question_answer

A point is moving in a straight line with uniform acceleration describes \[a\] and \[b\] feet in successive intervals of \[{{t}_{1}}\] and \[{{t}_{2}}\] seconds. Then, the acceleration is

A) \[\frac{2(b{{t}_{1}}-a{{t}_{2}})}{{{t}_{1}}{{t}_{2}}({{t}_{1}}+{{t}_{2}})}\]                 

B) \[\frac{b{{t}_{2}}-a{{t}_{1}}}{{{t}_{1}}{{t}_{2}}({{t}_{1}}+{{t}_{2}})}\]

C) \[\frac{b{{t}_{2}}-a{{t}_{1}}}{2{{t}_{1}}{{t}_{2}}({{t}_{1}}+{{t}_{2}})}\]                    

D)  None of these

Correct Answer: A

Solution :

Let the uniform acceleration be \[f\] and the initial velocity be\[u\], then using     \[s=ut+\frac{1}{2}f{{t}^{2}}\]                 \[OA=a=u{{t}_{1}}+\frac{1}{2}ft_{1}^{2}\] Now, initial velocity for the distance                 \[AB=u+f{{t}_{1}}\]              \[(\because \,\,v=u+ft)\] \[\therefore \]  \[AB=b=(u+f{{t}_{1}}){{t}_{2}}+\frac{1}{2}ft_{2}^{2}\]                                                 \[\left( by\,\,s=ut+\frac{1}{2}f{{t}^{2}} \right)\] \[\Rightarrow \]               \[\frac{b}{{{t}_{2}}}=u+f{{t}_{1}}+\frac{1}{2}f{{t}_{2}}\]                 ? (ii) On subtracting Eq. (i) from Eq. (ii), we get                 \[\frac{b}{{{t}_{2}}}-\frac{a}{{{t}_{1}}}=\frac{1}{2}f({{t}_{1}}+{{t}_{2}})\] \[\Rightarrow \]               \[\frac{b{{t}_{1}}-a{{t}_{2}}}{{{t}_{1}}{{t}_{2}}}=\frac{1}{2}f({{t}_{1}}+{{t}_{2}})\] \[\Rightarrow \]               \[f=\frac{2(b{{t}_{1}}-a{{t}_{2}})}{{{t}_{1}}{{t}_{2}}({{t}_{1}}+{{t}_{2}})}\]