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?

Guest Jun 18, 2019

#1**+2 **

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

#1**+2 **

Best Answer

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