Calculating Velocity Vector of Object in 2D Space

[justinpaul]

Homework Statement

Find the velocity of an object using measurements of radial velocity and distance to the target

Relevant Equations

tV^2 = rV1^2 + pV1^2. = rV2^2 + pV2^2

The question I am trying to solve is what is the velocity vector (direction and magnitude) of an object in 2 d space. We know the distance measured to the car from two different angles. We know the radial velocity of the car on both measurements. The radial velocity is the component of the velocity projected directly toward the origin. For example, if the car is traveling on a path vector perpendicular to the ray from the origin the radial velocity is 0.

I have tried a bunch of stuff that does not work. I think the amount of change in radial velocity, change in distance, the angle between the measurements could create a ratio of some kind. The velocity of the object is the same in each measurement so Total V1 = Total V2

(radial velocity -> rV, total Velocity -> tV, perpendicular V -> pV, distance -> d)

tV^2 = rV1^2 + pV1^2. = rV2^2 + pV2^2Any help or hints would be appreciated

 

[mfb]

With the angles and the distances you can determine the positions and find the direction of motion - this doesn't need the radial velocities. You can then find the velocity with a single radial velocity measurement.

 

[haruspex]

With the angles and the distances you can determine the positions and find the direction of motion - this doesn't need the radial velocities. You can then find the velocity with a single radial velocity measurement.

I don't think the angles are known.

@justinpaul:
You can make life a bit simpler by rotating your axes so that, say, the first distance lies along the y axis.
Suppose the second distance is at (x,y) in Cartesian and (r₂, θ) in polar.
Let the speed be v and the direction be at φ to the x axis.
What equations can you write relating x, y, r₁, r₂, θ and φ?
How about v₁, v₂, v, θ and φ?

I get five equations, and you have five unknowns.

 

[justinpaul]

OK, thanks for the hints. I will work on what you suggest. FYI, we do know the angles. I don't know if this is useful or not. It would actually be better if I can solve the problem using one radial velocity

 

[haruspex]

OK, thanks for the hints. I will work on what you suggest. FYI, we do know the angles. I don't know if this is useful or not. It would actually be better if I can solve the problem using one radial velocity

Ok, I was misled by your "homework statement".
Yes, if you know the angles you only need one radial velocity, and you are able to find the direction of movement (which was not possible in my version).

A pity - I liked my version of the question.

 

[mfb]

A pity - I liked my version of the question.

You would need to know the time between measurements if you don't know the angles.

 

[haruspex]

You would need to know the time between measurements if you don't know the angles.

Are you sure?
I had
$x=r_2\cos(\theta)$
$y=r_2\sin(\theta)=r_1+x\tan(\phi)$
$v_1=v\sin(\phi)$
$v_2=v\cos(\theta-\phi)$
Five independent equations, five unknowns.
Could be messy, though.

 

[mfb]

Hmm, you are right. And I found an explicit solution using Cartesian coordinates. I'll post it here once OP solved the original homework problem.

The final answer is much simpler than I expected.

 

[justinpaul]

I have these equations for the two points …

$x_1 = r_1 cos(\theta_1)$
$y_1 = r_1 sin(\theta_1)$

$x_2 = r_2 cos(\theta_2)$
$y_2 = r_2 sin(\theta_2)$

This is the slope

$\frac {(y_1 – y_2)}{ (x_1-x_2)}$From the point slope formula

$y – y_1 = \frac {(y_1 – y_2)}{ (x_1-x_2)} (x – x_1)$

We have all the values to plug into get an equation for the line traveled by the objectThe velocity of the object is …

$\left|{v_t}\right|= \sqrt {v_x^2 + v_y^2}$

Is ??

$v_{r1} = cos(\theta_1) v_x + sin(\theta_1) v_y$
$v_{r2} = cos(\theta_2) v_x + sin(\theta_2) v_y$

I think this would give me two equations with two unknowns.

I don’t think I know how to do it without both radial velocities. Which I guess is not required but I am curious to know how if someone would like to give me another hint.

(I am learning LaTeX along with trig. ... if you are a newbie edit your post in an editor so you will not loose your work)

 

[haruspex]

I have these equations for the two points …

$x_1 = r_1 cos(\theta_1)$
$y_1 = r_1 sin(\theta_1)$

$x_2 = r_2 cos(\theta_2)$
$y_2 = r_2 sin(\theta_2)$

This is the slope

$\frac {(y_1 – y_2)}{ (x_1-x_2)}$From the point slope formula

$y – y_1 = \frac {(y_1 – y_2)}{ (x_1-x_2)} (x – x_1)$

We have all the values to plug into get an equation for the line traveled by the objectThe velocity of the object is …

$\left|{v_t}\right|= \sqrt {v_x^2 + v_y^2}$

Is ??

$v_{r1} = cos(\theta_1) v_x + sin(\theta_1) v_y$
$v_{r2} = cos(\theta_2) v_x + sin(\theta_2) v_y$

I think this would give me two equations with two unknowns.

I don’t think I know how to do it without both radial velocities. Which I guess is not required but I am curious to know how if someone would like to give me another hint.

(I am learning LaTeX along with trig. ... if you are a newbie edit your post in an editor so you will not loose your work)

Looks good.

You have not used the radial distances. From those and the angles, as @mfb mentioned in post #2, you can find the coordinates of the two points. This gives you the relationship between the x and y components of velocity.

 

[mfb]

Looks good. Now we have three ways to solve this problem.

• Both radial distances, both angles and one radial velocity - what I was suggesting in post 2.
• Both angles and both radial velocities but no distances - post 9
• Both radial distances and both velocities but no angles - what haruspex and I developed

Here is the third approach in Cartesian coordinates:
Define the y-axis as the direction of the object at the first measurement. Let t be the unknown time between measurements. x and y are the coordinates at the time of the second measurement. We can then set up 5 equations:
$v_1 = v_y$
$x=tv_x$
$y=r_1+t v_y$
$r_2=\sqrt{x^2+y^2}$
$v_2 = \frac{x v_x+y v_y}{r_2}$
The last equation comes from $\displaystyle v_2 = v_{radial} = \frac{ \vec v \cdot \vec r}{|r|}$
vx,vy,x,y,t are unknown and we have 5 equations.

Eliminate $v_y$, $x$ and $y$ using the first three relations and square the fourth one:
$r_2^2=t^2 v_x^2+(r_1+t v_1)^2$
$v_2 r_2 = t v_x^2+(r_1+t v_1) v_1$ (eq1)
Multiply the second one by t and subtract the two equations from each other to get rid of $v_x$:
$$r_2^2 - t v_2 r_2 = (r_1+t v_1)^2 - t v_1 (r_1+t v_1)$$The $t^2$ term cancels, so this is a linear equation in t. $$r_2^2-r_1^2 = t(r_1 v_1 + v_2r_2)$$
This is really nice to cross-check. If $r_1=r_2$ then either t=0 and we are lost or $v_1=-v_2$ as expected if we observe two points on a circle (this is a degenerate case as well, we won't get the velocity then). Ignoring this case we can solve for t:
$$t = \frac{r_2^2-r_1^2}{r_1 v_1 + v_2r_2}$$
To get the velocity we can solve eq1 for $v_x^2$:
$$v_x^2=\frac{v_2 r_2 -(r_1+t v_1) v_1}{t} = \frac{v_2 r_2-v_1r_1}{t}-v_1^2$$
But instead of calculating $v_x^2$ separately this gives us a more direct way to the velocity:
$$v_x^2 + v_1^2 = \frac{v_2 r_2-v_1r_1}{t} = \frac{(v_2 r_2-r_1v_1)(r_1 v_1 + v_2r_2)}{r_2^2-r_1^2} = \frac{v_2^2 r_2^2-r_1^2v_1^2}{r_2^2-r_1^2}$$
And therefore
$$v^2 = \frac{v_1^2 r_1^2-r_2^2v_2^2}{r_1^2-r_2^2}$$