Questions relating to motion in a plane

[pharm89]

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

 

[SGT]

Your solution is right.

 

[m2003]

Your procedure for solving these problems is correct. In part (a), you correctly used the cosine function to find the north component of Suzanne's velocity vector. In part (b), you correctly used the Pythagorean theorem to find the magnitude of Tim's velocity vector, and then used the tangent function to find the direction. Your final answer of 8.64 m/s (S 42.3 degrees W) is correct.

Just a couple of suggestions for improvement:

- When using the Pythagorean theorem, it is customary to take the square root of the sum of squares to find the magnitude of the vector, rather than squaring the square root. So in part (b), it would be more standard to write V = √[(5.8)^2 + (6.4)^2] = 8.64 m/s. This will make it easier for others to understand your solution.
- In part (b), you correctly used the Pythagorean theorem to find the magnitude of the velocity vector, but then you used the tangent function to find the direction. This is not necessary, as you can use the inverse cosine function to find the angle. So the final answer could be written as V = 8.64 m/s, θ = cos^-1(5.8/8.64) = 42.3 degrees. This would be a more standard way to write the answer.

Overall, your solution is correct and well-explained. Keep up the good work!