An airplane starts from rest and accelerates at 11.5m/s^2. What is its speed at the end of a 655m runway?
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
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
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)
To find the speed of the airplane at the end of the runway, we can use the kinematic equation:
v^2 = u^2 + 2as
Where:
v is the final velocity
u is the initial velocity (which is 0 since the airplane starts from rest)
a is the acceleration
s is the displacement
In this case:
u = 0 (starting from rest)
a = 11.5 m/s^2 (acceleration)
s = 655 m (displacement)
Plugging these values into the equation, we can solve for v:
v^2 = 0^2 + 2 * 11.5 * 655
v^2 = 0 + 15,053
v^2 = 15,053
To find v, we need to take the square root of both sides:
v = √15,053
v ≈ 122.7 m/s
Therefore, the speed of the airplane at the end of the 655m runway is approximately 122.7 m/s.