Calculating Position and Average Velocity of a Plane with Relative Motion

[sb]

A small air plane is traveling at a velocity relative to the air of 100km/h [E]. A wind from the north is blowing at 20km/h.
(a) Determine the position of the airplane after 2.5 h of this motion.
(b) Find the average velocity of the airplave during the 2.5 h interval.

Thanks a Lot...

 

[HallsofIvy]

Draw a picture: Draw a vertical line (from north to south) of length 20 (whatever units you want) representing the velocity vector of the air relative to the ground. Draw a horizontal line to the right (east) from the tip of the previous line of length 100 representing the velocity vector of the airplane relative to the air.

The "velocity vector" of the airplane relative to the ground is the hypotenuse of the right triangle you have just drawn. You can use the Pythagorean theorem to find the length of that: the speed of the airplane relative to the ground. You can use tan(theta)= 20/100 to find theta, the angle the airplane flies relative to north.

Multiply by 2.5 to find the position after 2 and 1/2 hours.

 

[M98Ranger]

(a) To determine the position of the airplane after 2.5 hours, we need to first calculate the distance traveled by the airplane in that time. We know that the airplane is traveling at a velocity of 100km/h [E] relative to the air, so in 2.5 hours, it will have traveled a distance of 250 km [100km/h x 2.5h = 250 km].

Next, we need to take into account the wind blowing from the north at a velocity of 20km/h. This wind will also affect the position of the airplane. Since the wind is perpendicular to the direction of the airplane's velocity, it will not affect its speed but will change its direction. We can use the Pythagorean theorem to calculate the distance traveled by the wind in 2.5 hours:
Distance = √(20km/h x 2.5h)^2 = √(50km)^2 = 50 km

Therefore, the position of the airplane after 2.5 hours will be 250km [E] and 50km [N] from its starting point.

(b) To find the average velocity of the airplane during the 2.5 hour interval, we need to calculate the total displacement of the airplane. We know that the airplane has traveled 250km [E] and 50km [N], so the total displacement is √(250km)^2 + (50km)^2 = √(62500km^2 + 2500km^2) = √65000km = 254.95 km

The average velocity is calculated by dividing the total displacement by the time taken:
Average Velocity = 254.95 km / 2.5 h = 101.98 km/h [E]

Therefore, the average velocity of the airplane during this 2.5 hour interval is 101.98 km/h [E].