Question icon
Grade 9General Physics

A particle is moving in positive x– direction with initial velocity of 10 m/s and uniform retardation such that it reaches the initial position after 10s. The distance traversed by the particle in 6 seconds is

(A) 24 m(B) 25 m(C) 26 m(D) 27 m

Profile image of Krishnansu
9 Years agoGrade 9
Answers icon

1 Answer

Profile image of Aaryan Gupta
ApprovedApproved Tutor Answer9 Years ago
It is clear from the question that the particle stops and reverts its path after 5s. Therefore, to find the retardation we use,
v=u+at  ,  i.e.  ,  0=10 – a*5 ,  which gives us a=2.
Now distance travelled in 5s => s=ut+1/2at​2  , i.e. ,  s=10*5 – 1/2*2*5​2       , which gives s=25m    
After 5s velocity becomes 0 and particle starts to move in negative x-direction with the acceleration of 2.
Therefore, distance travelled in the 6th second => s=0*1 + 1/2*2*1​2         which gives us s=1m
Hence, total distance travelled = 25+1= 26m. So, the (C) option is correct.                               ​