Question

An airline route from San Francisco to Honolulu is S37E. A jet flying at 450 mph on that bearing runs into a wind blowing at 39 mph from a direction of 114 degrees. Find the resulting groundspeed and direction of the jet.

Answer 1

The resulting groundspeed and direction of the jet are approximately 475.53 mph and 50.89°, respectively.

Step-by-step explanation:

To find the resulting groundspeed and direction of the jet, we need to break down the velocity components. The velocity of the jet can be broken down into two components: the northward component and the eastward component. The northward component is given by the groundspeed multiplied by the sine of the angle (37°) with respect to the north. The eastward component is given by the groundspeed multiplied by the cosine of the angle.

So, the northward component of the jet's velocity is 450 mph * sin(37°) ≈ 270.68 mph and the eastward component is 450 mph * cos(37°) ≈ 359.17 mph.

To find the resulting groundspeed, we need to subtract the wind speed from the jet's velocity components. The northward component of the wind is 39 mph * sin(114°) ≈ -33.86 mph and the eastward component is 39 mph * cos(114°) ≈ -15.69 mph.

Therefore, the resulting northward groundspeed is 270.68 mph - (-33.86 mph) = 304.54 mph and the eastward groundspeed is 359.17 mph - (-15.69 mph) ≈ 373.86 mph.

The resulting groundspeed is the magnitude of the resultant vector, which can be found using the Pythagorean theorem as follows:

Resulting groundspeed = √(northward groundspeed^2 + eastward groundspeed^2) ≈ √(304.54^2 + 373.86^2) ≈ 475.53 mph.

The direction of the resulting groundspeed can be found using the inverse tangent function:

Direction = arctan(eastward groundspeed / northward groundspeed) ≈ arctan(373.86 / 304.54) ≈ 50.89°.