Q:

A ferry leaves Nanaimo to make the 22 km trip to Vancouver at the same time as a forry leaves Vancouver for Nanaimo. The ferry leaving Nanaimo travels 2 km/h faster than the other ferry How far are they from Vancouver when they meet 45 minutes later?

Accepted Solution

A:
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.