Respuesta :
Answer:
The resultant velocity is 12.21 m/s.
Step-by-step explanation:
We are given that the water from a fire hose follows a path described by y equals 2.0 plus 0.9 x minus 0.10 x squared (units are in meters).
Also, v Subscript x is constant at 10.0 m/s.
The water from a fire hose follows a path described by the following equation below;
[tex]y=2.0 + 0.9x-0.10x^{2}[/tex]
The velocity of the [tex]x[/tex] component is constant at = [tex]v_x=10.0 \text{ m/s}[/tex]
and the point at which resultant velocity has to be calculated is (9.0,2.0).
Let the velocity of x and y component be represented as;
[tex]v_x=\frac{dx}{dt} \text{ and } v_y=\frac{dy}{dt}[/tex]
Now, differentiating the above equation with respect to t, we get;
[tex]y=2.0 + 0.9x-0.10x^{2}[/tex]
[tex]\frac{dy}{dt} =0 + 0.9\frac{dx}{dt} -(0.10\times 2)\frac{dx}{dt}[/tex]
[tex]\frac{dy}{dt} = 0.9\frac{dx}{dt} -0.2\frac{dx}{dt}[/tex]
[tex]v_y = 0.9v_x -0.2v_x[/tex]
[tex]v_y = 0.7v_x[/tex]
Now, putting [tex]v_x=10.0 \text{ m/s}[/tex] in the above equation;
[tex]v_y = 0.7 \times 10.0[/tex] = 7 m/s
Now, the resultant velocity is given by = [tex]v=\sqrt{v_x^{2}+v_y^{2} }[/tex]
[tex]v=\sqrt{10^{2}+7^{2} }[/tex]
= [tex]\sqrt{149}[/tex] = 12.21 m/s
