If,
[tex]d \: = \: \frac{1}{2} a {t}^{2} [/tex]
then, with 3x time t, (suppose, new distance is h)
[tex]h \: = \: \frac{1}{2} a {(3t)}^{2} [/tex]
[tex] = \frac{1}{2} a9 {t}^{2} [/tex]
[tex] = 9 \: \frac{1}{2} a{t}^{2}[/tex]
[tex] = 9d[/tex]
Therefore, new distance h will be 9 times bigger than distance d.
answer: c