Velocity of moving object at x and y axis

I want to calculate the velocity and acceleration of a moving object. I have the centroid value and i am calculating the distance covered by using euclidean distance formula on current and previous centroid. I am confused on how do we calculate velocity and acceleration of a moving object at x and y-axis. Do we simply divide the euclidean distance formula into x(centroid x) and y(centroid y) parts?

[KSSV]

Velocity is defined as displacement over time. You have distance now, do you have time?

[BlueBee77]

Yes, i have displacement and time. Just confused on how to get velocity x and velocity y

Answer 1

[Jan]

You can get the velocities in X and Y direction independently.

Vx = gradient(X, TimeStep);
Vy = gradient(Y, TimeStep);

Or:

S = blobCentroid - PrevCent;
V = gradient(S, TimeStep);

Now V contains teh X and Y component.

[BlueBee77]

Just to be sure that i am doing things correctly, I am calculating the distance covered as:

EucDist= sqrt(sum((blobCentroid - PrevCent) .^ 2))

then i am calculating velocity as:

V= EucDist/Time; and Acceleration= V/Time

Am i supposed to use V= diff(EucDist)/Time; and Acceleration= diff(V)/Time rather then the above one??

Using gradient, i can skip the euclidean distance??

[Jan]

@BlueBee77: Start with defining, what you want to know: If you want the velocities in X and Y direction separately, you do not need the Euclidean distance. If you want the total velocity independent from the direction, the Euclidean distance will solve this. Or you can calculate the velocities in X and Y direction at first, and then V_total = sqrt(V_x ^ 2 + V_y ^ 2).

It might get clear with an example: The object travels from the point (1, 2) to (2, 4) in 1 time unit h (e.g. seconds, but we can omit the units here). Then:

V_x = (2 - 1) / h = 1
V_y = (4 - 2) / h = 2
V_total = sqrt(1^2 + 2^2) = 2.236

Or you use the Euclidean distance:

V_total = Dist / h = sqrt((2 - 1)^2 + (4 - 2)^2) / h = 2.236

Both calculations are equivalent. In fact, the same numbers appear in the same equations.

[BlueBee77]

@Jan Simon: Thanks alot for such a nice explanation and solving my problem. I need the velocities and acceleration in X and Y direction :)

Answer 2

[Image Analyst]

You can use diff(distances) to get "instantaneous" velocity. Then again on velocity to get acceleration. Or use diff(x) and diff(y) to get velocity components along the two directions separately.

[BlueBee77]

Thanks for your response but how to get the x and y components of displacement??

[BlueBee77]

Can i write it like this for the displacement and then use the for velocity_x and Velocity-y :

     d_x= sqrt(x2-x1)^2 and d_y= sqrt(y2-y1)^2 % code