275 km/hr
275 km/hr
At full speed an airplane can travel 3500 miles with the wind in 5 hours but it requires 7 hours to travel the same distance against the wind?
Let the speed of the airplane in still air be represented by V.Velocity = Distance ÷ Time, thus Distance = Velocity x Time.D = 4(V + 20) = 5(V - 20), then 4V + 80 = 5V - 100 : V = 180 kphThe speed in still air is 180kph
To calculate the average speed of the airplane, you divide the total distance traveled by the total time taken. In this case, the airplane flies 1000 km in 4 hours. Therefore, the average speed is 1000 km ÷ 4 hours = 250 km/h.
Air speed or ground speed?
275 km/hr
317
At full speed an airplane can travel 3500 miles with the wind in 5 hours but it requires 7 hours to travel the same distance against the wind?
275 km/hr
Let the speed of the airplane in still air be represented by V.Velocity = Distance ÷ Time, thus Distance = Velocity x Time.D = 4(V + 20) = 5(V - 20), then 4V + 80 = 5V - 100 : V = 180 kphThe speed in still air is 180kph
The average speed if an airplane travels 1364 miles in 5.5 hours is 248 miles/hr.
To calculate the average speed of the airplane, you divide the total distance traveled by the total time taken. In this case, the airplane flies 1000 km in 4 hours. Therefore, the average speed is 1000 km ÷ 4 hours = 250 km/h.
Air speed or ground speed?
To find out how far an airplane can fly in 3 hours at an average speed of 800 km/h, you can use the formula: distance = speed × time. By substituting the values, the calculation is: distance = 800 km/h × 3 hours = 2400 km. Therefore, the airplane can fly 2400 kilometers in 3 hours.
u said airspeed remains the same, so 840 mi in 3 hrs the groundspeed is 280, for the return trip groundspeed is 240, therfore the wind speed is 40
The airplane travels 720 miles to the city at a speed of 295 mph, taking approximately 2.44 hours (720 mi ÷ 295 mph). The total round trip time is 5 hours, so the return flight must take about 2.56 hours (5 hrs - 2.44 hrs). If we let the wind speed be ( w ), the return speed becomes ( 295 - w ). Thus, the equation for the return flight is ( 720 = (295 - w) \times 2.56 ). Solving for ( w ) gives a wind speed of approximately 34.84 mph.
191.05882 mph