Aeroplane: express the horizontal distance in terms of t

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A light aeroplane flies horizontally at a steady speed in a straight line above horizontal ground; its position at time 't' is (30t - 300, 10t + 500). On the ground there are two lookouts, A and B, at positions (7100,2800) and (125700,42500), respectively.

Question:

a) Write down an expression in terms of 't' for the square of the horizontal distance between the aeroplane and lookout A at time 't', and simplify your answer.
 
interval said:
On the [horizontal] ground there are two lookouts, A and B, at positions (7100,2800) and (125700,42500), respectively.

Why are the y-coordinates of each lookout not zero?
 
Hello, interval!

\(\displaystyle \text{A plane flies horizontally at a steady speed in a straight line above horizontal ground.}\)

\(\displaystyle \text{Its position at time }t\text{ is: }\;\begin{array}{ccc}x&=& 30t - 300 \\ y &=& 10t + 500\end{array}\)

\(\displaystyle \text{On the ground there are two lookouts: }\;A(7100,2800)\,\text{ and }\,B(125700,42500).\)

\(\displaystyle \text{a) Write an expression in terms of }t\text{ for the square of the horizontal distance between the plane and lookout }A\text{ at time }t.\)

\(\displaystyle \text{I assume that the }xy\text{-plane is the ground, and the plane is flying above it.}\)

\(\displaystyle \text{For part(a), we can ignore lookout }B.\)


Code:
   (7100,2800)                        *
      A o                       *
          *               *
            *       *
              o P
          * (30t-300, 10t+500)
    *

\(\displaystyle \text{The "horizontal distance" is the length of segment }AP.\)


\(\displaystyle \overline{AP} \;=\;\left[(30t-300) - 7100\right]^2 + \left[(10t+500) - 2800\right]^2\)

. . . \(\displaystyle =\;(30t - 7400)^2 + (10t - 2300)^2\)

. . . \(\displaystyle =\;900t^2 - 444,\!000t + 54,\!760,\!000 + 100t^2 - 46,\!000t + 5,\!290,\!000\)

. . . \(\displaystyle =\;1000t^2 - 490,\!000t + 60,\!050,\!000\)
 
soroban said:
\(\displaystyle \text{I assume that the }xy\text{-plane is the ground, and the plane is flying above it.}\)

I knew that. :wink:



\(\displaystyle \overline{AP} \;=\;\left[(30t-300) - 7100\right]^2 + \left[(10t+500) - 2800\right]^2\)


The lefthand side should be \(\displaystyle \overline{AP} \cdot \overline{AP}\), yes?
 
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