A jet plane lands with a speed of 110 m/s and can accelerate with a maximum magnitude of 7.00 m/s2 as it comes to rest.
(a) From the instant the plane touches the runway, what is the minimum time interval needed before it can come to rest?
(b) Can this plane land on a small tropical island airport where the runway is 0.800 km long?
(c) What is the minimum distance the plane requires to land?
a) Well, I guess it depends on how impatient the passengers are! But let's do some math. We know the initial velocity is 110 m/s and the acceleration is -7.00 m/s² (negative because it's decelerating to come to rest). To find the time needed, we can use the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
So, substituting the values, we have 0 = 110 + (-7)t. Solving for t, we get t = 110/7 = 15.71 seconds.
b) Ah, so you want to know if the runway is long enough for this jazzy jet, huh? Well, let's see. The runway length is given as 0.800 km, which we can convert to meters by multiplying by 1000. So we have 0.800 km * 1000 m/km = 800 m. Since the plane needs 15.71 seconds to come to rest, we can find the distance covered during this time using the formula d = ut + (1/2)at², where d is the distance, u is the initial velocity, t is the time, and a is the acceleration.
Substituting the values, we have d = 110 * 15.71 + (1/2)(-7)(15.71)². Plugging this into a calculator, we find d ≈ 1727.5 m.
Oops! Looks like this little tropical island airport won't be able to accommodate this jet. It needs a minimum length of approximately 1727.5 meters to land safely.
c) The minimum distance the plane requires to land is approximately 1727.5 meters, as calculated above. So make sure you find an airport with a runway that can handle it, unless you want your jet to belly flop into the ocean!
To find the answers to these questions, we'll use the equations of motion.
First, let's note down the given information:
Initial speed of the plane (u) = 110 m/s
Maximum acceleration (a) = 7.00 m/s^2
(a) To find the minimum time interval needed for the plane to come to rest, we can use the equation of motion:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Since the plane comes to rest, its final velocity (v) will be 0. Substituting the values into the equation, we have:
0 = 110 + 7t
Rearranging the equation to solve for t, we get:
7t = -110
t = -110/7
t ≈ -15.71 s
Since time cannot be negative, we discard the negative value. Therefore, the minimum time interval needed for the plane to come to rest is approximately 15.71 seconds.
(b) To determine if the plane can land on a 0.800 km long runway, we need to find the distance covered by the plane as it comes to rest.
We can use the equation of motion:
v^2 = u^2 + 2as
where s is the distance covered.
Since the final velocity is 0, we have:
0 = (110)^2 + 2(7)s
Simplifying the equation, we get:
0 = 12100 + 14s
Rearranging the equation to solve for s, we have:
14s = -12100
s = -12100/14
s ≈ -864.29 m
Again, since distances cannot be negative, we disregard the negative value. Therefore, the minimum distance required for the plane to land is approximately 864.29 meters.
(c) Since the runway on the small tropical island airport is 0.800 km (or 800 meters) long, we compare this with the minimum distance required for the plane to land.
The minimum distance required for the plane to land is approximately 864.29 meters, which is greater than the length of the runway (800 meters). Therefore, the plane would not be able to land on the small tropical island airport with a runway of 0.800 km length.
To answer these questions, we will use the equations of motion and kinematics. Let's break down each part of the problem:
(a) From the instant the plane touches the runway, what is the minimum time interval needed before it can come to rest?
To determine the minimum time interval, we need to find the time it takes for the plane to decelerate from its initial speed to zero. We can use the equation of motion:
v = u + at
where:
v = final velocity (0 m/s, since the plane comes to rest)
u = initial velocity (110 m/s)
a = acceleration (-7.00 m/s^2, because it's decelerating)
Re-arranging the equation, we have:
t = (v - u) / a
Substituting the given values into the equation:
t = (0 - 110) / -7.00
Simplifying, we find:
t = 15.71 seconds
Therefore, the minimum time interval needed for the plane to come to rest is approximately 15.71 seconds.
(b) Can this plane land on a small tropical island airport where the runway is 0.800 km long?
To determine whether the plane can land on the small tropical island airport, we need to compare the minimum distance it requires to come to rest with the length of the runway.
The distance required for an object to come to rest can be calculated using the equation:
s = (u^2 - v^2) / (2a)
where:
s = distance
u = initial velocity (110 m/s)
v = final velocity (0 m/s)
a = acceleration (-7.00 m/s^2)
Substituting the given values into the equation:
s = (110^2 - 0^2) / (2 * -7.00)
Simplifying, we get:
s = 850 m
Since the length of the runway is given as 0.800 km (800 m), we can see that the plane requires a distance of 850 m to come to rest. Therefore, the runway is not long enough for the plane to land without overshooting.
(c) What is the minimum distance the plane requires to land?
Using the same equation as before:
s = (110^2 - 0^2) / (2 * -7.00)
Simplifying, we find:
s = 850 m
Therefore, the minimum distance the plane requires to land is 850 meters.