Answer: D
Rs = 10.0 m/s
The speed of the boat relative to an observer standing on the shore as it crosses the river is 10.0m/s
Explanation:
Since the boat is moving perpendicular to the current of the river, the speed of the boat has two components.
i. 8.0m/s in the direction perpendicular to the current
ii. 6.0m/s in the direction of the current.
So, the resultant speed can be derived by using the equation;
Rs = √(Rx^2 + Ry^2)
Taking
Ry = 8.0m/s
Rx = 6.0m/s
Substituting into the equation, we have;
Rs = √(6.0^2 + 8.0^2)
Rs = √(36+64) = √100
Rs = 10.0 m/s
The speed of the boat relative to an observer standing on the shore as it crosses the river is 10.0m/s
D. 10m/s
Given,
speed of boat ([tex]v_{B}[/tex]) = 8.0m/s
speed of river current ([tex]v_{R}[/tex]) = 6.0m/s
Since the boat travels perpendicular to the current;
The resultant speed (v) of the boat, relative to an observer, will be the vector sum of the individual speeds of the boat and the current. i.e
v = [tex]\sqrt{v_{B} ^{2} + v_{R} ^{2} }[/tex] -------------------------(i)
Substitute the values of [tex]v_{B}[/tex] and [tex]v_{R}[/tex] into equation (i)
v = [tex]\sqrt{8.0^{2} + 6.0^{2} }[/tex]
v = [tex]\sqrt{64 + 36}[/tex]
v = [tex]\sqrt{100}[/tex]
v = 10m/s
Therefore, the speed of the boat relative to an observer standing on the shore as it crosses the river is 10m/s.
