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 \[\frac{b}{{{t}_{2}}}-\frac{a}{{{t}_{1}}}=\frac{1}{2}f({{t}_{1}}+{{t}_{2}})\] \[\Rightarrow \] \[\frac{b{{t}_{1}}-a{{t}_{2}}}{{{t}_{1}}{{t}_{2}}}=\frac{1}{2}f({{t}_{1}}+{{t}_{2}})\] \[\Rightarrow \] \[f=\frac{2(b{{t}_{1}}-a{{t}_{2}})}{{{t}_{1}}{{t}_{2}}({{t}_{1}}+{{t}_{2}})}\]You need to login to perform this action.
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