Need help ASAP!!! Determine the MAGNITUDE of the bird watcher's (a) displacement and (b) average velocity.?
A bird watcher meanders through the woods, walking 0.219 km due east, 1.42 km due south, and 4.76 km in a direction 75.4 ° north of west. The time required for this trip is 1.616 h. Determine the MAGNITUDE of the bird watcher's (a) displacement and (b) average velocity.
I've tried this a couple of times but i seem to be getting the wrong answer.. i think it is to do with the magnitude ('+ve or -ve') .. thanks for any help!!
- 9 years agoBest Answer
Displacement and velocity are both vectors so only the initial and final points count, not distance traveled.
Here is what you have:
+0.219 km right (positive x axis)
-1.42 km south (negative y axis)
And 4.76 km north of west which must be broken down into x and y components:
The x component is cosine of the angle (75.4) multiplied by 4.76, while the y component is sine of the angle multiplied by 4.76
Since it is northwest the x component will be negative and y will be positive.
Now you need to add your x components together and your y components together and draw it on a piece of paper (be careful with signs).
Now using pythagoreans theorem (x^2 + y^2 = z^2) to find the total displacement.
If you have to be in SI units (meters and seconds) divide displacement by 1000 and multiply time 1.616 by 3600).
Then just solve for velocitySource(s): If anything is unclear, edit your question and let me know... I don't give solutions onY! Answers, I just provide the required inforation to allow them to be found.
- 4 years ago
West component of 3rd. leg = (cos 44) x 2.2, = 1.5825km. North component = (sin 44) x 2.2, = 1.52825km. (1.5825 - 0.3) = 1.2825km. west of origin, and (1.52825 - 0.75) = 0.77825km. north of origin. Displacement = sqrt. (0.77825^2 + 1.2825^2) = 1.5km., and angle = arctan (0.77825/1.2825) = 31.25 deg. N of W.