Question

Suppose a particular jet needs to attain a speed of 220 mph to take off. If it can accelerate from 0 to 220 mph in 30 seconds, how long must the runway be? Assume constant acceleration. Note: 1 mph = 22/15 ft/sec.

Answer 1

so your acceleration, or a, is: a=(220mph-0)/30s but you have to change hours into seconds so 220mph=220/3600=.06111 so a=.06111/30=.002037mi/s^2

the time it takes to reach this spead is given by Vf = Vi +at or final velocity equals initial velocity plus acceleration times time

so (.06111mi/s-0)/.002037mi/s^2=t=30 seconds.

so the distance is x=Vit + .5at^2
or x=0+(.5)(.002037)(30)^2 = .91655 miles which equals 4839.9 feet

Answer 2

4839.9 feet is the length of the runaway

Answer 3

What D:< lol

Answer 4

and someone's homework was done without that person learning at all

Answer 5

ok heres a better solution for him

you are given:
initial speed, u = 0
final speed, v = 220
acceleration = (v-u)/t ===> t= time taken for the plane to reach final velocity(in hours)

then you use the formula:
displacement, s = u.t + (1/2).a.t^2