(1). An Aeroplane leaves an airport, flies due North for 2 hrs at 500km/h. It then flies on a bearing of 053 degree at 300km/h for 90 mins. How far is the plane from the airport correct to 1 dp
(2). Find the bearing of the plane from the Airport correct to 1 dp.
30 points
(3). If it takes the Aeroplane 150 mins to fly back to the airport. Find the average speed of the aeroplane for the whole flight correct to 1dp.
(1) you fly on a heading, not a bearing. The law of cosines is the way to find this value. Since distance = speed*time,
z^2 = 1000^2 + 450^2 - 2(1000)(450)cos53°
(2) Here, "bearing" is correctly used. It is NθE where
tanθ = (450sin53°) / (1000+450cos53°)
(3) avg speed = distance/time
= (1000+450+z)/(2 + 1.5 + 2.5)
I think you first answer is wrong because it supposed to be 1320.7 km.
And also your solution seems to be wrong
Who ever responds first and correctly gets $100
To solve these questions, we can use the concept of vectors and trigonometry.
(1) The first step is to convert the speeds and times into distances. We know that speed = distance/time. So, the first leg of the journey (due North for 2 hours at 500 km/h) will cover a distance of 2 hours * 500 km/h = 1000 km.
Next, we need to calculate the distance traveled during the second leg of the journey. We can use the formula distance = speed * time. The speed is 300 km/h and the time is 90 minutes = 1.5 hours. So, the distance traveled during the second leg is 1.5 hours * 300 km/h = 450 km.
To find the total distance from the airport, we need to use vector addition. We have a right-angled triangle formed by the 1000 km distance traveled due North and the 450 km distance traveled at a bearing of 053 degrees. We can use the Pythagorean theorem to find the hypotenuse (total distance from the airport).
Using the Pythagorean theorem: c^2 = a^2 + b^2
where c is the hypotenuse, a is the distance due North, and b is the distance at the bearing.
c^2 = 1000^2 + 450^2
c^2 = 1,000,000 + 202,500
c^2 = 1,202,500
c = √1,202,500
c ≈ 1097.43 km (correct to 1 decimal place)
So, the plane is approximately 1097.4 km from the airport.
(2) To find the bearing of the plane from the airport, we can use trigonometry.
We have a right-angled triangle formed by the 1000 km distance due North and the 450 km distance at a bearing of 053 degrees. We can use the inverse tangent (arctan) function to find the angle.
tan(angle) = opposite/adjacent
tan(angle) = 450/1000
angle = arctan(450/1000)
angle ≈ 24.4 degrees (correct to 1 decimal place)
However, the question asks for the bearing, which is measured from the North direction in a clockwise manner. Since the plane is flying on a bearing of 053 degrees (clockwise from North), we need to add this angle to the angle we just calculated.
Bearing = 053 degrees + 24.4 degrees
Bearing ≈ 77.4 degrees (correct to 1 decimal place)
So, the bearing of the plane from the airport is approximately 77.4 degrees.
(3) To find the average speed of the plane for the whole flight, we need to find the total distance traveled and divide it by the total time taken.
We already calculated the total distance as approximately 1097.4 km. The total time taken is the sum of the times for each leg of the journey: 2 hours + 1.5 hours + 150 minutes (which is equivalent to 2.5 hours).
Total time = 2 hours + 1.5 hours + 2.5 hours
Total time = 6 hours
Average speed = total distance / total time
Average speed = 1097.4 km / 6 hours
Average speed ≈ 182.9 km/h (correct to 1 decimal place)
So, the average speed of the airplane for the whole flight is approximately 182.9 km/h.