Question 1201561
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Pascal and his friend Fermat live on opposite sides of a river that is 1 km wide. 
Fermat lives 2 km downstream from Pascal on the opposite side of the river. 
Pascal can swim at a rate of 3 km/h and the river’s current has a speed of 4 km/h. 
Pascal swims from his cottage directly across the river.
a) What is Pascal’s resultant velocity?
b) How far away from Fermat’s cottage will Pascal be when he reaches the other side?
c) How long will it take Pascal to reach the other side?
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<pre>
Pascal participates in two movements, simultaneously.

He swims at the rate 3 km/h in the direction perpendicular to the banks of the river 
(and perpendicular to the river' stream).

At the same time, he moves at the rate of 4 km/h along the river, together with the stream
and the current.

The vector of the combined speed is the hypotenuse of a right angle triangle with the legs 
of 3 km/h and 4 km/h.


So, the magnitude of his combined rate is  v = {{{sqrt(3^2 + 4^2)}}} = {{{sqrt(9+16)}}} = {{{sqrt(25)}}} = 5 km/h.


It is the answer to question (a).



To answer question (c), notice that Pascal will cross the river in 

    {{{the_river_width/the_rate_of_swimming_perpendicular_to_the_river}}} = {{{1_km/3_km_per_hours}}} = {{{1/3}}} of an hour = 20 minutes.


It is the answer to question (c).



At that time, the river stream will drift Pascal the distance 

    4 km/h * (1/3) hour = 4/3 km = 1{{{1/3}}} km.


Hence, Pascal will reach the opposite bank of the river at the distance  4 - 1{{{1/3}}} = 2{{{2/3}}} km from the Fermat's house.

It is the answer to question (b).
</pre>

Solved. &nbsp;&nbsp;// &nbsp;&nbsp;All questions are answered.


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Notice that I should change the order of my answers comparing with the order of questions,
in order for the solution could follow the normal/regular human logic.