Ok so I did the first question of this and the second question up till 2c). Do we solve the question using position vector equals original position vector plus time times the velocity vector? If so, how do we figure out the time?
I have attached the image of the questionenter image description here
For question 2c, the first thing to do is to express the Californica's position relative to the Titanica as a function of time. At $t$ hours past midnight the position of the Californica is, as you say, its original position plus its velocity vector times time, so it is
$(25,-10) + (-10,17)t = (25-10t, -10+17t)$
The Titanica is at position $P=(0,11)$, so at time $t$ the position of the Californica relative to the Titanica is:
$(25-10t, -10+17t) - (0,11) = (25-10t, -21+17t)$
At time $t$ the distance of the Californica from the Titanica is:
$\sqrt{(25-10t)^2 + (-21+17t)^2}$
You need to find the value of $t$ that minimises this distance.