A) \[15m/{{s}^{2}}\]
B) \[~16m/{{s}^{2}}\]
C) \[17\text{ }m/{{s}^{2}}\]
D) \[20\text{ }m/{{s}^{2}}\]
Correct Answer: C
Solution :
Distance described in a particular second \[s=u+\frac{1}{2}\,f\,(2t-1)\] According to the question \[100=u+\frac{1}{2}f.\,9\] \[\Rightarrow \] \[100-\frac{9}{2}f=u\] ?.. (i) and \[151=u+\frac{1}{2}\,f.\,15\] \[\Rightarrow \] \[151-\frac{15}{2}\,f=u\] ?.. (ii) From Eqs. (i) and (ii), \[100-\frac{9}{2}\,f=151-\frac{15}{2}f\] \[\Rightarrow \] \[\left( \frac{15}{2}-\frac{9}{2} \right)\,f=51\] \[f=\frac{51\times 2}{6}\] \[\Rightarrow \] \[f=17m/{{s}^{2}}\]
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