Answer to Question #146691 in Mechanics | Relativity for Daga-ang, Al Francis H.

Question

The takeoff speed for this plane will be 70 m/s. Assuming this minimum acceleration, what is the minimum

allowed length for the runway?

Answer 1

Taking the minimum acceleration to be 5m/s²

Given the initial velocity, u=0m/s and final velocity to be v=70m/s.

Acceleration =5m/s²

^

If $d$ be the allowed length of the runway.

Using the formula ;$v^{2} - u^{2} =2aS$

$=(70) ^{2} - (0)^{2} =2(5)S$

4900=10S

$S=\frac{4900}{10}$

$d=490m$