A plane flying at 70.0 m/s suddenly stalls with an acceleration of 9.8 m/s² directed downward. If the stall lasts 5.0 seconds, we can analyze the situation as follows:
- The acceleration of 9.8 m/s² downward corresponds to the acceleration due to gravity, indicating the plane is in free fall or losing lift during the stall.
- Given the initial velocity v0=70.0 m/sv_0=70.0,m/sv0=70.0m/s and constant downward acceleration a=9.8 m/s2a=9.8,m/s^2a=9.8m/s2, the velocity change during the stall can be calculated using the equation:
v=v0−atv=v_0-atv=v0−at
where t=5.0 st=5.0,st=5.0s.
- Substituting values:
v=70.0 m/s−(9.8 m/s2)(5.0 s)=70.0 m/s−49.0 m/s=21.0 m/sv=70.0,m/s-(9.8,m/s^2)(5.0,s)=70.0,m/s-49.0,m/s=21.0,m/sv=70.0m/s−(9.8m/s2)(5.0s)=70.0m/s−49.0m/s=21.0m/s
So after 5 seconds of stalling with downward acceleration, the plane's speed decreases to 21.0 m/s. This reflects the loss of lift due to the stall, where the wing exceeds its critical angle of attack, causing airflow separation and a sharp reduction in lift
. The stall condition means the plane cannot maintain level flight, and gravity causes it to accelerate downward at 9.8 m/s² until recovery or impact
. In summary:
- Initial speed: 70.0 m/s
- Downward acceleration during stall: 9.8 m/s²
- Stall duration: 5.0 s
- Speed after stall: 21.0 m/s downward
This simplified analysis assumes constant acceleration due to gravity and neglects other aerodynamic forces during the stall phase.
