Questions relating to motion in a plane

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SUMMARY

This discussion focuses on solving motion problems in a plane using vector analysis. For Suzanne's skiing velocity of 18.0 m/s at an angle of 30 degrees west of north, the northward component of her velocity is calculated to be 15.6 m/s. Tim's change in velocity, transitioning from 6.4 m/s south to 5.8 m/s west, results in a magnitude of 8.64 m/s directed at 42.3 degrees south of west. The calculations employ trigonometric functions and the Pythagorean theorem to derive the necessary components and resultant vectors.

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Homework Statement



I just would like to know if I'm using the right procedure to solve these problems.

(a) Suzanne is skiing with a velocity of 18.0 m/s (N30.0 degrees W). What is the (N) component of her velocity vector?

(b) TIm is running cross-country at 6.4 m/s (S) when he completes a wide turn and continues at 5.8 m/s (W). What is his change in velocity?



Homework Equations


Cos(theta) = adjacent / hypotenuse

velocity = V2 - V1
use of Pythagorean theorum formula


The Attempt at a Solution



(a) Given:
velocity = 18.0 m/s
30 degrees

Required: Vn

Analysis:
cos(theta) = adjacent/hypotenuse
Cos30 degrees = Vn / 18 m/s
Vn = (18.0 m/s) (Cos30 degrees)
=15.6 m/s

(b) Given:
V1 = 6.4 m/s (E)
V2 = 5.8 m/s W

Required: change in velocity

Analysis: velocity = V2 - V1
=V2 + (-V1)
=V2 - V1

use the pythagorean theorum to determine the magintude of the velocity vector.
V^2 = (5.8)^2 + (6.4)^2
=33.64 + 40.96
=74.6 m^2 s^2
=8.64 m/s

Then determine the direction of the velocity vector.
Tan (theta) = 5.8 m/s / 6.4 m/s =0.91
=tan-10.91
=42.3 degrees

Therfore his change in velocity is 8.64 m/s (S 42.3 degrees W)

Any help is always appreciated.
Thanks
Pharm 89:confused:
 
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