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Kinematics Physics Question


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#1 SophiaRivera007

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Posted 16 January 2017 - 06:18 AM

Can anyone help me to solve this?

 

A lifeguard who can swim at 1.2 m/s in still water wants to reach a dock positioned perpendicularly directly across a 550 m wide river.

a) If the current in the river is 0.80 m/s, how long will it take the lifeguard to reach the dock?

b) If instead she had decided to swim in such a way that will allow her to cross the river in a minimum amount of time, where would she land relative to the dock?


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#2 Klaynos

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Posted 16 January 2017 - 06:23 AM

What have you done to try and answer this?

Drawn a diagram?
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#3 AshBox

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Posted 17 January 2017 - 10:03 AM

My attempt at solving it:

a)
1.2 sin(α)= 0.8
sin(α)= 0.8 / 1.2
sin(α)= 2/3
α= sin^-1(2/3)
α= 41.8 degrees

So his crossing velocity is:
1.2 cos41.80= 0.894 m/s

And the time taken is:
distance = speed * time
550= 0.894 * t
t= 550 / 0.894
t=615.21 seconds

b)
550 / 1.2 = 458.3 seconds
distance = speed x time
distance = 0.80 m/s x 458.3s = 366.6 m


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#4 SophiaRivera007

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Posted 17 January 2017 - 10:13 AM

Thank You AshBox, This will help me.


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#5 Klaynos

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Posted 17 January 2017 - 11:25 AM

My attempt at solving it:
a)
1.2 sin(α)= 0.8
sin(α)= 0.8 / 1.2
sin(α)= 2/3
α= sin^-1(2/3)
α= 41.8 degrees

So his crossing velocity is:
1.2 cos41.80= 0.894 m/s

And the time taken is:
distance = speed * time
550= 0.894 * t
t= 550 / 0.894
t=615.21 seconds

b)
550 / 1.2 = 458.3 seconds
distance = speed x time
distance = 0.80 m/s x 458.3s = 366.6 m


We try not to give direct worked answers here but to lead people through doing it themselves correcting and pushing where necessary.
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Klaynos - share and enjoy.

#6 AshBox

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Posted 19 January 2017 - 07:39 AM

We try not to give direct worked answers here but to lead people through doing it themselves correcting and pushing where necessary.

I agree, but in this case we just have to put values in related equations which I have guided here. 


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