Answer:

Since the distance traveled is S =\frac {V _ 1 ^ 2-V _ 0 ^ 2} {2\times a}S=

2×a

V

1

2

−V

0

2

, where V_1V

1

is the final speed, V_0V

0

is the initial, aa is the acceleration, then S =\frac {72 ^ 2-54 ^ 2} {2\times 2} = 567S=

2×2

72

2

−54

2

=567 m;

Since by another formula the distance is S = V _ 0\times t +\frac {a\times t ^ 2} {2}S=V

0

×t+

2

a×t

2

, we will express from this equation tt :

\frac {a\times t^2}{2}+V_0\times t - S=0

2

a×t

2

+V

0

×t−S=0 ,

\frac {2\times t^2}{2}+54\times t - 567=0

2

2×t

2

+54×t−567=0 ,

t=9t=9 s.