A) 21 m
B) 24 m
C) 27 m
D) 30 m
Correct Answer: A
Solution :
a) In \[v=u+at\] or \[v=u-at,\] if \[\left| \,a\, \right|\] is doubled, the time t is halved. |
Hence, if t is the time for acceleration, then \[\frac{t}{2}\] is the time taken for retarded motion/ |
\[\therefore \] total time duration for travels is\[t+t+\frac{t}{2}=\frac{5t}{2}\] |
\[\frac{5t}{2}=5s(given)\] |
\[\therefore t=2s\] |
(i) Distance covered in accelerated motion |
\[{{S}_{1}}=ut+\frac{1}{2}a{{t}^{2}}\] |
\[=0\,\,\times \,\,t+\frac{1}{2}\,\,\times \,\,3\,\,\times \,\,{{2}^{2}}=6m\] |
(ii) Distance covered in uniform motion |
\[v=u+at=0+3\times 2=6\,m{{s}^{-1}}\] |
\[\therefore {{S}_{2}}=vt=6\times 2=12\,m\] |
(iii) Distance travelled in retarded motion |
\[{{S}_{3}}=ut-\frac{1}{2}\,\,a{{t}^{2}}\] |
\[=6\times 1-\frac{1}{2}\,\,\times \,\,6\,\,\times \,\,{{1}^{2}}=3m\] |
\[\therefore \]total distance traversed |
\[S={{S}_{1}}+{{S}_{2}}+{{S}_{3}}=6+12+3=21\,\,m\] |
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