Answer:
Let\[{{t}_{1}}\]and\[{{t}_{2}}\]be the times of acceleration and retardation respectively.
Acceleration part:
\[u=0,\,v=8m/s,\,\,a=+a,\,\,t={{t}_{1}}\]
We know,\[v=u+at\]
\[\Rightarrow 8=a{{t}_{1}}\,\,or\,\,{{t}_{1}}=\frac{8}{a}\] ……(1)
Retardation part:
\[u=8m/s,\,\,v=0,\,\,a=-a,\,\,t={{t}_{2}}=4-{{t}_{1}}\]
We know,\[v=u+at\]
(or) \[0=8-a(4-{{t}_{1}})\]
\[\therefore 8=a\left( 4-\frac{8}{a} \right)\] [from (1)]
\[8=4a-8\,\,or\,\,a=4\,\,and\,\,{{t}_{1}}=8/4=2s\]
Now, \[{{s}_{1}}=0\times 2+\frac{1}{2}\times 4{{(2)}^{2}}\,\,or\,\,{{s}_{1}}=8m\]
\[{{s}_{2}}=8\times 2+\frac{1}{2}\times 4{{(2)}^{2}}\,\,or\,\,{{s}_{2}}=8m\]
\[\therefore {{s}_{1}}+{{s}_{2}}=16m.\]
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