Answer:
1 mile
Explanation:
We can use the following equation of motion to solve for this problem:
[tex]v^2 - v_0^2 = 2a\Delta s[/tex]
where v m/s is the final take-off velocity of the airplane, [tex]v_0 = 0[/tex] initial velocity of the can when it starts from rest, a is the acceleration of the airplanes, which are the same, and [tex]\Delta s[/tex] is the distance traveled before takeoff, which is minimum runway length:
[tex]v^2 - 0^2 = 2a\Delta s[/tex]
[tex]\Delta s = \frac{v^2}{2a}[/tex]
From here we can calculate the distance ratio
[tex]\frac{\Delta s_1}{\Delta s_2} = \frac{v_1^2/2a_1}{v_2^2/2a_2}[/tex]
[tex]\frac{\Delta s_1}{\Delta s_2} = \left(\frac{v_1}{v_2}\right)^2\frac{a_2}{a_1}[/tex]
Since the 2nd airplane has the same acceleration but twice the velocity
[tex]\frac{\Delta s_1}{\Delta s_2} = 0.5^2* 1[/tex]
[tex]\Delta s_2 = 4 \Delta s_1 = 4*(1/4) = 1 mile[/tex]
So the minimum runway length is 1 mile
