a jet airplane lands with a speed of 122 m/s and can accelerate uniformly at a max rate of - 6.2m/s2 as it comes to a rest. can this airplane land in 1.0 km runway

average speed during stop = 122/2 = 61 m/s

to get time to stop
v = 0 = 122 - 6.2 t
t = 122/6.2 = 19.7 seconds
so
distance = v average * t = 61*19.7 = 1200 meters

Off the end of the runway and 200 meters into the bleak ocean

To find out if the airplane can land in a 1.0 km runway, we need to determine the distance it takes for the airplane to come to a rest.

We can use the equation of motion:

V^2 = U^2 + 2as

Where:
V = final velocity (0 m/s, as the airplane comes to a rest)
U = initial velocity (122 m/s)
a = acceleration (-6.2 m/s^2)
s = distance

Rearranging the equation to solve for distance:

s = (V^2 - U^2) / (2a)

Substituting the given values:

s = (0^2 - 122^2) / (2 * -6.2)

s = (-14884) / (-12.4)

s = 1200 meters

So, the airplane requires 1200 meters to come to a rest. Since the given runway is 1.0 km or 1000 meters long, the airplane will not be able to land safely within the given runway.

To determine whether the jet airplane can land in a 1.0 km runway, we can use the equations of motion.

First, we need to find the time it takes for the airplane to come to a rest. We can use the equation:

v = u + at

where:
- v is the final velocity (0 m/s, since the airplane comes to a rest)
- u is the initial velocity (122 m/s)
- a is the acceleration (-6.2 m/s^2, with a negative sign because it's deceleration in this case)
- t is the time it takes to come to a rest

Rearranging the equation, we have:

t = (v - u) / a

Plugging in the values:

t = (0 - 122) / -6.2
t = 19.68 seconds

Now we know it takes 19.68 seconds for the airplane to come to a rest.

Next, we need to calculate the distance the airplane travels during this time. We can use the equation:

s = ut + (1/2)at^2

where:
- s is the distance traveled
- u is the initial velocity (122 m/s)
- t is the time (19.68 s)
- a is the acceleration (-6.2 m/s^2)

Plugging in the values:

s = (122 * 19.68) + (1/2) * (-6.2) * (19.68^2)
s = 1205.536 m

Therefore, the airplane would require 1205.536 meters to come to a rest. Since the runway has a length of 1.0 km (which is equal to 1000 meters), the airplane cannot land within this distance.