Question

Самолет отрывается от земли при скорости 216км в час.с каким ускорением он должен двигаться по взлетной полосе длина которой 1,2км

Answer 1

В этой задаче необходимо использовать 2формулы кинематики)одна с S другая с t..Итак:

v=216км/ч=*1000/3600=60м/с

a=?м/с

S=1.2км=1200м

S=(a*t^2)/2 т к V0=0м/с

a=2S/t^2

t=(v-v0)/a

a=v^2/2S=1,5 м/с^2

Ответ:с ускорением 1,5 =)

Answer 2

Calculating Acceleration for Takeoff

To calculate the acceleration required for the plane to take off, we can use the following kinematic equation:

v^2 = u^2 + 2as

Where: - v = final velocity (which is 216 km/h or 60 m/s, as 1 km/h = 1/3.6 m/s) - u = initial velocity (which is 0 m/s as the plane starts from rest) - a = acceleration - s = distance (which is 1.2 km or 1200 m)

First, let's convert the final velocity to m/s: - 216 km/h = 216 * (1000 m / 1 km) / (3600 s / 1 h) = 60 m/s

Now, we can rearrange the equation to solve for acceleration:

a = (v^2 - u^2) / (2s)

Substitute the values: a = (60^2 - 0) / (2 * 1200) a = 3600 / 2400 a = 1.5 m/s^2

So, the plane must accelerate at 1.5 m/s^2 along the runway to take off at a speed of 216 km/h.