Explanation:
It is given that, the water from a fire hose follows a path described by equation :
[tex]y=3+0.8x-0.4x^2[/tex]........(1)
The x component of constant velocity, [tex]v_x=5\ m/s[/tex]
We need to find the resultant velocity at the point (2,3).
Let [tex]\dfrac{dx}{dt}=v_x[/tex] and [tex]\dfrac{dy}{dt}=v_y[/tex]
Differentiating equation (1) wrt t as,
[tex]\dfrac{dy}{dt}=0.8\times \dfrac{dx}{dt}-0.8x\times \dfrac{dx}{dt}[/tex]
[tex]v_y=0.8\times v_x-0.8x\times v_x[/tex]
[tex]v_y=0.8v_x(1-x)[/tex]
When x = 2 and [tex]v_x=5\ m/s[/tex]
So,
[tex]v_y=0.8\times 5\times (1-2)[/tex]
[tex]v_y=-4\ m/s[/tex]
Resultant velocity, [tex]v=\sqrt{v_x^2+v_y^2}[/tex]
[tex]v=\sqrt{5^2+(-4)^2}[/tex]
v = 6.4 m/s
So, the resultant velocity at point (2,3) is 6.4 m/s. Hence, this is the required solution.
