How do I solve this physics problem dealing with 1-dimensional motion with constant acceleration?

distance = 0.5 * A * T^2

Velocity = A * T

===>

100 m/s = 5.00 m/s^2 * T

solve for T

800 = 0.5 * 5.0 * T (from above) ===> Min Landing distance if less than 800 meters you are golden  greater than  800 meters you are SOL to land on the island

QED

MINIMUM TIME NEEDED TO COME TO REST - 20 SECONDS

RUNWAY LENGHT REQUIRED - 1050 m i.e 1.05km

thus the plane cannot land and come to rest in 0.8KM when its approach speed is 100m/s.

given

initial velocity of the plane , U = 100 m/s
acceleration, a = -5.00 m/s^2
since tthe plane comes to rest after landing,its final velocity is , V = 0 m/s

use the equation V x V = U X U - 2 X a X S

where S is the distance the plane will cover after landing to come to complete rest

substituting the above values in the equation we get

 0 x 0 = 100 x 100 - 2 x 5 x S
 0 = 10000 - 10 x S
 - 10000 = -10 x S

 S = ( -10000) / ( - 10 )
 S = 1000 meters

 so this plane cannot land on a  small tropical island airport where the runway is 0.800 km long as the plane requires a minimum of 1 km runway length to come to a halt.

trust this has clarifiend your doubt