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.



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक