What is the velocity of the passenger relative to the water?

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To determine the velocity of the passenger relative to the water, vector addition is necessary, combining the boat's velocity and the passenger's walking speed. The boat moves at 1.13 m/s, while the passenger walks up the stairs at 0.50 m/s at a 45° angle. The velocity vectors should be expressed in horizontal and vertical components for accurate calculation. A diagram illustrating these vectors can aid in visualizing the problem. The correct approach will yield both the magnitude and direction of the passenger's velocity relative to the water.
thschica
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A passenger on a boat moving at 1.13 m/s on a still lake walks up a flight of stairs at a speed of 0.50 m/s, Fig. 3-44. The stairs are angled at 45° pointing in the direction of motion as shown. What is the velocity of the passenger relative to the water? :cry:
I have to get the magnitude of it and I get -123 which isfor sure not right
I also have to get the direction in degrees---- how do I do that.
My teacher hasn't taught us all of this yet!
 
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It's best to draw a picture showing the velocity vectors. This is vector addition.
 
\vec{V}_{person/water} = \vec{V}_{person/boat} + \vec{V}_{boat/water}

Hint: express the velocity vectors in terms of vertical and horizontal components.
 
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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