Quick swimmer's motion question

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A swimmer with a speed of 1.15 m/s aims to cross a 150 m wide river with a current of 0.80 m/s. The motion involves vector addition of her swimming speed across the river and the downstream current. To determine how far downstream she lands, one must first calculate the time it takes to cross the river without current, then use that time to find the distance carried downstream by the current. This approach allows for separate analysis of the two motion components. The discussion emphasizes understanding the interplay between swimming speed and river current for accurate calculations.
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A swimmer is capable of swimming 1.15 m/s in still water.
(a) If she aims her body directly across a 150 m wide river whose current is 0.80 m/s, how far downstream (from a point opposite her starting point) will she land?
:cry: I have tried this twice and I have gotten the wrong answer does anyone know how I am supposed to do this?
 
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The swimmer's motion will have 2 components, one across the river and one downsteam (carried by the current). The resulting motion will be the vector addition of the 2 components.
 
The nice thing about the "components" hotvette mentioned is that you can work with them completely separately.

If there were no current at all, how long would it take this swimmer to cross the river?

Now include the current- in the time you just calculated, how far downstream will the current carry her?
 
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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