1. May 12, 2013

### Googl

I understand how to work with vectors but I am not yet very confident on how they work or think about them. I am working with mechanics A level.

For example how would you imagine the velocity vector (2i +5j) m/s. So I am assuming that the object is moving at 2m/s east or along the x-axis and 5m/s along the y-axis or north. Am I right?

So suppose an object is moving in a direction $$3i-4j$$ and has speed 10m/s. Could you explain how finding the velocity in that direction works.

I know how to work it out but not exactly how to think about it.

I know I would find the magnitude of $$3i-4j$$ which is; $$√(3^2+4^2)$$ then;

Velocity is equal to: $$10χ1/5(3i-4j) = (6i-8j)$$ I have trained myself to work that out but I am not sure about the logic behind it. Please explain. More bout vectors in general.

Thank you.

2. May 12, 2013

### VantagePoint72

If you identify the positive y axis with "north" then yes. Unless you are specifically doing a problem with cardinal directions, it would be better to keep it in the language of x, y, and z axes.

The magnitude of the vector is, as you computed, 5. That means that $(1/5)(3i - 4j)$ is a unit vector. If you explicitly calculate its magnitude, you will see it is 1.

In general, multiplying a vector by some scalar $\alpha$ changes its length by a factor of $\alpha$. Picture a vector, and now imagine, say, doubling each of its components. This is what you do when you multiply a vector by $\alpha = 2$. It shouldn't be hard to see that this doubles the length the of vector. You might find it helpful to draw a few sketches if this isn't clear.

So, now you can see why dividing a vector by its length gives you a unit vector: if it originally has length $L$ and you multiply it by $1/L$, its new length is $(1/L)L = 1$. Once you have your unit vector, you just multiply it by the length you want to have (in this case 10).

3. May 12, 2013

### Googl

I now understand the unit vector part. How can you relate a unit vector that you have found to a single component of the vector. So for example. A unit vector in the x-axis. So suppose that I have worked out the unit vector factor to be 1/5. Would I be right in saying a unit vector in the x-axis is 1/5x10i ? (which is equal to 2i in the x-axis) alone.

How come multiplying a unit vector by speed provides the velocity vector?

4. May 12, 2013

### VantagePoint72

No, $(1/5)10i$ is not a unit vector. As you can see by direct calculation, its length is 2, not 1. The components of a unit vector are not, in general, unit vectors themselves. You know how to calculate the length of a vector. If you are wondering if something is a unit vector, then try calculating its length yourself before asking if it is a unit vector. If it's length isn't 1 then by definition it is not a unit vector.

Multiplying a unit vector by a speed provides a velocity vector because velocity is, by the definition, a speed with a direction. If you have a dimensionless unit vector and you multiply it by a speed then, the result is a vector in a particular direction whose magnitude is a speed. Thus, the vector is a velocity.

5. May 12, 2013

### voko

Any vector is "something" with a direction. Displacement is the obvious case. Velocity is a little bit more complex: its direction is the direction of motion (which is a lot like in the displacement case), but its magnitude is the magnitude of velocity, i.e., the speed. The same is true for acceleration and force. Things get somewhat more bizarre when vectors are used to quantify rotational phenomena. In those cases, the direction is usually not the direction of motion as one might think intuitively, but the orientation of the axis of rotation, and the magnitude is the measure of the entity involved, such as the angular speed, torque, etc.

In any case, however, the direction is worked out by finding the corresponding unit vector, which is simply the original vector divided by its magnitude.

6. May 12, 2013

### tiny-tim

Hi Googl!
Velocity vectors work exactly the same way as position vectors.

If you had a position (strictly, a displacement) vector that was 2i + 5j,

you'd use Pythagoras, and say that the total distance was √(22 + 52) …

same with velocities.

7. May 12, 2013

### Googl

Thanks for the reply you've cleared up a few gaps.

I am unable yet to imagine when the question asks; ...an object is moving in the direction vector; $$3i -4j$$. What does that mean about the position of the object? Has the object moved? or Is it the distance to be covered? I have trained myself to calculate its velocity when speed of object is given which is finding the unit of the direction vector and multiplying it with the speed.

8. May 12, 2013

### VantagePoint72

The velocity by itself does not tell you anything about the position of the object. The position at a particular time needs to be given separately.

9. May 12, 2013

### tiny-tim

the position (displacement) of the object has changed, in the direction 3i -4j
that is correct

(and the displacement vector will then be the velocity vector times the time)

10. May 12, 2013

### VantagePoint72

Plus the initial displacement. Somewhere, your first calculus teacher felt a shiver and didn't know why...

11. May 12, 2013

### Googl

Thanks that helped a lot!