a plane travels 1.5 km at angle of 38 degrees to the ground, then changes direction...

a plane travels 1.5 km at angle of 38 degrees to the ground, then changes direction and travels 6.9 km at an angle of 11 degrees to the ground. what is the magnitiude of the plane's total displacement? answer in units of km. What angle above the horizontal if the plane's total displacement? Answer in units of degrees

Solutions

Expert Solution

d1 = 1.5 km, d2 = 6.9 km, q1 = 38 deg, and q2 = 11 deg.

Use the formula cos q = x / d and plug in: cos (38 deg) = x1 / 1.5 km, so x1 = (1.5 km)(cos (38 deg)) = 1.18201613041 km, and plug in: cos (11 deg) = x2 / 6.9 km, so x2 = (6.9 km)(cos (11 deg)) = 6.77322756579 km.

Use the formula sin q = y / d and plug in: sin (38 deg) = y1 / 1.5 km, so y1 = (1.5 km)(sin (38 deg)) = 0.92349221298 km, and plug in: sin (11 deg) = y2 / 6.9 km, so y2 = (6.9 km)(sin (11 deg)) = 1.3165820681 km.

Add Vector components, so x = x1 + x2 = 1.18201613041 km + 6.77322756579 km = 7.9552436962 km and y = y1 + y2 = 0.92349221298 km + 1.3165820681 km = 2.24007428108 km.

Use the formula Dd = (Dx^2+Dy^2)^0.5 and plug in Dd = ((7.9552436962 km )^2 + (2.24007428108 km)^2)^0.5 = 8.26461342415 km = 8.3 km. Use the formula q = arctan ( y / x ) and plug in:

q = arctan (2.24007428108 km / 7.9552436962 km) = 15.7264032 deg = 15.73 deg.

Yielding 8.3 km at 15.73 deg to the horizontal.

genius_generous answered 1 year ago

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