I HAVE AN EXAM REVIEW QUESTION AND I REALLY NEED HELP.
An airplane has an air speed of 330km/h and is heading in a direction of N30E when it encounters a wind from the north. The resultant ground velocity has a direction of N40E. How long does it take for the plane to travel 960km.
Did you make a diagram?
My vector diagram has angles 10°, 30°, and 140° with the longest sides as 330.
so the side across the 30° represents the ground velocity of the plane, let it be x
By the sine law
x/sin30 = 330/sin140
x = 256.69
so the actual groundspeed is 269.69 km/h, then to go 960 km would take
960/269.69 hours or 3.74 hours
or 3 hours and 44 minutes and 23.5 seconds
To solve this problem, we can break down the airplane's velocity into two components: the air speed component and the wind velocity component. Let's assume the wind velocity is W km/h.
Given:
Air speed of airplane = 330 km/h
Heading direction = N30E
Wind velocity = W km/h
Ground velocity = N40E
Distance to travel = 960 km
Step 1: Determine the horizontal component of the airplane's air speed.
The horizontal component of the air speed is given by the formula:
horizontal component = air speed × cosine(angle)
The angle N30E can be broken down into a north component and an east component. The north component is N30W, and the east component is E30N.
Since we are interested in the east component, use the E30N angle.
So, the horizontal component of the air speed is:
horizontal component = 330 km/h × cosine(30°)
Step 2: Determine the horizontal component of the wind velocity.
The horizontal component of the wind velocity is given by the formula:
horizontal component = wind velocity × cosine(180° - angle)
Since the wind is from the north, the angle is 90°.
So, the horizontal component of the wind velocity is:
horizontal component = W km/h × cosine(180° - 90°)
Step 3: Find the resultant horizontal component.
The resultant horizontal component is the sum of the horizontal components of the air speed and the wind velocity.
resultant horizontal component = horizontal component of air speed + horizontal component of wind velocity
Step 4: Calculate the time taken to travel.
The time taken to travel can be found by dividing the distance to travel by the resultant horizontal component.
time taken = distance to travel / resultant horizontal component
Now, let's calculate the values:
Step 1: Determine the horizontal component of the airplane's air speed.
horizontal component = 330 km/h × cosine(30°)
Step 2: Determine the horizontal component of the wind velocity.
horizontal component = W km/h × cosine(180° - 90°)
Step 3: Find the resultant horizontal component.
resultant horizontal component = horizontal component of air speed + horizontal component of wind velocity
Step 4: Calculate the time taken to travel.
time taken = 960 km / resultant horizontal component
Please provide the wind velocity (W) so that I can proceed with the calculations.
To solve this problem, we can break it down into two components: the motion of the airplane in still air and the effect of the wind on the airplane's motion.
First, let's determine the airplane's ground speed in still air. We can use the concept of vector addition to find this velocity.
The airplane has an airspeed of 330 km/h in a direction N30E. We can think of this as a vector with a magnitude of 330 km/h and an angle of 30 degrees measured from the north direction towards the east.
Using trigonometry, we can break down this vector into its northward and eastward components. The northward component is 330 * sin(30°) and the eastward component is 330 * cos(30°).
The northward component is given by:
Northward component = 330 km/h * sin(30°) = 165 km/h
The eastward component is given by:
Eastward component = 330 km/h * cos(30°) = 285 km/h
Now, let's consider the effect of the wind. The wind is coming from the north, so it will have a northward velocity component. Let's call this component "w".
The resultant ground velocity has a direction of N40E. This tells us that the airplane's motion due to the wind alone has a northward component equal to "w" and an eastward component equal to "285 km/h" (from the airplane's velocity in still air).
Using trigonometry again, we can determine the value of "w". The northward component of the resultant ground velocity is given by:
Northward component = w
Since the direction is N40E, the angle from the north direction towards the east is 40 degrees. Therefore, the eastward component of the resultant ground velocity is:
Eastward component = 285 km/h
We can now use trigonometry to find the magnitude of the wind velocity "w". The magnitude of the resultant ground velocity can be found using the Pythagorean theorem:
Resultant ground velocity = sqrt((Northward component)^2 + (Eastward component)^2)
Since the resultant ground velocity is known to be 330 km/h, we can use the above equation to find the magnitude of the wind velocity "w".
Now that we know the magnitude of the wind velocity "w", we can find its direction relative to the north direction. This can be determined using trigonometry as well.
With the northward and eastward components of the wind velocity known, we can now calculate the time it takes for the airplane to travel 960 km.
Time = Distance / Ground speed
Since the ground speed is the resultant velocity (330 km/h), we can substitute the given values into the equation to find the time it takes for the plane to travel 960 km.
I hope this breakdown helps you solve the problem step by step.