Respuesta :
As we know that position is given as
[tex]x = 4 cost , y = 6 sint[/tex]
now in order to find the velocity
[tex]\frac{dx}{dt} = v_x[/tex]
so here we will have
[tex]v_x = \frac{d}{dt} 4cost = -4sint[/tex]
similarly in y direction
[tex]v_y = \frac{dy}{dt}[/tex]
[tex]v_y = \frac{d}{dt}(6sint) = 6 cost[/tex]
now the net velocity is given as
[tex]v = \sqrt{v_x^2 + v_y^2}[/tex]
[tex]v = \sqrt{(-4sint)^2+(6cost)^2}[/tex]
[tex]v = \sqrt{16sin^2t + 36 cos^2t}[/tex]
so above is the velocity as a function of time
The velocity of an object depends on its displacement within a given time interval.
The velocity of the object is [tex]v(t)= \sqrt{16 sin^2t+36cos^2t}[/tex] .
What is the velocity?
Velocity can be defined as the change in the displacement x of an object with respect to a given time interval t.
[tex]v = \dfrac {\Delta x}{\Delta t}[/tex]
The displacement of object in x-direction is x = 4 cos(t) and the displacement of object in y-direction is y = 6 sin(t).
The velocity in the x-direction is calculated as given below.
[tex]v_x = \dfrac {dx}{dt}[/tex]
[tex]v_x = \dfrac {d}{dt} (4cos (t))[/tex]
[tex]v_x = - 4sin (t)[/tex]
The velocity in the y-direction is calculated as given below
[tex]v_y = \dfrac {dy}{dt}[/tex]
[tex]v_y = \dfrac {d}{dt} (6 sin (t))[/tex]
[tex]v_y = 6cos (t)[/tex]
The magnitude of the net velocity can be calculated as given below.
[tex]v(t) = \sqrt{v_x^2 + v_y^2}[/tex]
[tex]v(t) = \sqrt{(-4sin(t))^2 + (6cos(t))^2}[/tex]
[tex]v(t)= \sqrt{16 sin^2t+36cos^2t}[/tex]
Hence we can conclude that the velocity of the object is [tex]v(t)= \sqrt{16 sin^2t+36cos^2t}[/tex] .
To know more about the velocity, follow the link given below.
https://brainly.com/question/862972.
