An aeroplane starts from rest with an acceleration of 3 ms-2 and takes a run for 35s before taking off. What is the minimum length of the runway with what velocity plane take off?
vf=1/2 a t^2=1/2 * 3*35^2 m/s
and
vf^2=2a*Length
or lengthrunway=vf^2/6 m =
To find the minimum length of the runway and the velocity of the plane during takeoff, we can use the equations of motion.
First, let's find the final velocity of the plane during takeoff. We can use the equation:
v = u + at
where:
v - final velocity
u - initial velocity (which is 0, as the plane starts from rest)
a - acceleration
t - time
Plugging in the values, we have:
v = 0 + (3 m/s^2) * 35 s
v = 105 m/s
Therefore, the final velocity of the plane during takeoff is 105 m/s.
Now, let's find the minimum length of the runway. We can use the equation:
s = ut + (1/2)at^2
where:
s - distance
u - initial velocity (0)
t - time
a - acceleration
Plugging in the values, we have:
s = (0 * 35 s) + (1/2)(3 m/s^2)(35 s)^2
s = (1/2)(3 m/s^2)(1225 s^2)
s = 1837.5 m
Therefore, the minimum length of the runway is 1837.5 meters.