Question

A small propeller airplane can comfortably achieve a high enough speed to take off on a runway that is 1/4 mile long. A large, fully loaded passenger jet has about the same acceleration from rest, but it needs to achieve twice the speed to take off.What is the minimum runway length that will serve? Hint: You can solve this problem using ratios without having any additional information.

Answer 1

Answer:

1 mile

Explanation:

We can use the following equation of motion to solve for this problem:

$v^2 - v_0^2 = 2a\Delta s$

where v m/s is the final take-off velocity of the airplane, $v_0 = 0$ initial velocity of the can when it starts from rest, a is the acceleration of the airplanes, which are the same, and $\Delta s$ is the distance traveled before takeoff, which is minimum runway length:

$v^2 - 0^2 = 2a\Delta s$

$\Delta s = \frac{v^2}{2a}$

From here we can calculate the distance ratio

$\frac{\Delta s_1}{\Delta s_2} = \frac{v_1^2/2a_1}{v_2^2/2a_2}$

$\frac{\Delta s_1}{\Delta s_2} = \left(\frac{v_1}{v_2}\right)^2\frac{a_2}{a_1}$

Since the 2nd airplane has the same acceleration but twice the velocity

$\frac{\Delta s_1}{\Delta s_2} = 0.5^2* 1$

$\Delta s_2 = 4 \Delta s_1 = 4*(1/4) = 1 mile$

So the minimum runway length is 1 mile