Respuesta :
Answer:
They are 11.7475km away from Vancouver when they meet.
Step-by-step explanation:
The first step to solve this problem is modeling the position of each ferry. The position can be modeled by a first order equation in the following format:
[tex]S(t) = S_{0} + vt[/tex], in which [tex]S_{0}[/tex] is the initial position of the ferry, t is the time in hours and v is the speed in km/h.
I am going to say that the positive direction is from Nanaimo to Vancouver, and that Nanaimo is the position 0 and Vancouver the position 22.
First ferry:
Leaves Nanaimo, so [tex]S_{0} = 0[/tex]. It is 2km/h faster than the second ferry, so i am going to say that [tex]v = v + 2[/tex]. It moves in the positive direction, so v is positive. The equation of the position of this train is modeled as:
[tex]S_{1}(t) = 0 + (v+2)t[/tex],
Second ferry:
Leaves Vancouver, so [tex]S_{0} = 22[/tex]. It has a speed of v, that is negative, since it moves in the negative direction. So
[tex]S_{2}(t) = 22 - vt[/tex]
The problem states that they meet in 45 minutes. Here we have to pay attention. Since the speed is in km/h, the time needs to be in h. So 45 minutes = 0.75h.
They meet in 0.75h. It means that
[tex]S_{1}(0.75) = S_{2}(0.75)[/tex]
With this we find the value o v, and replace in the equation of [tex]S_{2}[/tex] to see how far they are from Vancouver when they meet.
[tex]S_{1}(0.75) = S_{2}(0.75)[/tex]
[tex]0.75(v+2) = 22 - 0.75v[/tex]
[tex]0.75v + 1.50 = 22 - 0.75v[/tex]
[tex]1.50v = 20.50[/tex]
[tex]v = \frac{20.50}{1.50}[/tex]
[tex]v = 13.67[/tex]km/h.
[tex]S_{2}(t) = 22 - vt[/tex]
[tex]S_{2}(0.75) = 22 - 13.67*0.75 = 11.7475[/tex]
They are 11.7475km away from Vancouver when they meet.
