projecting vectors with different coordinate planes

[JustinPaul]

I am trying to solve this problem. I need the cars velocity vector (direction and magnitude). A car is traveling with a unknown velocity . 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 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.

Any help or hints would be appriciated

thanks

 

[skeeter]

Thought I would make an attempt to provide some guidance for this query ... I believe my approach is valid, but if not, I'm sure someone will point out any mistakes or come up with a simpler approach.

let [MATH]r[/MATH] be the distance (what you labeled as D) from the origin to any point [MATH](x,y)[/MATH] on the plane

in general, [MATH]r^2 = x^2 + y^2[/MATH]
using calculus ...

[MATH]r \cdot \dfrac{dr}{dt} = x \cdot \dfrac{dx}{dt} + y \cdot \dfrac{dy}{dt}[/MATH]
[MATH]\dfrac{dr}{dt} = v_r[/MATH] is the radial velocity of the object

[MATH]\dfrac{dx}{dt} = v_x[/MATH] is the horizontal component of the object's velocity

[MATH]\dfrac{dy}{dt} = v_y[/MATH] is the vertical component of the object's velocity

the object's speed would be [MATH]|v| = \sqrt{v_x^2 + v_y^2}[/MATH]
the object's direction w/respect to the x-axis would be [MATH]\phi = \arctan\left(\dfrac{v_y}{v_x}\right)[/MATH], paying close attention to the signs of both velocity components to get the correct quadrant direction.

so, the object is to find those two components of velocity.
note that each position coordinate is [MATH]x = r\cos{\theta}[/MATH] and [MATH]y = r\sin{\theta}[/MATH]
using your diagram for the two positions ...

[MATH]r_1 \cdot v_{r_1} = r_1\cos{\theta_1} \cdot v_x + r_1 \sin{\theta_1} \cdot v_y [/MATH]
[MATH]r_2 \cdot v_{r_2} = r_2\cos{\theta_2} \cdot v_x + r_2 \sin{\theta_2} \cdot v_y [/MATH]
note [MATH]r_1 \text{ and } r_2[/MATH] cancel in each equation ...

[MATH]v_{r_1} = \cos{\theta_1} \cdot v_x + \sin{\theta_1} \cdot v_y [/MATH]
[MATH]v_{r_2} = \cos{\theta_2} \cdot v_x + \sin{\theta_2} \cdot v_y [/MATH]
convert these two equations to a matrix format ...



solving for [MATH]v_x \text{ and } v_y[/MATH] requires the use of the first matrix's inverse ...



where ...