We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
141
2
avatar

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

Best Answer 

 #1
avatar+19724 
+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

 Jun 18, 2019
 #1
avatar+19724 
+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

31 Online Users

avatar
avatar