Aferry 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?

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:

, in which 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 . It is 2km/h faster than the second ferry, so i am going to say that . It moves in the positive direction, so v is positive. The equation of the position of this train is modeled as:

,

Second ferry:

Leaves Vancouver, so . It has a speed of v, that is negative, since it moves in the negative direction. So

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

With this we find the value o v, and replace in the equation of to see how far they are from Vancouver when they meet.

km/h.

They are 11.7475km away from Vancouver when they meet.

i believe it's a.

i'm not sure on this one, most likely correct.