Answer:
Here, \[\text{u}=0\];
\[\text{t}=\text{ 1}0\text{s}\];\[\text{a }=\text{ a}\left( \text{say}
\right)\]; \[\text{s}=\text{ }{{\text{S}}_{\text{1}}}\]
As, \[\text{s =}-\text{ ut }+\frac{1}{2}\text{a}{{\text{t}}^{\text{2}}}\]
\[{{\text{S}}_{\text{1}}}=0\times
\text{1}0+\,\frac{1}{2}\text{a }\times
\text{1}{{0}^{\text{2}}}=\text{5}0\text{a}\] ..(i)
Taking motion of particle for 10 s + 10 s = 20 s, we have
\[\left(
{{\text{S}}_{\text{1}}}+\text{ }{{\text{S}}_{\text{2}}} \right)=0\times
\text{2}0+~\frac{1}{2}\text{a }\times \text{2}{{0}^{\text{2}}}=\text{2}00\text{
a}\] ..(ii)
\[{{\text{S}}_{\text{2}}}=\left(
{{\text{S}}_{\text{1}}}+{{\text{S}}_{\text{2}}} \right){{\text{S}}_{\text{1}}}=\text{2}00\text{a}\text{5}0\text{a}=\text{15}0\text{a}\].
\[\frac{{{S}_{2}}}{{{S}_{1}}}=\frac{150a}{50a}=\text{ 3}\]or\[{{\text{S}}_{\text{2}}}=\text{3}{{\text{S}}_{\text{1}}}\]
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