question_answer

A particle is moving in a straight line and passes through a point \[O\] with a velocity of 6 \[m{{s}^{-1}}\] .The particle moves with a constant retardation of 2 \[m{{s}^{-2}}\] for 4 s and there after moves with constant velocity. How long after leaving \[O\] does the particle return to \[o\]?

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.