A plane landing on a small tropical island has just 80 m of runway on which to stop. If its initial speed is 65 m/s, what is the maximum acceleration of the plane during landing, assuming it to be constant? Answer in units of m/s 2
V^2 = Vo^2 + 2a*d = 0.
a = -(Vo^2)/2d. It will be negative.
To find the maximum acceleration of the plane during landing, we can use the equation:
\(v^2 = u^2 + 2as\)
Where:
\(v\) is the final velocity of the plane (which is 0 m/s since it stops),
\(u\) is the initial velocity of the plane (65 m/s),
\(a\) is the acceleration of the plane,
\(s\) is the distance covered (runway length of 80 m).
Rearranging the equation, we have:
\(a = \frac{{v^2 - u^2}}{{2s}}\)
Substituting the given values, we get:
\(a = \frac{{0^2 - 65^2}}{{2 \cdot 80}}\)
Calculating this expression gives:
\(a = \frac{{-4225}}{{160}}\)
Therefore, the maximum acceleration of the plane during landing is approximately -26.41 m/s². Note that the negative sign indicates deceleration.
To find the maximum acceleration of the plane during landing, we can use the equation of motion:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance.
In this case, we want to find the maximum acceleration, so we assume the final velocity is zero, since the plane needs to stop. The initial velocity u is given as 65 m/s. The distance s is given as 80 m.
Rearranging the equation, we get:
a = (v^2 - u^2) / (-2s)
Since v is zero, the equation becomes:
a = -u^2 / (2s)
Substituting the given values, we have:
a = -65^2 / (2 * 80)
Calculating this expression, we find:
a ≈ -33025 / 160
Simplifying further, we get:
a ≈ -206.41 m/s^2
Since acceleration is a vector quantity, the magnitude of the maximum acceleration would be:
|a| ≈ 206.41 m/s^2
Therefore, the maximum acceleration of the plane during landing, assuming it to be constant, is approximately 206.41 m/s^2.