Kinematics question on 2 planes moving relatively with each other

In summary, the conversation discusses finding components and determining the correct approach for a problem involving a moving origin. The equations of motion for plane B with respect to plane A are given, and expressions for radius and angle are derived. There is some confusion about handling acceleration, but the conversation ends with clarification on this and progress being made on part B of the problem.
  • #1
Santilopez10
81
8
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.
Relevant 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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  • #2
Santilopez10 said:
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:
  • #3
PeroK said:
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.
 
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  • #4
Santilopez10 said:
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?
 
  • #5
PeroK said:
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??
 
  • #6
Santilopez10 said:
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.
 
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  • #7
PeroK said:
You have plane B accelerating away from A.
oh you are right, that was a typo mistake actually, thanks a lot for the help!
 

1. How do you calculate the relative velocity between two planes?

To calculate the relative velocity between two planes, you need to first determine the velocities of each plane and then use the formula Vrel=V1-V2, where Vrel is the relative velocity, V1 is the velocity of the first plane, and V2 is the velocity of the second plane.

2. What is the difference between velocity and speed?

Velocity and speed are often used interchangeably, but they actually have different meanings. Velocity is a vector quantity that describes the rate of change of an object's position and includes both direction and magnitude. Speed, on the other hand, is a scalar quantity that describes only the magnitude of an object's motion.

3. How does relative motion affect the distance between two planes?

Relative motion refers to the movement of one object in relation to another. In the case of two planes moving relatively with each other, the distance between them will change depending on their velocities and the direction they are moving in. If the planes are moving in the same direction, the distance between them will decrease, and if they are moving in opposite directions, the distance between them will increase.

4. Can two planes have the same velocity but different speeds?

Yes, two planes can have the same velocity but different speeds. Velocity takes into account the direction of an object's motion, whereas speed only considers the magnitude. This means that two planes can be moving at the same speed but in different directions, resulting in different velocities.

5. How does acceleration factor into relative motion between two planes?

Acceleration is the rate of change of velocity and can have a significant impact on the relative motion between two planes. If one plane is accelerating faster than the other, their relative velocity will change. Additionally, acceleration can also affect the relative distance between the two planes, as it can cause them to move closer together or further apart.

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