Respuesta :
Answer: The magnitude of her velocity ( v ) with respect to the shore is 5.70 km/h.
Explanation:
Magnitude of Velocity of the boat = Q
Magnitude of Velocity of the river flowing = P
R = Resultant velocity due to velocity of boat and velocity of river.
Applying Law of triangles of vector addition :
[tex]R=\sqrt{P^2+Q^2+2PQCos\theta}[/tex]
From the figure attached:
P = 3.5 k/h, Q = 4.5 km/h
[tex]\theta= 90^o[/tex]
[tex]R=\sqrt{P^2+Q^2+2PQCos\theta}=\sqrt{(3.5)^2+(4.5)^2+3.5\times 4.5\times cos90^o}=5.70[/tex]
[tex](Cos90^o=0),(sin 90^o=1)[/tex]
[tex]\alpha =tan^{-1}\frac{Qsin\theta}{P+Qcos\theta}=tan^{-1}\frac{4.5 sin 90^o}{3.5+4.5 cos90^o}=tan^{-1}\frac{4.5}{3.5}=52.12^o[/tex]
The magnitude of her velocity ( v ) with respect to the shore is 5.70 km/h.
The magnitude of her velocity with respect to the shore will be 5.70 Km/h. The change of displacement with respect to time is defined as velocity.
What is velocity?
Velocity is defined as the change in displacement with respect to time. The quantity of velocity is a vector quantity. It is a component that is time-based. It is measured in meters per second.
The given data in the problem is;
Q is the magnitude of Velocity of the boat = 4.5 km/hr
P is the magnitude of Velocity of the river flowing = 3.5 km/hr
R is the resultant of velocity P and Q=?
θ is the angle between the two velocities = 90°
From the law of vector addition;
[tex]\rm R= \sqrt{P^2+Q^2+2PQcos\theta} \\\\ \rm R= \sqrt{(3.5)^2+(4.5)^2+2\times 3.5 \times 4.5 cos90^0} \\\\ \rm R=5.70 \ m/sec[/tex]
Hence the magnitude of her velocity with respect to the shore will be 5.70 Km/h.
To learn more about the velocity refer to the link;
https://brainly.com/question/862972
