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Grade 9Mechanics

The distance travelled by a body in the nth second is given by the expression (2+3n). Find the initial velocity and acceleration . Also find its velocity at end of 2 seconds.

Profile image of Razz kumar
6 Years agoGrade 9
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1 Answer

Profile image of Arun
ApprovedApproved Tutor Answer6 Years ago
we know,
   Sn = u + (2n-1)a/2
where Sn is distance traveled in nth second.
         u is initial velocity 
         a is acceleration.
given,
  Sn = 2 + 3n
        = 2 + 3(2n-1+1)/2
        =2 +3/2 + (2n-1)3/2 
        = 7/2 + (2n-1)3/2
compare above equation to this equation 
  initial velocity = 7/2 m/s
acceleration = 3 m/s^2
now, velocity after 2sec = u + at 
                           = 7/2 + 3*2 = 3.5 + 6 = 9.5 m/s