An airplane pulls into a vertical loop, and has horizontal and vertical velocities Vx=90m/s and Vy=120m/s. If the radius of curvature of the flight path is R=900m, and the airplane has a vertical acceleration Ay=11m/s^2, determine its tangential and horizontal accelerations at this instant.
Answers:
(At= - 5m/s^2, Ax= - 23m/s^2)
To determine the tangential and horizontal accelerations of the airplane at this instant, we need to break down the given information and use certain formulas.
First, let's understand the given information:
- The horizontal velocity of the airplane is Vx = 90 m/s.
- The vertical velocity of the airplane is Vy = 120 m/s.
- The radius of curvature of the flight path is R = 900 m.
- The vertical acceleration of the airplane is Ay = 11 m/s^2.
To find the tangential acceleration (At), we can use the formula:
At = sqrt((Vx^2) + (Vy^2)) / R
Substituting the given values into the formula, we get:
At = sqrt((90^2) + (120^2)) / 900
= sqrt(8100 + 14400) / 900
= sqrt(22500) / 900
= 150 / 900
= 1/6
= 0.1667 m/s^2
Hence, the tangential acceleration (At) of the airplane at this instant is approximately 0.1667 m/s^2.
To find the horizontal acceleration (Ax), we need to use the formula for horizontal acceleration in a vertical loop:
Ax = -(Vy^2) / R
Substituting the given values into the formula, we get:
Ax = - (120^2) / 900
= - 14400 / 900
= - 16 m/s^2
Hence, the horizontal acceleration (Ax) of the airplane at this instant is -16 m/s^2.
Therefore, the tangential acceleration (At) is approximately -0.1667 m/s^2, and the horizontal acceleration (Ax) is -16 m/s^2.