[tex]d = v_{i}t + \frac{1}{2}at^{2}[/tex]
where :
d-distance
vi-initial velocity
t-time
a-acceleration
d=110m
vi=0 since it is stated it started from rest
t=5.21s
[tex]110 = 0*5.21 + \frac{1}{2}a(5.21)^{2} \\ 110 = 0 + \frac{1}{2}a(5.21)^{2} \\ 110 =\frac{1}{2}a(5.21)^{2} \\ 2*110 = a(5.21)^{2} \\ 220 = 27.1441a \\ \frac{220}{27.1441} = \frac{27.1441a}{27.1441} \\ 8.10 \frac{m}{s^{2}} = a[/tex]
