A jet plane lands with a speed of 120 m/s and can decelerate uniformly at a maximum rate of 5.8 m/s2 as it comes to rest.
Can this plane land at an airport where the runway is 0.89 km long? Answer this by calculating.
Answer in units of km.
Vf^2=Vi^2+2*a*d
solve for d, remember the acceleration is -5.8m/s2
gbfgn
To determine if the plane can land on a runway, we need to calculate the distance it will take to come to a stop.
We can use the equation:
vf^2 = vi^2 + 2ad
where:
vf = final velocity (0 m/s as the plane comes to a stop)
vi = initial velocity (120 m/s)
a = acceleration (deceleration) (-5.8 m/s^2 since the plane is slowing down)
d = distance
Rearranging the equation, we have:
d = (vf^2 - vi^2) / (2a)
Substituting the given values, we have:
d = (0^2 - 120^2) / (2 * -5.8)
Simplifying this equation, we find:
d = (-14400) / (-11.6)
d ≈ 1241.38 m
The distance required for the plane to come to a stop is approximately 1241.38 m.
Now, we need to convert this distance to km. There are 1000 meters in 1 kilometer, so:
d = 1241.38 m / 1000
d ≈ 1.24138 km
Therefore, the plane requires approximately 1.24138 km to come to a stop. Since the runway length is 0.89 km, the plane can successfully land at the airport.
To find out if the plane can land on a runway that is 0.89 km long, we need to calculate the distance required for the plane to come to a stop.
First, let's convert the speed from m/s to km/h, as it will make it easier to work with. We can do this by multiplying the speed by 3.6:
120 m/s * 3.6 = 432 km/h
Now, we need to calculate the distance required to decelerate from 432 km/h to 0 km/h with a deceleration rate of 5.8 m/s^2. We can use the following equation:
v^2 = u^2 - 2as
where:
v = final velocity (0 km/h)
u = initial velocity (432 km/h)
a = acceleration/deceleration (-5.8 m/s^2)
s = distance
Plugging in the values, we can solve for s:
0^2 = (432)^2 - 2 * (-5.8) * s
0 = 186,624 - 11.6s
11.6s = 186,624
s ≈ 16,079 m
Now we need to convert this distance from meters to kilometers by dividing it by 1000:
16,079 m / 1000 = 16.079 km
Since the distance required to decelerate is approximately 16.079 km, and the runway is 0.89 km long, we can conclude that the plane can land on this runway with some extra distance to spare.