# A-Level Mechanics - Vectors

This is Part 2, of the tutorial on vectors, aimed at A-level Mechanics students.

These are the 'exam style questions', I will be covering by the end of this tutorial.

1. At noon, boat B has the position i+2j km and velocity, 3i+2j km/hr. A lighthouse, L, has the position 5i + j. Find vector LB at 2pm.
2. Find LB at 3pm.
3. What is the distance between the lighthouse and the boat at 2pm?
4. Find LB, at time t hours after noon?
5. At what time, will the boat be north of the lighthouse, and what is vector LB at that time?
6. Find the distance between the boat and the lighthouse, at t hours after noon.
7. At what time will the distance between the boat and the lighthouse be 7km?

## Magnitude of a vector

Before getting on the hard questions, I'd like to clarify some things first.

Say I have a vector, 3i + 4j. What is the magnitude of this vector?

The magnitude just means the length of the vector (the length of the arrow). Notice you have a right angle triangle here. The base has a length of 3 units, and the height is 4 units. So the hypotenuse would be 5 units (${\sqrt {3^{2}+4^{2}}}$ ). Therefore the magnitude of the vector 3i +4j, is 5 units. This statement can be mathematically written as: |3i+4j| = 5. (The vertical lines mean magnitude).

## Displacement vectors and distances

A displacement vector, is essentially an arrow that goes from one co-ordinate to another co-ordinate. Example: Lighthouse A and B have co-ordinates (2,3) and (5,7) respectively. What is the vector AB?

First I shall plot the points:

Answer: Well, AB is a vector going from A to B. From the drawing below you can see that AB = 3i +4j.

Question: What is BA?

Answer: $BA=-3i-4j$ If a question asks for the displacement between A and B, it doesn’t specify direction, and so you could write out $AB=3i+4j$ , or $BA=-3i-4j$ . They would both be correct.

Now, before I continue, I want you to have an appreciation for the fact that a position vector is essentially a co-ordinate. So for example, if you say that object A has co-ordinates (2,3), it’s the same as saying that object A has a position vector (2i+3j). This can be mathematically written as $r_{a}=(2i+3j)$ .

Knowing this, you can now take advantage of a simple formula for figuring out displacements vectors.

This formula is: $AB=r_{b}-r_{a}$ So, $AB=(5i+7J)-(2i+3j)=3i+4j$ Also $BA=r_{a}-r_{b}=(2i+3j)-(5i+7J)=-3i-4j$ Also, don’t confuse position vectors with displacement vectors.

Now, what is the distance between points A and B?

So you could do this without knowing anything about vectors.

Just applying Pythagoras' theorem, you can calculate the distance between A and B to be 5 units (${\sqrt {3^{2}+4^{2}}}$ ).

Or...you could have just figured out the length/magnitude of vectors AB or BA:

$|3i+4j|={\sqrt {3^{2}+4^{2}}}=5$ , or

$|-3i-4j|={\sqrt {(-3)^{2}+(-4)^{2}}}=5$ Another important thing to know, is that when you are asked for something like “find the position of B relative to boat A”, you are basically being asked for vector AB (NOT vector BA), so the answer would be 3i + 4j. Write down the box below on a separate piece of paper, as it will help with the rest of the questions I’ll be giving you on this page.

B relative to A = $AB$ = $r_{b}-r_{a}$ ## Exam Style Questions

At noon, boat B has the position i+2j km and velocity, 3i+2j km/hr. A lighthouse, L, has the position 5i + j.

### 1. Find vector LB at 2pm.

Draw it first...

In order to answer the question, you need to figure out where the boat will be at 2pm. For that you use the “position equation”, as I have shown in part 1 of this tutorial.

$R_{b}=R_{0}+v\cdot t$ $=(i+2j)+(3i+2j)t$ $=(1+3t)i+(2+2t)j$ At 2pm, t=2,

$R_{b}=(1+3x2)i+(2+2x2)j$ $=7i+6j$ $LB=R_{b}-R_{l}$ $=(7i+6j)-(5i+j)$ $=(2i+5j)$ Looking at the diagram, you can see that this is correct...

### 2. Find LB at 3pm.

From before: $R_{b}=(1+3t)i+(2+2t)j$ Thus at 3pm, $R_{b}=(1+3x3)i+(2+2x3)j=10i+9j$ So, $LB=(10i+8j)-(5i+j)=5i+7j$ .

Notice how the vector LB is NOT constant. It changes with time.

### 3. What is the distance between the lighthouse and the boat at 2pm?

Simply work out the magnitude of vector LB at 2pm:

$|2i+5j|={\sqrt {2^{2}+5^{2}}}={\sqrt {29}}=5.39km$ ### 4. Find LB, at time t hours after noon?

Well, from before we know that at any given time:

$R_{b}=(1+3t)i+(2+2t)j$ and $R_{l}=(5i+j)$ Therefore:

$LB=R_{b}-R_{l}$ $=[(1+3t)i+(2+2t)j]-(5i+j)$ $=(-4+3t)i+(1+2t)j$ This equation tells you what the vector LB is, as a function of time.

Test it out. Question 1 (from before): Find vector LB at 2pm.

At 2pm t=2, $LB=(-4+3x2)i+(1+2x2)j$ $=2i+5j$ (Same answer as you got before)

### 5. At what time, will the boat be north of the lighthouse, and what is vector LB at that time?

Draw it out...

You can see that when the boat is north of the lighthouse, vector LB has a zero i component.

From before, $LB=(-4+3t)i+(1+2t)j$ ,

We can see that the i component is zero when -4+3t = 0, which rearranges to give t = 4/3. This is equivalent to 1 hour, and 20 minutes. So the answer is 1.20pm.

At that time vector LB would be:

$LB=(-4+3x{\frac {4}{3}})i+(1+2x{\frac {4}{3}})j$ $=0i+3{\frac {2}{3}}j$ Looking at the diagram above, you can see this about right. Also notice how a similar question was answered in a different way in Part 1 of the tutorial.

### 6. Find the distance between the boat and the lighthouse, at t hours after noon.

Once again, from before, vector LB at any time is LB = (-4+3t)i + (1+2t)j.

To find the distance between L and B, find the magnitude/length of vector LB.

$|LB|=|(-4+3t)i+(1+2t)j|$ $={\sqrt {(-4+3t)^{2}+(1+2t)^{2}}}$ $={\sqrt {16+9t^{2}-24t+1+4t^{2}+4t}}$ $={\sqrt {13t^{2}-20t+17}}$ Distance = ${\sqrt {13t^{2}-20t+17}}$ This equation tells you the distance between the boat and the lighthouse as a function of time. Again, you can test it out. Question 3 (from before): What is the distance between the lighthouse and the boat at 2pm?

At 2pm, t=2,

Distance = ${\sqrt {13\cdot 2^{2}-20\cdot 2+17}}$ =${\sqrt {29}}$ = 5.39km. (Same answer as you got before)

### 7. At what time will the distance between the boat and the lighthouse be 7km?

Substitute distance = 7,

$7={\sqrt {13t^{2}-20t+17}}$ $49=13t^{2}-20t+17$ $13t^{2}-20t-32=0$ $t={\frac {-(-20)\pm {\sqrt {(-20)^{2}-4\cdot 13\cdot (-32)}}}{2\cdot 13}},$ $t={\frac {20\pm 45.43}{2\cdot 13}}$ t = 2.51 hours, and t = - 0.978

Ignore negative time solution.

To convert 2.51 hours to clock time: 2 hours + 60x${\frac {51}{100}}$ minutes

= 2 hours 31 minutes.

This is equivalent to 2.31pm.

## Final notes

1. That's it! If you have found this article useful, please comment in the discussion section (at the top of the page). Its just a nice ego boost really...
2. Also please comment if there are other questions you want covered, or if there something in this tutorial you did not understand.
3. I’d strongly advise you to do all the questions I’ve posted here again by yourself, as it will show you whether you understood it all.
4. DRAW!!! It doesn't have to be on square paper. Just a quick sketch (like I've done in part 1 of the tutorial), will do for most questions. When you are able to draw/visualise vectors, things will become a lot easier...

## Something extra...

Someone has left me a question on how to work out angles between vectors. Here it is...

Question: What is the angle between vector -2i+3j and the unit vector 'i'?

Firstly draw it out:

To find this angle out, you may need to find out a simpler angle first:

$\tan {\alpha }=2/3$ $\alpha =33.7^{\circ }$ $\theta =180-33.7$ $\theta =146.3^{\circ }$ Thats it!

Also, make sure not to draw it incorrectly/ work out the wrong angle, like below: