Distance formula word problem?

[DasRabbit]

You're flying from Joint Base Lewis-McChord (JBLM) to an undisclosed location 39 km south and 222 km east. Mt. Rainier is located approximately 56 km east and 40 km south of JBLM. If you are flying at a constant speed of 800 km/hr, how long after you depart JBLM will you be the closest to Mt. Rainier?

______minutes?

I got 16, but I'm wrong. Can someone help me?

 

[mmm4444bot]

Welcome to the boards. We are a tutoring site. Please show your work, or at least explain how you set up the problem and then post your equation. Cheers :cool:

 

[DasRabbit]

We choose our coordinate system so the +x direction is east, and the -y direction is south. JBLM is at the origin (0, 0). All distance units are kilometers.

You are flying on line
.. 39x + 222y = 0, which can be reduced to
.. 13x + 74y = 0

The slope of the line ax+by=c is given by -a/b, which suggests we can form the equation of a perpendicular line by swapping x and y coefficients and negating one of them. Thus
.. 74x - 13y = c ... is perpendicular to the flight path. This can also be written as
.. 74x - 13y - c = 0 ... we can find c so that this is true at the coordinates of Mt. Rainier.

The perpendicular line through (56, -40) is
.. 74x - 13y - (74(56) - 13(-40)) = 0
.. 74x - 13y - 4664 = 0
The distance in km from JBLM (0, 0) to this line is given by
.. |74(0) - 13(0) - 4664|/√(74^2+13^2) = 4664/√5645 ≈ 62.076

At 800 km/hr, you will have flown that 62.076 km in 62.076/800 hr ≈ 4 min 39 sec
_____
Your total travel time is 16 minutes, 54 seconds. Your closest approach is 29.7 km.

 

[soroban]

Hello, DasRabbit!

I'll outline the game plan I used.

You're flying from Joint Base Lewis-McChord (J) to a location (L) 39 km south and 222 km east.
Mt. Rainier (R) is located approximately 56 km east and 40 km south of JBLM.
If you are flying at a constant speed of 800 km/hr, how long after you depart J will you be the closest to R?

This is my diagram.

      |J
(0,39)o
      |   *
      |       *
      |           *    P
      |               o
      |                   *
      |              *        *
      |                           *   L
  - - + - - - - - - * - - - - - - - - o - -
      |                           (222,0) 
      |            o R
      |         (56,-1)
      |

We want to locate point \(\displaystyle P\) so that \(\displaystyle RP \perp JL.\)

Line \(\displaystyle JL\) has equation: .\(\displaystyle y \:=\:\text{-}\frac{74}{13}x + 39\) .[1]

Line \(\displaystyle RP\) contains \(\displaystyle R(56,\text{-}1)\) and has slope \(\displaystyle \frac{13}{74}\)
. . Its equation is: .\(\displaystyle y \:=\:\frac{13}{74}x - \frac{401}{37}\) .[2]

Equate [1] and [2], and solve for \(\displaystyle x.\)
Substitute into [1] or [2], and solve for \(\displaystyle y.\)
. . We have located point \(\displaystyle P.\)

Determine the distance \(\displaystyle JP.\)
Divide by 800 and get the time in hours.
Multiply by 60 to get the time in minutes.