Questions relating to motion in a plane

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Suzanne's skiing velocity of 18.0 m/s at an angle of N30.0 degrees W results in a northward component of 15.6 m/s, calculated using the cosine function. Tim's change in velocity, transitioning from 6.4 m/s south to 5.8 m/s west, is determined to be 8.64 m/s at an angle of S42.3 degrees W, using the Pythagorean theorem and tangent function for direction. The calculations for both scenarios correctly apply trigonometric principles and vector analysis. The discussion confirms the accuracy of the solutions provided. Understanding these concepts is essential for solving motion in a plane problems effectively.
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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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Your solution is right.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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