- #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!
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!