Question

1. 0 / 1 points an airplane maintains a speed of 663 km/h relative to the air it is flying through as it makes a trip to a city 815 km away to the north. (assume north is the positive y - direction and east is the positive x - direction.) (a) what time interval is required for the trip if the plane flies through a headwind blowing at 34.2 km/h toward the south? (b) what time interval is required if there is a tailwind with the same speed? (c) what time interval is required if there is a crosswind blowing at 34.2 km/h to the east relative to the ground?

Explanation:

Step1: Recall the speed - distance - time formula

The formula is $t=\frac{d}{v}$, where $t$ is time, $d$ is distance and $v$ is speed.

Step2: Calculate time without wind

Given $d = 815$ km and $v=663$ km/h. Using the formula $t=\frac{d}{v}$, we have $t_1=\frac{815}{663}\approx1.23$ h.

Step3: Calculate time with head - wind

The effective speed with a head - wind of $34.2$ km/h is $v_2=663 - 34.2=628.8$ km/h. Then $t_2=\frac{815}{628.8}\approx1.30$ h.

Step4: Calculate time with tail - wind

The effective speed with a tail - wind of $34.2$ km/h is $v_3=663 + 34.2=697.2$ km/h. Then $t_3=\frac{815}{697.2}\approx1.17$ h.

Step5: Calculate time with cross - wind

The cross - wind does not affect the north - south speed of the plane. The speed of the plane relative to the ground in the north - south direction is still $v = 663$ km/h. So $t_4=\frac{815}{663}\approx1.23$ h.

Answer:

(a) Approximately $1.30$ h
(b) Approximately $1.17$ h
(c) Approximately $1.23$ h