Answer:
From second newton's equation of motion:
[tex]S = ut + \frac{1}{2} a {t}^{2} [/tex]
s » displacement or distance
u » initial speed, u = 0 ( at rest )
a » acceleration
t » time
[tex]s = (0 \times 10) + ( \frac{1}{2} \times 3 \times 10) \\ \\ s = 5 \times 3 \\ \\ { \underline{ \underline{ \: \: distance = 15 \: meters}}}[/tex]
