PLEASE HELP ME! it's a physics problem again, due in an hour!!!?

Question

A plane landing on a small tropical island has just 62 m of runway on which to stop. If its initial speed is 55 m/s, what is the maximum acceleration of the plane during landing, assuming it to be constant? Answer in units of m/s.

Answer 1

v^2 = u^2 + 2as

0 = 55^2 + 2a*62

0 = 3025 + 124a

a = -3025/124 = -24.4 m/s^2

v = u + at

0 = 55 - 24.4 * t

55 = 24.4 * t

t = 55/24.4 = 2.25 s

Ans: Acceleration = -24.4 m/s^2, time = 2.25 s

Answer 2

problems where you are given distances, initial speeds and you are asked to find accelerations are often best addressed with the equation:

vf^2=v0^2+2ad where vf is the final speed (here it is zero); v0 is the initial speed (here it is 55 m/s), a is the acceleration (which we need to find) and d is the distance traveled (here 62 m)

our equation becomes:

0=55^2+2xax62

a=-55^2/(2x62)=-24.4m/s/s

the minus sign indicating the acceleration is negative, in other words, the plane is slowing down

acceleration is change in speed/time, so if the plane changes its speed by 55 m/s at a rate of 24.4m/s/s, then the time in which this occurs is

55m/s/24.4m/s/s = 2.25s  

Answer 3

acceleration as a function of dist.:

(x1-x2)a = (V1^2-V2^2)/2

x1 = 0

x2 = 62m

V1 = 55 m/s

V2 = 0

plug n play, solve for a

to find time, v = at

v = V1

solve for t