a piolet wishes to fly to a city 1.20*10^3 km/h(N30E) of her present position .There is a wind of 72km/h(E),what direction (heading) should she fly and how long the trip takes?
To determine the heading she should fly and the time it will take, we need to consider the combined effect of the pilot's desired heading and the wind.
Let's break down the given information:
- The pilot wants to fly to a city located 1.20 * 10^3 km away to the North 30 degrees East (N30E) of her present position.
- There is a wind blowing at 72 km/h to the East.
First, we need to determine the actual direction the plane needs to head in order to account for both the desired destination and the wind. To do this, we'll calculate the resultant (combined) vector of the desired heading and the wind vector.
1. Calculate the components of the desired heading vector:
- North component = 1.20 * 10^3 * sin(30°)
- East component = 1.20 * 10^3 * cos(30°)
2. Calculate the components of the wind vector:
- North component = 0 (no wind in the north direction)
- East component = 72 km/h
3. Add the corresponding components together to get the resultant vector:
- North component (resultant) = desired heading (north component) + wind (north component)
- East component (resultant) = desired heading (east component) + wind (east component)
4. Calculate the magnitude and direction of the resultant vector:
- Magnitude = sqrt(North component^2 + East component^2)
- Direction = arctan(East component / North component)
The resultant vector will give us the heading the pilot should follow to account for the wind. Now, let's calculate the values:
North component = 1.20 * 10^3 * sin(30°) = 1.20 * 10^3 * 0.5 = 600 km
East component = 1.20 * 10^3 * cos(30°) = 1.20 * 10^3 * 0.866 = 1.039 * 10^3 km
Wind North component = 0
Wind East component = 72 km/h
Resultant North component = 600 km + 0 = 600 km
Resultant East component = 1.039 * 10^3 km + 72 km/h = 1.039 * 10^3 km
Magnitude = sqrt(600^2 + 1.039 * 10^3^2) = sqrt(360,000 + 1.080,321) ≈ sqrt(1,440,321) ≈ 1,199.77 km
Direction = arctan(1.039 * 10^3 km / 600 km) ≈ 59.82°
Therefore, the pilot should fly at a heading of approximately N30°W (north 30 degrees west) and it will take approximately 1,199.77 km/hour to complete the trip.