Yet another AP Physics question?

Question

An airplane starting from rest reaches its takeoff velocity of 202 mi/h over a runway of 6000 ft. How long did this take if the plane rolled with a constant acceleration?

I got 20.25 seconds, but I'm too irritated to explain why. Thanks for the help.

Answer 1

d= 6000 ft

vf= 200 mi/h( 1h/3600 s) ( 5280 ft/ 1 mile)= 293.33 ft/s

find t

vf^2= vi^2 +2ad

since vi= 0

vf^2=  2a d

a= vf^2/  2d

a= (293.33 ft/s)^2/ 2( 6000 ft

a= 7.17 ft/ s^2

t= vf/a

t= 293.33 ft/s/  7.17 ft/s^2

t= 40.9 sec or 41 sec

Answer 2

Starting from rest we have for velocity and distance:

v = at

s = 1/2(at^2)

at = 202 mi/hr = 296.267 ft/sec

a = 296.267/t

1/2(at^2) = 6000 ft

1/2(296.267/t)t^2 = 6000

t = 6000/148.333 = 40.504 seconds <<--- answer

(all numbers rounded)

The rest of the problem:

a = 296.267/40.504 =  7.3145 ft/sec^2

Check 1/2(at^2) = 1/2(7.3145)(40.504^2) = 6000.0