Kinematics Question 170
Question: 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 $ ?
Options:
A) 3 s
B) 8 s
C) Never
D) 4 s
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Let the particle moves toward right with velocity 6 m/s. due to retardation, after time $ t _{1} $ , it’s 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}=ut _{1}-\frac{1}{2}at _{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{D it’s\tan ce}{Velocity}=\frac{8}{2}=4s $
Total time taken by particle to return at point 0 it’s $ T=t _{0A}+t _{AB}+t _{BC}=3+1+4=8 $ s.
BETA
