Question

Самолет заходит на посадку со скоростью 200 км ч и до полной остановки пробегает 1 км по полосе.определите ускорение самолета во время посадки и время через которое он останавливается.РЕШИТЬ ЗАДАЧУ ЧЕРЕЗ КИНЕМАТИКУ

Answer 1

{S=V0*t-(a*t^2)/2   (1)
{V1=V0-a*t           (2)
где S-путь торможения равный 1000м
V0-скорость перед посадкой 200к/ч=55.5м/с
V1-скорость в момент остановки равна V1=0

из (2) 
0=V0-a*t   => V0=a*t   =>   t=V0/a
поставим значение t в уравнение (1)

S=V0^2/a - V0^2/2*a=(2V0^2-V0^2)/2a=V0^2/2a 
a=V0^2/S=3080/2000 =1.54 м/с^2

тогда t=V0/a=55.5/1.54=36c

Answer 2

Problem Analysis

We are given that an airplane is landing with a speed of 200 km/h and travels 1 km before coming to a complete stop. We need to determine the acceleration of the airplane during landing and the time it takes for the airplane to come to a stop.

Solution

To solve this problem, we can use the equations of motion from kinematics. The equation that relates initial velocity (v0), final velocity (v), acceleration (a), and displacement (s) is:

v^2 = v0^2 + 2as

We are given the initial velocity (v0 = 200 km/h) and the displacement (s = 1 km). We need to find the acceleration (a) and the time it takes for the airplane to come to a stop.

Let's convert the given values to SI units (m/s) for consistency:

Initial velocity (v0) = 200 km/h = (200 * 1000) / 3600 m/s = 55.56 m/s Displacement (s) = 1 km = 1000 m

Now we can substitute the values into the equation and solve for acceleration (a):

v^2 = v0^2 + 2as

0^2 = (55.56)^2 + 2a(1000)

Simplifying the equation:

0 = 3086.0736 + 2000a

-3086.0736 = 2000a

a = -3086.0736 / 2000

a ≈ -1.543 m/s^2

The negative sign indicates that the acceleration is in the opposite direction of the initial velocity, which makes sense since the airplane is decelerating during landing.

To find the time it takes for the airplane to come to a stop, we can use the equation:

v = v0 + at

Since the final velocity (v) is 0, we can solve for time (t):

0 = 55.56 + (-1.543)t

-55.56 = -1.543t

t = -55.56 / -1.543

t ≈ 36 seconds

Therefore, the acceleration of the airplane during landing is approximately -1.543 m/s^2 and it takes approximately 36 seconds for the airplane to come to a stop.