Confused about negative velocities (beginner physics)

[mujadeo]

Book says that velocitys must have speed AND direction.
going down or to the left on your coordinates, is negative.

So if I am walking left along the x-axis at 2m/s, my velocity is -2m/s
But If I walk down along the y-axis will I also have a neg velocity?

If so, then what is my velocity if I walk diagonally left and up,
or right and down?

Can someone explain this in simple terms? Thanks!

 

[Danger]

Welcome to PF, Mujadeo.
To start with, speed is a scalar measurement. Velocity is a vector, which involves both speed and direction.
You cannot be 'walking left'. You walk in a particular direction, which might occassionally change. Or you walk in the same direction, but change your pace. Either one imparts an accelaration, since it changes the vector.

 

[rcgldr]

A velocity vector as you have described has two components, an x component and a y component. So while either or both components may be negative, it doesn't really translate into meaning the velocity vector is negative. The magnitude will always be a positive value.

Polar or spherical cordinates can be used for describing vectors, which translates into a direction and a magnitude, but the math isn't as easy in most cases. However classic navigation uses a spherical like system, latitude, longitude, and altitude for current position, then heading and speed for velocity vector (heading is usually restricted so that altitude is constant, but could be generalized to be spherical instead of polar like).

 

[olgranpappy]

You cannot be 'walking left'.

Perhaps what Mujadeo meant to say is that if one orients a piece of graph paper relative to one's face in the typical manner and similarly one draws in a pair of x and y axes in the typical manner, then the direction of the negative x-axis is indeed "to the left." And "positive y" is "up", et cetera--you get the picture.

Thus if we now place a small ladybug at the origin of the coordinates and instruct her to walk to the point (-2,0) then she will certainly "walk left."

An example of a velocity that is both "up" and "to the left" is given in x and y components by:

v=(-42,+69)

The magnitude of this velocity
(sqrt[(-42)^2+69^2]=sqrt[42^2+69^2]=80.78)
is still positive as it must be.

 

[Danger]

Good point, Olgranpappy. Having no formal education, I tend to think in 'real world', rather than theoretical, terms. The original question really didn't make any sense to me until your clarification. Thanks.

 

[olgranpappy]

No problem. Cheers.

 

[Mr Virtual]

If you have considered your right as positive x axis, and your front as positive y axis, then:

1. if you walk right, you have +ve velocity in x-axis and zero velocity in y axis.
2. if you walk left, you have -ve velocity in x-axis and zero velocity in y axis.
3. if you walk forward (+ve y-axis), you have +ve velocity in y-axis and zero velocity in x axis.
4. if you walk backward, you have -ve velocity in y-axis and zero velocity in x axis.
5. if you walk diagonally right-forward, you have +ve velocities in both axes.
6. if you walk diagonally right-backward, you have +ve velocity in x-axis but
-ve velocity in y axis.
7. if you walk diagonally left-forward, you have -ve velocity in x-axis and +ve velocity in y axis.
8. if you walk diagonally left-backward, you have -ve velocities in both axes.

So, you have to consider both axes in case of 2D plane. You may have positive velocity in one axis but negative velocity in the other. The magnitude of net velocity is always positive, though I am not sure about the sign we should put against the net velocity (in vector form, of course!).
It is like you are in a helicopter which is flying to the east. If suddenly the fan on its head breaks down ( the tail fan is working), then the helicopter will plummet to the land, though it is still going eastward. You cannot say that your net velocity is +ve or -ve: it is negative in case of vertical velocity and positive in case of horizontal velocity.
Now suppose both the fans break down in the midst of a strong storm blowing in the opposite direction. In this case, you are falling down and moving westwards ( the storm is carrying you in that direction). So both your velocities are negative.
I hope my post helped you. If I am wrong, please correct me.

Mr V

 

[cristo]

So, you have to consider both axes in case of 2D plane. You may have positive velocity in one axis but negative velocity in the other. The magnitude of net velocity is always positive, though I am not sure about the sign we should put against the net velocity (in vector form, of course!)

You would write a two dimensional velocity vector in the form v=ai+bj. Like you say, the magnitude of the vector-- $\sqrt{a^2+b^2}$ --is always positive. However, it does not make sense to give the overall direction as being either positive or negative; what we can do is say that the velocity is, say, a in the positive x direction and b in the positive y direction.

Of course, if we are considering real life situations, for example me driving in my car, then it does make sense to define that whenever I am moving forwards, my velocity is positive and whenever I am moving backwards, my velocity is negative.

 

[Mr Virtual]

Thanks cristo for clearing my confusion.

regards
Mr V

 

[timman_24]

You can set whatever direction you want to be negative and go from there. You could use the classic X,Y coordinate system. In which (looking straight down at the graph) right is +X, left is -X, up is +Y, and down is -Y. The number denotes the magnitude and the sign tells you the direction of travel. This only works for vectors however, you would have two separate velocities to totally describe the motion of the particle in the typical X,Y system. These are called vectors.

2m/s in the the X direction and -4m/s in the Y direction would be written as:
v=<2,-4> or 2i-4j