hey you keep asking the same thing !
a = 11.5
v = 11.5 t + Vi but Vi = 0 so
v = 11.5 t
x = 0 + (11.5/2) t^2
so
655 = 5.75 t^2
solve for t = time
v = 11.5 t
An airplane starts from rest and accelerates at 11.5m/s^2. What is its speed at the end of a 655m runway?
4 answers
Sorry , I'm new user to jiskha
LOL
well for constant acceleration = a
v = Vi + a t
x = Xi + Vi t + (1/2) a t^2
on earth due to gravity a = -9.8 m/s^2 approximately
well for constant acceleration = a
v = Vi + a t
x = Xi + Vi t + (1/2) a t^2
on earth due to gravity a = -9.8 m/s^2 approximately
when you get to projectiles
same equations for vertical problem
if angle up at start = A and speed = s
Vi = s sin A at start
and v = Vi - 9.8 t etc
H = Hi + Vi t - 4.9 t^2
and u = s cos A forever (no horizontal force)
same equations for vertical problem
if angle up at start = A and speed = s
Vi = s sin A at start
and v = Vi - 9.8 t etc
H = Hi + Vi t - 4.9 t^2
and u = s cos A forever (no horizontal force)
1 answer