This series of caches is reaching the end of its run.
I will be archiving them in August 2022
It is recommended, but not necessary, that you complete the puzzles in order as each puzzle will have an increasing level of difficulty. The puzzles in the series are
Physics Puzzle #1 (GC6XBGC)
Physics Puzzle #2 (GC6XDN0)
Physics Puzzle #3 (GC6XDXM)
Physics Puzzle #4 (GC6XK5Q)
Physics Puzzle #5 (GC6XMW4)
Physics Puzzle #6 (GC6XP3J)
Physics Puzzle #7 (GC6XXY3)
Physics Puzzle #8 (GC6XY1H)
Physics Puzzle #9 (GC6XY50)
The Lesson
There is an old logic problem related to this puzzle:
A hiker leaves his camp and walks 1.5 km south, 1.5 km west, and 1.5 km north ending up at his camp site. He sees a bear near his camp. What colour is the bear? See the hint below for the answer!
This time, you are starting to look at vectors which are measured quantities that require a direction. The measured quantity is called the magnitude. You are already well versed in direction of a vector as you use it whenever you go geocaching! The above problem makes use of vectors. A measurement that does not need a direction is called a scalar. For comparison, examples of both are shown:
Some scalar quantities are temperature, time, volume, energy, speed, and distance.
Some vector quantities are forces (which way should you push?), acceleration, velocity, position, and displacement.
To show a vector in a diagram, we use a "directed line segment" (more commonly called an "arrow"). The length of the arrow shows the magnitude of the vector while the arrow points in the vector's direction. For the vector shown in the diagram, the magnitude is 325 m and the direction is a bearing of 345° (assuming north is UP on the diagram). In the usual notation, this would be simply written as 325 m [345°]. Note the use of square brackets to separate the direction from magnitude.
Terminology:
(Remember: There is more terminology in GC6XBGC)
position - a location including distance and direction from a fixed reference point
displacement - a change in position (distance and direction in moving from one position to another)
velocity - rate of change in displacement (essentially speed with direction)
The Puzzle
We are going to concern ourselves with distance, displacement, and position. For the purposes of this puzzle, we are going to treat distance as the actual "How far did I go?" and displacement is going to be "As the crow flies". Round off to the nearest ten metres or the nearest one degree.
Take a trip from Lindsay Street (N44°21.332' W078°44.087') along Kent Street to Victoria Avenue (N44°21.246' W078°44.453')
a displacement of __ __ 0 m [ __ __ __ °] (A B 0 [ C D E°]). Next head north-ish along Victoria Avenue finishing up outside the Old Jail (N44°21.541' W078°44.580')
a displacement of __ __ 0 m [ __ __ __ °] (F G 0 [ H J K°]) from Victoria and Kent. The final displacement on the other hand is from start to finish - never going from finish to start unless you walk backwards
but that is a different puzzle! The displacement is __ __ 0 m [ __ __ __ °] (L M 0 [ N P Q°]) while the distance travelled was __ __ __ 0 m. (R S T 0)
Note: For consistency, I have been measuring to/from the centre of the nearest intersection.
I found some handy tools http://www.geolifeline.com
A = __ B = __ C = __ D = __ E = __ F = __ G = __ H = __ J = __
K = __ L = __ M = __ N = __ P = __ Q = __ R = __ S = __ T = __
The cache can be found at N 44° a b . c d e’ and W 078° f g . h j k’
There is no connection between the upper case letters above and the lower case letters here. Follow the rules of BEDMAS!
a = (A)(E) - (C)(J): _____; b = T ÷ J + M ÷ K: _____; c = (S)(H) + (P)(L): _____;
d = (B)(M) + (N)(R): _____; e = (C)(K) - (E)(Q): _____; f = (F + H) ÷ (B + R): _____;
g = S ÷ (E + R): _____; h = A (D - N) - Q: _____.
j = S x T: _____; k = N (Q + B): _____.
The cache is at: N 44° __ __ . __ __ __ ' and W 078° __ __ . __ __ __'
You can check your answers for this puzzle on GeoChecker.com.