4 – 0.4 m = 3.6 m.  In this way, the first Music Therapy lecturer computed how a tennis player should interact with his opponent by computing how to shake one’s opponent’s hand after a tennis match.
The Communications Law lecturer showed how to maintain items with a particular density.  He did this by showing how to clean a square coin that had a solid density.  He did this by marking the half way point between the top left and bottom left corners with a pencil, e.g. 0.04 m / 2 = 0.02 m.  Then, he marked the point on the right edge that was 90° from the line between the mid-left point to the mid-right corner of the coin using a protractor.  Then, he measured half-way along the line between the mid-left and mid-right points, 0.04 m / 2 = 0.02 m, to wipe the coin with a cloth to the left and to the right of.  In this way, the Communications Law lecturer showed how to maintain items with a particular density by showing how to clean a square coin that had a solid density.

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