Question

The water from a fire hose follows a path described by y equals 3.0 plus 0.8 x minus 0.40 x squared ​(units are in​ meters). If v Subscript x is constant at 5.0 ​m/s, find the resultant velocity at the point left parenthesis 2.0 comma 3.0 right parenthesis .

Answer 1

Explanation:

It is given that, the water from a fire hose follows a path described by equation :

$y=3+0.8x-0.4x^2$........(1)

The x component of constant velocity, $v_x=5\ m/s$

We need to find the resultant velocity at the point (2,3).

Let $\dfrac{dx}{dt}=v_x$ and $\dfrac{dy}{dt}=v_y$

Differentiating equation (1) wrt t as,

$\dfrac{dy}{dt}=0.8\times \dfrac{dx}{dt}-0.8x\times \dfrac{dx}{dt}$

$v_y=0.8\times v_x-0.8x\times v_x$

$v_y=0.8v_x(1-x)$

When x = 2 and $v_x=5\ m/s$

So,

$v_y=0.8\times 5\times (1-2)$

$v_y=-4\ m/s$

Resultant velocity, $v=\sqrt{v_x^2+v_y^2}$

$v=\sqrt{5^2+(-4)^2}$

v = 6.4 m/s

So, the resultant velocity at point (2,3) is 6.4 m/s. Hence, this is the required solution.