Answer:
Distance travelled by A in5th second,
\[{{S}_{A}}=0+\frac{{{a}_{1}}}{2}(2\times 5-1)=\frac{9}{2}{{a}_{1}}\] ….(1)
Distance travelled by B in 3rd second,
\[{{S}_{B}}=0+\frac{{{a}_{2}}}{2}(2\times 3-1)=\frac{5}{2}{{a}_{2}}\]
Given,\[{{S}_{A}}={{S}_{B}}\]
\[\frac{9}{2}{{a}_{a}}=\frac{5}{2}{{a}_{2}}\,\,or\,\,\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{5}{9}\]
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