A steady wind blows a kite due to west. The kite height above ground from horizontal position x=0 to x=25 given by y=50-0.1(x-15)^2. Find the distance travel by the kite?

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2 Answers
Nathan L.
Jul 4, 2017

#"Distance" = 47# #"m"#

Explanation:

To solve this problem, we can plug in #25# in for #x# in the #y#-position equation to find the #y#-position then:

#y = 50 - 0.1((25)-15)^2 = color(red)(40# #color(red)("m"#

The #x#-position here is #25# #"m"# (what we plugged in)

The distance it traveled was

#r = sqrt(x^2 + y^2) = sqrt((25color(white)(l)"m")^2 + (color(red)(40)color(white)(l)color(red)("m"))^2) = color(blue)(47# #color(blue)("m"#

rounded to #2# significant figures.

A08
Jul 27, 2017

I got a different answer!

Explanation:

Given for horizontal position #x=0" to "x=25#
Height above ground is
#y=50-0.1(x-15)^2#

  1. For #x_1=0#
    #y_1=50-0.1(0-15)^2#
    #=>y_1=27.5#
  2. For #x_2=25#
    #y_2=50-0.1(25-15)^2#
    #=>y_2=40#

The kite moves from location #(0,27.5) " to "(25,40)#
Distance traveled by kite between two points #(x_1, y_1) and (x_2, y_2)#
#d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)#
#d=sqrt((25-0)^2+(40-27.5)^2)#
#d=sqrt(625+156.25)#
#d=sqrt(781.25) =27.95#, rounded to two decimal places.

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