Solve Easy Vector Problem: Find Displacement of Clock Hands from 3pm-6pm

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SUMMARY

The discussion focuses on calculating the displacements of the clock hands from 3 PM to 6 PM, specifically the minute hand (0.45 m) and the hour hand (0.23 m). The minute hand remains at the 6 o'clock position, resulting in a displacement of zero. The hour hand, moving from the 3 o'clock to the 6 o'clock position, requires the application of vector displacement calculations. The user attempted to use the Pythagorean theorem but encountered errors in their calculations, indicating a need for clarity on angular displacement and vector quantities.

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1. A wall clock has a minute hand with a length of 0.45 m and an hour hand with a length of 0.23 m. Take the center of the clock as the origin, and use a Cartesian coordinate system with the positive x-axis pointing to 3 o'clock and the positive y-axis pointing to 12 o'clock. Find the displacements of the tip of each hand (that is, ΔAarrowbolditalic and ΔBarrowbolditalic) when the time advances from 3:00 P.M. to 6:00 P.M. (Aarrowbolditalic and Barrowbolditalic represent the tip of the hour hand and the tip of the minute hand, respectively.)

Let the +y direction be straight up and the +x direction be to the right. Recall the definition of displacement, and remember that positions are vector quantities.

Homework Equations


dont know, tried Pythagorean theorem.

The Attempt at a Solution


the part dealing with the minute hand is obviously 0. but as far as the hour hand, I've tried using the pythagorean theorem to calculate the displacement which would just be sqrt((.23^2)+(.23^2)^2) but apparently that was wrong, or maybe i just put the wrong sign. only have 1 more submission so not about to guess it.
 
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can you calculate their angular displacements?
 

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