Question

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). If v Subscript x is constant at 10.0 ​m/s, find the resultant velocity at the point left parenthesis 9.0 comma 2.0 right parenthesis .

Answer 1

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;

$y=2.0 + 0.9x-0.10x^{2}$

The velocity of the $x$ component is constant at =  $v_x=10.0 \text{ m/s}$

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;

$v_x=\frac{dx}{dt} \text{ and } v_y=\frac{dy}{dt}$

Now, differentiating the above equation with respect to t, we get;

$y=2.0 + 0.9x-0.10x^{2}$

$\frac{dy}{dt} =0 + 0.9\frac{dx}{dt} -(0.10\times 2)\frac{dx}{dt}$

$\frac{dy}{dt} = 0.9\frac{dx}{dt} -0.2\frac{dx}{dt}$

$v_y = 0.9v_x -0.2v_x$

$v_y = 0.7v_x$

Now, putting $v_x=10.0 \text{ m/s}$ in the above equation;

$v_y = 0.7 \times 10.0$ = 7 m/s

Now, the resultant velocity is given by = $v=\sqrt{v_x^{2}+v_y^{2} }$

                            $v=\sqrt{10^{2}+7^{2} }$

                               = $\sqrt{149}$ = 12.21 m/s