Assignment 1

Assignment 1 - PHYS 230 Fall 2007 Assignment 1 1. At 12:00...

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PHYS 230 Fall 2007 Assignment 1 1. At 12:00 hours ship A is 10 km east and 20 km north of a certain port. It is steaming at 40 km/hr in a direction 30 east of north. At the same time ship B is 50 km east and 40 km north of the port, and is steaming at 20 km/hr in a direction 30 west of north. (a) Draw a diagram of this situation, showing positions and velocities measured in the frame of the Earth. The components of the velocities for A are v x = 40 cos 60 = 20 v y = 40 sin 60 = 20 3 and for B are v x = - 20 cos 60 = 10 v y = 20 sin 60 = 10 3 all in km/hr. East is ˆ x and North is ˆ y . - 6 North East t A B t ¶7 S S So (b) If the ships continue to move with the above velocities, what is their closest distance to one another and when does it occur? Work in the frame of the Earth. There are several ways to do this. One way is to write down the equations for the paths of A and B, as a function of time, and therefore to find the distance between them, also as a function of time. Calculus can then be used to find the minimum distance. The equation for the path of A is x A = 10 + 20 t y A = 20 + 20 3 t 1
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and that of B is x B = 50 - 10 t y B = 40 + 10 3 t Thus, the distance (squared) between them is d 2 = ( x B - x A ) 2 + ( y B - y A ) 2 or d 2 = (40 - 30 t ) 2 + (20 - 10 3 t ) 2 Differentiating (there is no need to multiply out, or to take the square root): 0 = 2(40 - 30 t )( - 30) + 2(20 - 10 3 t )( - 10 3) Collecting terms, dropping the zeroes and the factors of 2: 0 = 12 + 9 t - 2 3 + 3 t or t = 1 + 3 / 6 1 . 29 hr Substituting back: d = 20 - 10 3 2 . 7 km 2. Repeat the previous problem, but work in the inertial frame attached to ship A . Thus: (a) Draw a diagram of this situation, showing positions and velocities measured in the frame attached to ship A. To find the position of B in the frame of A, subtract x A from x B , and y A from y B . Set North to be ˆ y and East to be ˆ x . x B - x A = 50 - 10 = 40 km y B - y A = 40 - 20 = 20 km To find the velocity of B in the frame of A, subtract v A from v B . v A = 40 2 ˆ x + 40 3 2 ˆ y v B = - 20 2 ˆ x + 20 3 2 ˆ y v B - v A = - 30ˆ x - 10 y 2
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The magnitude of this vector is thus 900 + 300 = 20 3 km/hr. - 6 North East t t A B B 0 C B 00 ¡ ¡“ ? ? ? ? ¡ ¡ ¡ ¡ ¡ ¡ ¡ @ @ (b) If the ships continue to move with the above velocities, what is their closest distance to one another and when does it occur? Work in the frame attached to ship A. Once again, there are various ways to solve the problem. For example, we could use
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This note was uploaded on 04/29/2008 for the course PHIL 210 taught by Professor Hallet during the Fall '07 term at McGill.

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Assignment 1 - PHYS 230 Fall 2007 Assignment 1 1. At 12:00...

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