The displacement or distance travelled by the airplane during the given period is 526.5m
Given the data in the question;
constant acceleration; [tex]a = 3.0m/s[/tex]
Initial velocity; [tex]u = 21m/s[/tex]
Final velocity; [tex]v = 60m/s[/tex]
Displacement or distance; [tex]s = ?[/tex]
To find the displacement or distance the plane has travelled during the given period, we use the Third Equation of Motion:
[tex]v^2 = u^2 + 2as[/tex]
Where v is the final velocity, u is the initial velocity, a is the acceleration and s is the displacement.
Lets make "s" the subject of the formula
[tex]s = \frac{v^2-u^2}{2a}[/tex]
We substitute our given values into the equation
[tex]s = \frac{(60m/s)^2 - (21m/s)^2}{2 \ *\ 3m/s^2} \\\\s = \frac{(3600m^2/s^2) - (441m^2/s^2)}{6m/s^2}\\\\s = \frac{3159m^2/s^2}{6m/s^2}\\\\s = 526.5m[/tex]
Therefore, The displacement or distance travelled by the airplane during the given period is 526.5m
Learn more; https://brainly.com/question/18266310
