A) 3 s
B) 8 s
C) Never
D) 4 s
Correct Answer: B
Solution :
[b] Let the particle moves toward right with velocity 6 m/s. due to retardation, after time \[{{t}_{1}}\], its velocity becomes zero, From \[v=u-at\Rightarrow 0=6-2\times {{t}_{1}}\] \[\Rightarrow {{t}_{1}}=3\sec \] But retardation works on it for 4 sec. it means after reaching point A, direction of motion gets reversed and acceleration works on the particle for next one second. \[{{S}_{0A}}=u{{t}_{1}}-\frac{1}{2}a{{t}_{1}}^{2}=6\times 3-\frac{1}{2}(2){{(3)}^{2}}=18-9=9m\]\[{{S}_{AB}}=\frac{1}{2}\times 2\times {{(1)}^{2}}=1m\] \[\therefore {{S}_{BC}}={{S}_{0A}}-{{S}_{AB}}=9-1=8m\] Now velocity of the particle at point B in return journey \[v=0+2\times 1=2m/s\] In returen journey form B to C particle moves with constant velocity 2 m/s to cover the distance 8 m. Time taken = \[\frac{Dis\tan ce}{Velocity}=\frac{8}{2}=4s\] Total time taken by particle to return at point 0 is \[T={{t}_{0A}}+{{t}_{AB}}+{{t}_{BC}}=3+1+4=8\]s.You need to login to perform this action.
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