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It's long time I've had vectors. My friend asked me to help with this exercise given below:

The train stations for each of the following cities $A$, $B$, $C$, $D$ have the following coordinates: $A=(-1,-2)$, $B=(10,3)$, $C=(1,5)$ and $D=(7,-1)$ in a coordinate system. It is known that the $X$-axis is east and $Y$-axis is north. The distance between the cities are measured in kilometers. The points $A$ and $B$ are connected in a straight line and the points $C$ and $D$ are connected in a straight line. Both lines are intersecting at the point $E$. See the following image:

http://puu.sh/CWtnJ/cfd9427e2f.png

a) Compute unit vectors for $A$ to $B$ and $C$ to $D$.

Two trains at the same time leave the railway stations in $A$ and $C$. The train from $A$ to $B$ runs at $100\text{ km/h}$ and the train from $C$ to $D$ runs at $65\text{ km/h}$

b) Specify a vector describing the direction and speed of the movement of the train from $A$ to $B$

c) Specify a vector describing the direction and speed of the movement of the train from $C$ to $D$

d) Specify a parameter representation for the straight-line movement of the train from $A$ to $B$

e) Specify a parameter representation for the straight-line movement of the train from $C$ to $D$

f) Determine the coordinates of $E$. Will the two trains hit each other?

So my work is:

a) $\vec{e_1}=\binom{0.91036}{0.41381}$ and $\vec{e_2}=\binom{0.70710}{-0.70710}$

b) I don't really understand this question very well. Can anyone give me a hint here?

c) This is similar to b) so if I can solve b) after the hint, I can do c)

d) I believe it should be like this: $\binom{x}{y}=\binom{-1}{-2}+t\binom{11}{5}$

e) I believe it should be like this: $\binom{x}{y}=\binom{1}{5}+t\binom{6}{-6}$

f) And here I know how to calculate the intersection, but I don't know if the trains would hit each other. The intersection point $E$ is $E=(5.19,0.81)$

Note: I want hints. Thanks in advance

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  • $\begingroup$ $\vec{e1}$ is not correct. It should be the unit vector corresponding to $(10 - (-1), 3 - 2) = (11, 1)$ $\endgroup$ – ab123 Mar 7 at 14:47
  • $\begingroup$ I forgot the negative sign. A=(-1,-2). I correct it now. $\endgroup$ – Anders Jørgensen Mar 7 at 14:49
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(b), (c) You need the vectors in the direction of the vectors $\vec{AB}$ and $\vec{CD}$ with magnitude equal to the respective speeds.

(d), (e) Notice that the train should move $\color{blue}{speed \times time}$ distance along the vector direction.

The parametric representation can be done in terms of the position $(x, y)$ in terms of the time elapsed, say $t$.

For the first train, in $t$ (measured in hours), position is $(x, y) = (-1, -2 ) + 100 \times t \times {\text{unit vector in direction of $\vec{AB}$}} $.

$\implies (x, y)=(-1, -2)+(11, 5)\frac{100t}{\sqrt{121 + 25}}$. Similarly calculate for (e).

(f) You can do this in two different ways -

First, you can equate the two parametric representations to find if there exists a solution $t'$ (Equate the $x$ and $y$ coordinates and see if a solution of $t$ is possible). This corresponds to the trains reaching a point in space at the same time.

Second, if you have already found the point of intersection, calculate the value of time $t$ for both the trains. If they are the same, trains meet at a point, otherwise, they don't.

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  • $\begingroup$ Good, thank you for the hints. Regarding (b) and (c), if I find the magnitude and direction, then I have found direction and speed? $\endgroup$ – Anders Jørgensen Mar 7 at 15:14
  • $\begingroup$ Yes, I think the magnitude of that vector must be equal to the speed. Edited that in the answer $\endgroup$ – ab123 Mar 7 at 15:16
  • $\begingroup$ For (f) I did the calculations in Maple and obtained different values of $t$. I also try with differents speed (extra exercise) with 110km/h and 95km/h and they didn't hit either. $\endgroup$ – Anders Jørgensen Mar 7 at 15:38
  • $\begingroup$ Ohh, I did a mistake. They actually hit in the point E. Also for 110km/h and 95km/h $\endgroup$ – Anders Jørgensen Mar 7 at 16:04

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