Answer:
Let a be constant acceleration and
\[S=ut+\frac{1}{2}a{{t}^{2}},\]then \[{{S}_{1}}=0+\frac{1}{2}a\times 100=50a\]
Velocity after 10s, is \[v=0+10a\]
So, \[{{S}_{2}}=10a\times 10+\frac{1}{2}a\times 100=50a\]
\[\Rightarrow {{S}_{2}}=3{{S}_{1}}\]
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