A) \[\frac{aT}{4}\]
B) \[\frac{3aT}{2}\]
C) \[\frac{aT}{2}\]
D) \[aT\]
Correct Answer: C
Solution :
For First part, u = 0, t = T and acceleration = a \[\therefore v=0+aT=aT\]and \[{{S}_{1}}=0+\frac{1}{2}a{{T}^{2}}=\frac{1}{2}a{{T}^{2}}\] For Second part, \[u=aT,\] retardation=a1, \[v=0\] and time taken = T1 (let) \[\therefore \]\[0=u-{{a}_{1}}{{T}_{1}}\]\[\Rightarrow aT={{a}_{1}}{{T}_{1}}\] and from \[{{v}^{2}}={{u}^{2}}-2a{{S}_{2}}\] \[\Rightarrow {{S}_{2}}=\frac{{{u}^{2}}}{2{{a}_{1}}}=\frac{1}{2}\frac{{{a}^{2}}{{T}^{2}}}{{{a}_{1}}}\] \[{{S}_{2}}=\frac{1}{2}aT\times {{T}_{1}}\] \[\left( As\,\,{{a}_{1}}=\frac{aT}{{{T}_{1}}} \right)\] \[\therefore \] \[{{v}_{av}}=\frac{{{S}_{1}}+{{S}_{2}}}{T+{{T}_{1}}}=\frac{\frac{1}{2}a{{T}^{2}}+\frac{1}{2}aT\times {{T}_{1}}}{T+{{T}_{1}}}\] \[=\frac{\frac{1}{2}aT\ (T+{{T}_{1}})}{T+{{T}_{1}}}\] \[=\frac{1}{2}aT\]
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