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# help

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A blimp flew at 45 mph in the direction of 160 degrees. After two hours it changes course and flew 40 mph for two and a half hours in the direction of 305 degrees. How far was it from the starting point at the end of the four and a half hours and what was the direction from the start?

Jun 18, 2019

#1
+23681
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Add the x and the y compnets of the two vectors

x's : 45(2) cos 160   + 40(2.5) cos 305   =  -27.21 miles

y's :  45(2.0) sin 160  + 40(2.5) sin 305 =  -51.13 miles

The resultant magnitude is the distance from the starting point......the angle is the tangent y/x

resultant magnitude = sqrt (27.21^2 + 51.13^2) = 57.92 miles

arctangent   -51.13/-27.21 =61.98 degrees    (reference angle)

not that both x and y are negative....so the angle is in the third quadrant    180 + 61.98 = 241.98 degrees

Jun 18, 2019

#1
+23681
0

Add the x and the y compnets of the two vectors

x's : 45(2) cos 160   + 40(2.5) cos 305   =  -27.21 miles

y's :  45(2.0) sin 160  + 40(2.5) sin 305 =  -51.13 miles

The resultant magnitude is the distance from the starting point......the angle is the tangent y/x

resultant magnitude = sqrt (27.21^2 + 51.13^2) = 57.92 miles

arctangent   -51.13/-27.21 =61.98 degrees    (reference angle)

not that both x and y are negative....so the angle is in the third quadrant    180 + 61.98 = 241.98 degrees

ElectricPavlov Jun 18, 2019