A) \[2m{{s}^{-2}}\]
B) \[4m{{s}^{-2}}\]
C) \[12m{{s}^{-2}}\]
D) \[14m{{s}^{-1}}\]
Correct Answer: D
Solution :
We have, \[{{V}_{1}}(t=2s)=10+2\times {{(2)}^{2}}=18m{{s}^{-1}}\] and \[{{V}_{2}}(t=5s)=10+2\times {{(5)}^{2}}=60m{{s}^{-1}}\] \[{{a}_{au}}=\frac{{{V}_{2}}-{{V}_{1}}}{{{T}_{2}}-{{t}_{1}}}=\frac{42}{3}=14m{{s}^{-1}}\]
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