Physics - deceleration?

Question

A jetliner touches down at 200 km/h, reverses its engines to provide braking, and comes to a halt 35 s later. What is the shortest runway on which this aircraft can land, assuming constant deceleration starting at touchdown?

= m

Answer 1

v = u + at

0 (halt) = u + a *t

a = - u/t

v^2 = u^2 + 2 a s

0 = u^2 - 2[u/t] s

u^2 = 2us/t

s = u * t/2  

s = [200*5/18] * [35/2] = 972.22 meter

973 m long runway at least

Answer 2

V = U + AT

200km/h = 200x1000/3600 = 55.55m/s

AT  = 0-55.55

A = -55.55 / 35 = -1.587ms2(neg for deceleration)

d = UT + 1/2 AT(squared)

d = 55.55 x 35 + 0.5 x (-1.587) x 35 x 35

d = 1944.25 + (-972.04)

d = 972.21m

973 m long runway at least