Calculus: A plane reaches flight at velocity of 200mph can acel 200mph / 30sec what is min. dist of runway?

Question

Calculus: A plane reaches flight at velocity of 200mph can acel 200mph / 30sec what is min. dist of runway?

Answer 1

The motion of the plane on the runway is uniformly accelerated from rest:

s = (1/2) at^2

and its velocity :

v = at  

----> v^2 = a^2 t^2 = 2as

----> s = v^2 / 2a

when

v = 200 mph,  and

a = 200 mph / 30sec = 200 x 3600 / 30 mph/h

s = 200^2 / (2 x 200 x 3600 / 30) = (200 x 200 x 30) / (2 x 200 x 3600) = 0.83 mile

Answer 2

There is no Calculus involved here. Simply use this formula

Vf^2 - Vo^2 = 2as

where

Vf = final velocity of plane = 200 mph

Vo = initial velocity of plane = 0

a = acceleration = 200 mph in 30 sec.= 24000 m/hr^2

s = runway distance

Substituting appropriate values,

(200)^2 - 0 = 2(24000)(s)

Solving for "s",

s = 40000/(2 * 24000)

s = 0.83 mile (minimum distance of runway required given the conditions of the problem).