Answer:
a) The swimmer should travel perpendicular to the bank to minimize the spent in getting to the other side.
b) 133.33 m
c) 53.13°
d) 106.67 m
Explanation:
a) The swimmer should travel perpendicular to the bank to minimize the spent in getting to the other side.
b) velocity = distance * time
Let the velocity of the swimmer be
= 1.5 m/s
The separation of the two sides of the river, d = 80 m
The time taken by the swimmer to get to the other end of the river bank,
![t = \frac{d}{v_{s} }](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7Bd%7D%7Bv_%7Bs%7D%20%7D)
t = 80/1.5
t = 53.33 s
The swimmer will be carried downstream by the river through a distance, s
Let the velocity of the river be
= 2.5 m/s
![S = v_{r} t](https://tex.z-dn.net/?f=S%20%3D%20v_%7Br%7D%20t)
S = 53.33 * 2.5
S = 133.33 m
c) To minimize the distance traveled by the swimmer, his resultant velocity must be perpendicular to the velocity of the swimmer relative to water
That is ,
![cos \theta = \frac{v_{s} }{v_{r} } \\cos \theta = 1.5/2.5\\cos \theta = 0.6\\\theta = cos^{-1} 0.6\\\theta = 53.13^{0}](https://tex.z-dn.net/?f=cos%20%5Ctheta%20%3D%20%5Cfrac%7Bv_%7Bs%7D%20%7D%7Bv_%7Br%7D%20%7D%20%5C%5Ccos%20%5Ctheta%20%3D%201.5%2F2.5%5C%5Ccos%20%5Ctheta%20%3D%200.6%5C%5C%5Ctheta%20%3D%20cos%5E%7B-1%7D%200.6%5C%5C%5Ctheta%20%3D%2053.13%5E%7B0%7D)
d) Downstream velocity of the swimmer, ![v_{y} = v_{s} sin \theta\\](https://tex.z-dn.net/?f=v_%7By%7D%20%3D%20v_%7Bs%7D%20sin%20%5Ctheta%5C%5C)
![v_{y} = 1.5 sin 53.13\\v_{y} = 1.2 m/s](https://tex.z-dn.net/?f=v_%7By%7D%20%3D%201.5%20sin%2053.13%5C%5Cv_%7By%7D%20%3D%201.2%20m%2Fs)
The vertical displacement is given by, ![y = v_{y} t](https://tex.z-dn.net/?f=y%20%3D%20v_%7By%7D%20t)
80 = 1.2 t
t = 80/1.2
t = 66.67 s
the horizontal speed,
![v_{x} = 2.5 - 1.5cos53.13\\v_{x} = 1.6 m/s](https://tex.z-dn.net/?f=v_%7Bx%7D%20%3D%202.5%20-%201.5cos53.13%5C%5Cv_%7Bx%7D%20%3D%201.6%20m%2Fs)
The downstream horizontal distance of the swimmer, ![x = v_{x} t](https://tex.z-dn.net/?f=x%20%3D%20v_%7Bx%7D%20t)
x = 1.6 * 66.67
x = 106.67 m