A blimp is traveling due North at 80km/hr with respect to the air, while the wind is blowing from northeast at 25km/hr with respect to the ground. What is the velocity of the blimp with respect to the ground? Thanks.
v= v (blimp) – v(wind) •cosα = 80 - 25cos45o = 62.32 km/hr
To find the velocity of the blimp with respect to the ground, we can combine the velocity of the blimp with respect to the air and the velocity of the wind with respect to the ground.
The blimp is traveling due North at 80 km/hr with respect to the air. This means that its velocity vector points directly northward at a magnitude of 80 km/hr.
The wind is blowing from the northeast at 25 km/hr with respect to the ground. To determine the direction of the wind vector, we can imagine a right-angled triangle with the north-south direction as the vertical side and the east-west direction as the horizontal side. The northeast direction is halfway between north and east. Since the wind is blowing from northeast to southwest (opposite direction), the wind vector points towards the southwest.
Now, we have a vector pointing northward with a magnitude of 80 km/hr and a vector pointing southwestward with a magnitude of 25 km/hr. To find the resultant vector (the velocity of the blimp with respect to the ground), we can add these two vectors using vector addition.
Using Pythagorean theorem, we can find the magnitude of the resultant vector:
Resultant vector magnitude = sqrt((80 km/hr)^2 + (25 km/hr)^2)
Calculating this, we find:
Resultant vector magnitude = sqrt(6400 km^2/hr^2 + 625 km^2/hr^2)
Resultant vector magnitude = sqrt(7025 km^2/hr^2)
Resultant vector magnitude = 83.78 km/hr (rounded to two decimal places)
Now, to find the direction of the resultant vector, we can use trigonometry. The angle between the resultant vector and the north direction can be calculated as:
Angle = arctan(25 km/hr / 80 km/hr)
Calculating this, we find:
Angle = arctan(0.3125)
Angle ≈ 17.21 degrees (rounded to two decimal places)
Hence, the velocity of the blimp with respect to the ground is approximately 83.78 km/hr at an angle of 17.21 degrees south of due west.