# Kinematics question on 2 planes moving relatively with each other

#### Santilopez10

Homework Statement
Plane A flies horizontally at 12192 m respect to ground y has acceleration $\vec a= 1.22 \frac{m}{s^2}$. It is equipped with a radar that detects another plane B, that flies in the same direction at 18288m. If in the shown instant $\theta = 30º$, A's velocity is 965 km/h and B's is 1448 km/h *constant*, find:
A) the time variation of the radial component of the velocity vector as a function of time.
B) Find the angular acceleration.
Homework Equations
Kinematic equations on polar coordinates
So my problem isn't actually finding the components, but knowing if the initial approach I took is correct. So what I did was:
At first I found that at the same instant, $x_{B/A}=10500 m$ so then I wrote the equation of motion for plane B respect to A:
so $$\vec a_{B/O}- \vec a_{A/O}=\vec a_{B/A} \rightarrow \vec a_{B/A}=-1.22 \frac{m}{s^2}$$ Here O represents a fixed stationary origin for relativistic movement purpose.
$$\vec v_{B/A}= 1.22t+v_{0_{B/A}}=1.22t+135 [m/s]$$ then $$\vec x_{B/A}=0.6t^2+135t+10500 [m]$$
So my expressions for radius and angle would end being:
$r_{B/A}=\sqrt{(0.6t^2+135t+10500)^2+6100^2}$ and $\theta_{B/A}=\arctan{\frac{6100}{0.6t^2+135t+10500}}$
Is this correct? Thanks!

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#### PeroK

Homework Helper
Gold Member
2018 Award
At first I found that at the same instant, $x_{B/A}=10500 m$ so then I wrote the equation of motion for plane B respect to A:
so $$\vec a_{B/O}- \vec a_{A/O}=\vec a_{B/A} \rightarrow \vec a_{B/A}=-1.22 \frac{m}{s^2}$$ Here O represents a fixed stationary origin for relativistic movement purpose.
$$\vec v_{B/A}= 1.22t+v_{0_{B/A}}=1.22t+135 [m/s]$$ then $$\vec x_{B/A}=0.6t^2+135t+10500 [m]$$
So my expressions for radius and angle would end being:
$r_{B/A}=\sqrt{(0.6t^2+135t+10500)^2+6100^2}$ and $\theta_{B/A}=\arctan{\frac{6100}{0.6t^2+135t+10500}}$
Is this correct? Thanks!
I found it hard to follow what your doing. I think it looks okay, except you haven't been careful with your acceleration.

Simply taking plane A as the origin would have been simplified things. And, you need to introduce your variables better. And, you really need to show when you are converting from km/h to m/s.

Alternatively, you could have done it algebraically. That's always easier to follow than lots of numbers.

Last edited:

#### Santilopez10

I found it hard to follow what your doing. I think it looks okay, except you haven't been careful with your acceleration.

Simply taking plane A as the origin would have been simplified things. And, you need to introduce your variables better. And, you really need to show when you are converting from km/h to m/s.

Alternatively, you could have done it algebraically. That's always easier to follow than lots of numbers.
But if I took the plane as my origin then my frame of reference would be moving with acceleration, and I do not know how to handle that yet.

#### PeroK

Homework Helper
Gold Member
2018 Award
But if I took the plane as my origin then my frame of reference would be moving with acceleration, and I do not know how to handle that yet.
For kinematic equations it doesn't make any difference. If you want to use Newton's laws, then you must be careful.

In any case, how are you getting on with part b?

#### Santilopez10

For kinematic equations it doesn't make any difference. If you want to use Newton's laws, then you must be careful.

In any case, how are you getting on with part b?
part B isn't really hard considering angular acceleration is nothing but $$\ddot{\theta}(t)$$ but still the hard part is getting right the polar equations.
Let say I place my origin at A, then plane A should be seeing B deaccelerating at a rate $\vec a= -1.22$ right? then the formula for the horizontal distance is the one I already had, is my reasoning correct??

#### PeroK

Homework Helper
Gold Member
2018 Award
part B isn't really hard considering angular acceleration is nothing but $$\ddot{\theta}(t)$$ but still the hard part is getting right the polar equations.
Let say I place my origin at A, then plane A should be seeing B deaccelerating at a rate $\vec a= -1.22$ right? then the formula for the horizontal distance is the one I already had, is my reasoning correct??
You have plane B accelerating away from A.

#### Santilopez10

You have plane B accelerating away from A.
oh you are right, that was a typo mistake actually, thanks a lot for the help!

"Kinematics question on 2 planes moving relatively with each other"

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