Answer:
Minimum time = 6.177 min
Explanation:
We assume a reference system with the positive x-axis (from left to right) and also a positive y-axis (from bottom to top)
According to this system the velocity vector of the river and the hunter are :
[tex]V_{hunter/river}=16.9 \frac{km}{h}i[/tex]
[tex]V_{river/ground}=-2.68\frac{km}{h}j[/tex]
The velocity vector of the hunter relative to the ground is the sum of the previously mentioned velocities
[tex]V_{hunter/ground}=16.9\frac{km}{h}i-2.68\frac{km}{h}j[/tex]
This means that,for example,in an hour the hunter moves 16.9 km in the positive x direction and 2.68 km in the negative y direction
We are looking for a displacement of 1.74 km in the x direction ⇒ We will use only the ''i'' component of the velocity
[tex]speed=\frac{distance}{time} \\time=\frac{distance}{speed} \\time=\frac{1.74km}{16.9\frac{km}{h}} \\time = 0.102 h\\time = 6.177 min[/tex]
We multiply the time in hours by 60 to obtain the time in minutes
time T = 6.177 min
Answer:
7.32 mins
Explanation:
Data:
Let the length of the river be = 1.74 km.
The speed of water = 2.68 km/h
The maximum speed of the boat = 16.9 km
Therefore, the time will be:
The speed of the boat relative to water = 16.9 - 2.68
= 14. 22 km/h
the time = 1.74/ 14.22
= 0.122 h
= 7.32 mins
