Question

Самолет разгняясь перед взлетом , в течение некоторого промежутка времени двигался равноускоренно. каково было при этом ускорение самолета, если за 30 с его скорость возросла от 10 м/с до 55м/с

Answer 1

A=V-Vо/t
a=45м/с/30с=1,5 м/c^2

Answer 2

Calculation of the Acceleration of the Plane

To calculate the acceleration of the plane, we can use the formula for uniformly accelerated motion:

v = u + at

Where: - v is the final velocity of the plane (55 m/s) - u is the initial velocity of the plane (10 m/s) - a is the acceleration of the plane - t is the time taken for the velocity to change (30 seconds)

Rearranging the formula, we get:

a = (v - u) / t

Substituting the given values:

a = (55 m/s - 10 m/s) / 30 s

Calculating the acceleration:

a = 45 m/s / 30 s

Simplifying:

a = 1.5 m/s²

Therefore, the acceleration of the plane was 1.5 m/s² during the time it moved with uniform acceleration before takeoff.

Answer 3

Calculation of Acceleration

To calculate the acceleration of the airplane, we can use the formula:

Acceleration (a) = (Final Velocity (v) - Initial Velocity (u)) / Time (t)

Given that the initial velocity (u) is 10 m/s, the final velocity (v) is 55 m/s, and the time (t) is 30 seconds, we can substitute these values into the formula:

Acceleration (a) = (55 m/s - 10 m/s) / 30 s

Simplifying the equation:

Acceleration (a) = 45 m/s / 30 s

Acceleration (a) = 1.5 m/s²

Therefore, the acceleration of the airplane during this period was 1.5 m/s².