A plane takes off 200 miles east of its destination with a 50mph wind from the s
ID: 3090002 • Letter: A
Question
A plane takes off 200 miles east of its destination with a 50mph wind from the south. If the plane’s velocity is 250 mphand its heading is always toward its destination, its path can bedescribed by the equation below. If the location of theinitial airport is (0,0) and the location of the destinationairport is (200, 0), y represents the perpendicular distance fromthe straight line path between initial location and the destinationlocation. When x = 150 miles, how far from the straight linepath is the plane? That is, what is y?
y=0.5x0.8-.5x1.2 where 0 is less than oreq. to x less than or eq. to 200
Explanation / Answer
. 1. y = (0.5)*(150)0.8 -(0.5)*(150)1.2 = -176.8 miles .Related Questions
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