Answer:
[tex] t = t_1 +t_2 = 4 sec+6 sec = 10 sec[/tex]
Step-by-step explanation:
Let's call the initial point [tex]I= (x_i=8,y_i=10)[/tex] and the second point [tex]R=(x_r = 16, 20)[/tex] we can find the distance between these two points with the following formula:
[tex] d_1 = \sqrt{(x_r -x_i)^2 +(y_r -y_i)^2}[/tex]
And if we replace we got:
[tex] d_1 = \sqrt{(16 -8)^2 +(20 -10)^2} =2\sqrt{41}[/tex]
So since we know the time in order to reach the point R we can find the velocity like this:
[tex] v=\frac{d_1}{t_1} = \frac{2\sqrt{41}}{4}=\frac{\sqrt{41}}{2}[/tex]
And this velocity is constant along all the displacement.
Let's call the final point [tex] F =(x_f = 28, y_f = 35)[/tex]
And we can find the distance between the point R and F [tex] d_2[/tex] like this:
[tex] d_2 = \sqrt{(x_f -x_r)^2 +(y_f -y_r)^2}[/tex]
And if we replace we got:
[tex] d_2 = \sqrt{(28 -16)^2 +(35 -20)^2}=3\sqrt{41}[/tex]
Since the velocity is constant we can find the time between point R and F like this:
[tex] t_2 = \frac{3\sqrt{41}}{\frac{\sqrt{41}}{2}}=6 sec[/tex]
And we are interested on "When will it pick up the ball located at position (28,35)?" And then the total time would be:
[tex] t = t_1 +t_2 = 4 sec+6 sec = 10 sec[/tex]
The figure attached illustrate the problem.
