Question

A plane lands with a velocity of 120 m/s and accelerates at a maximum rate of -6.0 m/s/s.From the instant, the plane touches the runway, what is the minimum time needed before it can come to restCan this plane land in the school parking lot which is 800 meters long? please help with this problem

Answer 1

The minimum time needed for the plane to come to rest is 20 seconds. The plane requires a stopping distance of 1200 meters, which exceeds the 800-meter length of the school parking lot; therefore, it cannot land safely in the parking lot.

Step-by-step explanation:

The question involves calculating the minimum time for the plane to come to rest and determining if the plane can land on an 800-meter school parking lot. To solve for the minimum time needed for the plane to stop, 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, v = 0 m/s, the initial velocity is u = 120 m/s, and the acceleration is a = -6.0 m/s² (deceleration). Solving for time (t):

0 = 120 m/s + (-6.0 m/s²)(t)

∂0 ∅ t = 120 m/s / 6.0 m/s²

∂0 ∅ t = 20 seconds

It takes 20 seconds for the plane to come to rest.

To check if the plane can land in an 800-meter parking lot, we need the displacement formula: s = ut + ½ at². Plugging in the values with t = 20 seconds (calculated above), we find:

s = (120 m/s)(20 s) + ½ (-6.0 m/s²)(20 s)²

∂0 ∅ s = 2400 m - 1200 m

∂0 ∅ s = 1200 meters

The stopping distance, 1200 meters, is greater than the 800-meter length of the parking lot, meaning the plane cannot land safely there.

Answer 2

The minimum time needed for the plane to come to rest will happen if the maximum acceleration is the same throughout the landing; this means that for this situation we can look at the motion as an uniform accelerated motion and then we will be able to use the equations describing this type of motion.

For this kind of motion we know that:

`a=(v_f-v_0)/(t)`

we want the plane to stop, this means that we want the final velocity to be zero; plugging this value in the equation above and the values given by the problem we have that:

`\begin{gathered} -6=(0-120)/(t) \\ t=(-120)/(-6) \\ t=20 \end{gathered}`

Therefore the plane will need a minimum of 20 seconds to stop.

To determine the minimum distance needed for the plane to stop safely we have to remember that for this kind of motion:

`v^2_f-v^2_0=2ax`

Plugging the values we have that:

`\begin{gathered} 0^2-120^2=2(-6)x \\ -120^2=-12x \\ x_{}_{}=(-120^2)/(-12) \\ x_{}=1200 \end{gathered}`

Hence, we need at least 1200 m for the plane to safely land and therefore the plan can't land in the school parking lot.