How to determine the time and place of collision of 2 vectors.

Question

Object A moves in a straight line, starting at (2,0), After t seconds, its positions (x,y) is given by the equation:

${x \choose y} = {2 \choose 0}+t{0.7 \choose 1}$

Object B travels in a straight line. The equation of its path is:

$\vec r= {1 \choose 35}+p{1 \choose -1}$

where p is the time in minutes.

Object B starts moving 5 minutes after object . Will the 2 collide? If so, when and where?

Yeah I am clueless in how to start this. Equate both of the questions and solve them simultaneously? Can someone please show me how is it done? What are the things i should keep in mind in regard to different takeoff time?

Answer 1

I started to answer your question when I noticed some parts don't make much sense. For one thing, you say for $A$ that time $t$ is measured in seconds, but for $B$ it's value of time $p$ is in minutes. Well, after $5$ minutes, i.e., $300$ seconds, object $A$'s $x$ value would be at $2 + 300(0.7) = 212$. Also, every minute, it would be increasing by $60(0.7)=42$. However, $B$'s $x$ value would start at $1$ and only increase by $1$ every minute after that. As you can see, $A$'s $x$ value starts off larger and is increasing much faster, so there's no chance it would ever coincide with $B$'s $x$ value at some later time.

I believe your sheet has a mistake. If $t$ is in minutes instead, then the question becomes much more reasonable. Let us assume this. Then, since $B$ would start $5$ minutes after object $A$, this means $p$ would be $5$ less than $t$, i.e., $p = t - 5$. Using this, you equate the $x$ and $y$ coordinate equations for objects $A$ and $B$ to solve each for $t$. If they give the same value, then the $2$ objects collide at that time. Otherwise, with different values, it means the $x$ and $y$ coordinates never match at the same time between the $2$ objects and, thus, they never collide.

The equations become

$$\begin{equation}\begin{aligned} 2 + 0.7t & = 1 + t - 5 \\ 6 & = 0.3t \\ t & = 20 \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

and

$$\begin{equation}\begin{aligned} t & = 35 - (t - 5) \\ 2t & = 40 \\ t & = 20 \end{aligned}\end{equation}\tag{2}\label{eq2A}$$

This means they would collide at a time of $20$ minutes, with a position of $x = 16$ and $y = 20$.

Answer 2

Equate both of the questions and solve them simultaneously?

Yes, that's what you do.

The key piece of information is that Object B starts moving $5$ minutes after object A. When $t$ is measured in seconds, this means that $p=t+300$ ($300$ seconds that is, when $t=0$, what is $p$?).

You equate the $x$ and $y$ coefficients: $$2+0.7t=1 + \frac{t+300}{60}$$ $$t = 35-\frac{t+300}{60}$$

If you have an inconsistent solution (such as when both lines are parallel), then they will never meet and there will be no collision. Otherwise, you can solve for $t$ and substitute back to find $x$ and $y$.