Say you have a Hiker $H$ walking at a constant velocity of $1.2\vec i - 0.9\vec j.$ Now say you have a rectangular field $OABC$ (the one above) with $OC$ due north and $OA$ due east, and at $12$ noon hiker $H$ is at the point $Y$ with the position vector $100 \vec j$ relative to the fixed origin. What would be the position vector of $H$ at time $t$ seconds after noon?
Furthermore, now say that you have another hiker $K$ is at the point with the the position vector $9\vec i + 46\vec j$, and has a constant velocity of $0.75\vec i + 1.8 \vec j.$ How would you show that at time $t$ seconds after noon, $\vec{HK} = (9-0.45t)\vec i+(2.7t-54)\vec j?$
So far I have tried the following things:
- Used the equation $r = r0 + vt$, to calculate the new position vector for hiker $H$ and got the answer as $r = 1.2t\vec i + (100 + 0.9t)\vec j$, which wrong for the first question apparently.
- For the second question I have no idea on where to start.
For the first question, you just substituted into the the formula wrong. You want $\boxed{100\vec j + t(1.2\vec i - 0.9\vec j)}$
I'm not sure what your problem is with the second part of the question. You find the position of $K$ just like you find the position of $H$. Is the problem how to compute $\vec{HK}?$ Remember that $\vec{HK}$ is the vector that takes you from $H$ to $K,$ that is, $H + \vec{HK} = K.$ Now solve for $\vec{HK}.$