A jet plane lands with a speed of 110 m/s and
can decelerate uniformly at a maximum rate
of 4.1 m/s2 as it comes to rest.
Can this plane land at an airport where
the runway is 0.80 km long? Answer this by
calculating.
Answer in units of km
Since the question is yes/no, units don't matter much.
The first step is to see how long it takes to decelerate from 110 to 0 at -4.1/sec: 110/4.1 = 26.829 sec
Now, the distance traveled is 1/2 at^2 = 1/2 * 4.1 * 26.829^2 = 1475.58 m
Looks like it's gonna end up in the mud.
Or maybe in the terminal.
To determine if the plane can land in a runway that is 0.80 km long, we need to find out if the distance required for the plane to stop is greater or smaller than the length of the runway.
Let's calculate the distance required for the plane to stop using the formula:
distance = (initial velocity)^2 / (2 * acceleration)
Given:
Initial velocity (v) = 110 m/s
Deceleration (a) = -4.1 m/s^2 (negative sign indicates deceleration)
First, let's convert the initial velocity to km/s:
110 m/s * 1 km/1000 m = 0.11 km/s
Next, let's calculate the distance required for the plane to stop:
distance = (0.11 km/s)^2 / (2 * -4.1 m/s^2)
distance = 0.0121 km / -8.2 m/s^2
distance ≈ -0.0015 km
The calculated distance is negative, indicating that the plane requires a smaller distance to stop than the length of the runway. Therefore, the plane can safely land on the 0.80 km long runway.
To determine if the plane can land on a runway that is 0.80 km long, we need to calculate the distance it will take for the plane to come to a complete stop.
We are given the initial speed of the plane (110 m/s) and the maximum deceleration rate (-4.1 m/s^2). The plane will decelerate until its final speed is 0 m/s.
We can use the following kinematic equation to find the distance traveled during deceleration:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s)
u = initial velocity (110 m/s)
a = acceleration (deceleration rate -4.1 m/s^2)
s = distance traveled
Rearranging the equation, we have:
s = (v^2 - u^2) / (2a)
Substituting the given values:
s = (0 - (110)^2) / (2 * -4.1)
s = -12100 / -8.2
s ≈ 1475.61 m
Now, we need to convert this distance from meters to kilometers by dividing it by 1000:
s ≈ 1475.61 m / 1000 = 1.47561 km
Therefore, the plane needs approximately 1.48 km to come to a complete stop. Since the runway is only 0.80 km long, it is too short for the plane to land.