Calculate Acceleration using relative distances

[Ash_Sdr]

Homework Statement

There are two cars traveling in a same direction. The Second car has a camera mounted on it and has ability to give relative X and Y (in meters) distance of first car with respect to it. Given the Speed of Second car , Calculate the Speed and Acceleration of first car. The First car gives relative x and y distances for every 0.01 seconds.

Homework Equations

Differentiation of distance gives velocity and double differentiation of distance gives me acceleration

The Attempt at a Solution

Vx = (x2-x1)/(t2-t1) ; t2-t1 is 0.01 seconds
; x2 is the X distance at second time instant and x1 is at first time instant
Vy = (y2-y1)/(t2-t1)

Speed = Sqrt(Vx^2 + Vy^2)

Ax = (Vx2 – Vx1)/ (t2-t1)

Ay = (Vy2 – Vy1)/ (t2-t1) I am not considering the Speed of Second vehicle at all... I think that's where I am getting wrong results. As camera gives me the relative x and y , and since camera itself is moving , Do I need to add the Speed to Vx and then compute acceleration >?

 

[NFuller]

Since they are traveling in the same direction, you can just add the speed of the first car to the speed of the second car. If they were traveling in different directions, then you would have to be more careful and add up the components of velocity.

 

[NFuller]

Also, make sure your signs are right for the speed. If the second car is going faster than the first car, then you should subtract the relative speed of the first car from the speed of the second car.

 

[Ash_Sdr]

Thanks NFuller for clarifying it.

Let me explain what I understood ...

let V_s be the speed of Second car which has camera mounted . V_s = sqrt(V_sx² + V_sy² )

For two time instants , 0.01 second and 0.02 Second ... I get relative (x1,y1) and (x2,y2) as (5.2,0.4) and (6.8,0.5) in meters

Speed of First vehicle in X direction is Vsx + (x2-x1)/0.01 (or) Since they are traveling in straight direction , can i say Vs+ (x2-x1)/0.01 ??

 

[NFuller]

Vsx + (x2-x1)/0.01

This is correct, since you are adding the x component of the second car's velocity to the x component of the first car's velocity.

Vs+ (x2-x1)/0.01

This is incorrect, since you are adding the total speed of the second car (which contains both x and y dependence) to only the x component of the first car's velocity.

The trick here is finding the components of velocity $V_{x}$ and $V_{y}$ of the second car using the speed of the second car and the direction of the first car.