What is the magnitude of the acceleration, in \' g \' units (where 1 g = 9.8 m/s
ID: 582347 • Letter: W
Question
What is the magnitude of the acceleration, in 'g' units (where 1 g = 9.8 m/s2), of a pilot whose aircraft enters a horizontal circular turn with a velocity of v0 = (400 x-direction + 500 y-direction) m/sand 24.0 s later leaves the turn with a velocity of vf = (400 x-direction 500 y-direction) m/s. What is the magnitude of the acceleration, in 'g' units (where 1 g = 9.8 m/s2), of a pilot whose aircraft enters a horizontal circular turn with a velocity of v0 = (400 x-direction + 500 y-direction) m/sand 24.0 s later leaves the turn with a velocity of vf = (400 x-direction 500 y-direction) m/s.Explanation / Answer
We assume the turn is made with uniform circular motion. Then the pilot’s acceleration is centripetal and has
magnitude a given by a = V^2/R
where R is the circle’s radius
time required to complete a full circle is the period given by T = 2*PI*R/v
Because we do not know radius R, we solve for R
=>R=T*V/(2*PI)
a=V^2/R =V^2/(T*V/(2*PI)) =2*PI*V/(T)
Speed v is the (constant)
magnitude of the velocity during the turning V=sqrt(Vx^2+Vy^2) =sqrt(400^2+500^2) =640.3m/s
As Vf =-Vi => Aircraft completed half revolution in t=24s =>T=48s
=>a =2*PI*640.3/(48) =83.815m^2/s =8.54g m/s^2
Related Questions
Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.