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Physics HELP: Easy 10 Points!: A bird watcher meanders through the woods, walking 0.40 km due east..?
A bird watcher meanders through the woods, walking 0.40 km due east, 0.75 km due south, and 2.95 km in a direction 36.0° north of west. The time required for this trip is 2.50 h. Determine the magnitude and direction (relative to due west) of the bird watcher's:
A: Displacement in km and direction north of west in degrees.
B: Average Velocity and direction north of west in degrees.
There are many ways to do vector displacement problems, and you should use the methods suggested by your teacher. One method is to make a scale drawing, showing each displacement as an arrow. The resultant displacement will be an arrow starting at the beginning of the first arrow drawn and ending at the end of the last arrow drawn. Use the same scale to find the value and a protractor to find the direction.
A more mathematical method is to convert each vector to its X and Y components, add all the X and all the Y components to determine the X and Y components of the resultant and then use Pythagorean theorem and the tangent formula to find the length and direction of the resultant.
In this case, I am setting North as 0° and angles are + clockwise. 36° N of W will be W = 270° + 36° = 306°
A = 0.4 km @ 90° = (0.4 km, 0)
B = 0.75 km @ 180° = (0, -0.75 km)
C = 2.95 km @ 306° = (2.95 km•sin(306°), 295 km•cos(306°)
C = (-2.387 km, 1.734 km)
A+B+C = (-1.987 km, 0.984 km)
A+B+C magnitude = sqr rt(-1.987^2 + 0.984^2) = 2.217 km
dir = arctan(X/Y) = arctan(-1.987/0.984) = arctan (-2.019)
dir = -63.65°
Since we chose N = 0° and angles + CW, the direction is 360° - 63.65 or 296.35° or 26.35° N of W (subtract W = 270°)
---
The displacement is 2.217 km @ 26.35° N of W.
Average velocity is distance/time = 2.217 km/2.5 h
Vav = 0.887 km/h @ 26.35° N of W
I don't know how your teacher wants you to handle significant figures, but the data contain values with one, two and three sig fig, so your teacher may only be looking for process and not precision.
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