An airplane lands with a velocity of 100 m.s. The maximum rate at which it can slow down is 5,0 m.s
1. What is the minimum time needed to come to rest, starting from when it first touches the ground?
2. Can this plane safely land at a small airport where the landing strip is 0,80 km long?
I assume you mean Vi = 100 m/s and a = -5 m/s^2
v = Vi + a t
0 = 100 -5 t
t = 20 seconds
to go from 100 to 0 at -5m/s^2
d = Vi (20) - (1/2)(5)(20)^2
d = 100 (20) - 2.5(400)
= 2000 - 1000
= 1000 meters to stop, runs 200 meters beyond the end of the runway into the harbor where he strikes a passenger ferry.
Another approach for question #2 (aside from Damon's answer) is to find the final velocity of the plane from the given 800 meters landing strip, that is, if the positive and not equal to zero, then it cannot safely land that small airport. So,
Vi= 100m/s
a= -5 m/s^2
x= 800 meters
Vf^2=Vi^2+2ax
Vf^2=(100)^2+2(-5)(800)
Vf^2=10000-8000=2000
Vf=44.72 m/s
It means that if the plane was to land the 800 landing strip, it will have a velocity of 44.72 m/s (not stopping) and cannot land the given landing strip/airport.
1. Well, if the maximum rate at which the airplane can slow down is 5 m/s, it would take 20 seconds to come to a complete stop. But let's hope there's a pilot on board who knows that this isn't ideal and they'll use the brakes more efficiently!
2. Ah, the small airport with a 0.80 km landing strip question. Sounds like it's time for the pilot to channel their inner NASCAR driver! With a velocity of 100 m/s and needing to come to a stop, it might be a bit of a tight squeeze. They might have to rely on some expert braking skills and pray that there's enough runway length to safely stop. But hey, who needs sleep, right?
To answer these questions, we need to use the equations of motion. Let's address each question one by one.
1. What is the minimum time needed to come to rest, starting from when it first touches the ground?
To find the minimum time needed to come to rest, we need to calculate the deceleration (negative acceleration) of the airplane.
Given:
Initial velocity (u) = 100 m/s
Maximum rate of deceleration (a) = -5.0 m/s² (negative because it represents deceleration)
Using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken, we can rearrange the equation to solve for time (t):
0 = 100 + (-5.0)t
Now, we can solve for t:
5.0t = 100
t = 100 / 5.0
t = 20 seconds
Therefore, the minimum time needed for the airplane to come to rest, starting from when it first touches the ground, is 20 seconds.
2. Can this plane safely land at a small airport where the landing strip is 0.80 km long?
To determine if the plane can safely land, we need to calculate the distance it will travel while decelerating until it comes to rest. If this distance is less than or equal to the length of the landing strip (0.80 km), the plane can safely land.
Using the equation v² = u² + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled, we can rearrange the equation to solve for distance (s):
0 = 100² + 2(-5.0)s
Simplifying further:
0 = 10,000 - 10s
10s = 10,000
s = 10,000 / 10
s = 1,000 meters
Since the distance traveled while decelerating is 1,000 meters, which is greater than the length of the landing strip (0.80 km or 800 meters), the plane cannot safely land at this small airport. The landing strip is not long enough to accommodate the required deceleration distance.