Question

A body 'A' starts moving with a uniform acceleration of 4 m/s^2 from a place .Another body 'B' starts from the same place 4 seconds later and moves in the same direction with uniform acceleration of 16 m/s^2.Where and when will 'B' meet 'A'?

Answer 1

(a)  The displacement of a particle at time t is given s = ut + 1/2at2At time (t - 1), the displacement of a particle is given byS' = u (t-1) + 1/2a(t-1)2 So, Displacement in the last 1 second is,St = S - S'= ut + 1/2 at2 – [u(t-1)+1/2 a(t-1)2 ]= ut + 1/2at2 - ut + u - 1/2a(t - 1)2= 1/2at2 + u - 1/2 a (t+1-2t) =  1/2at2 + u - 1/2at2 - a/2 + atS = u + a/2(2t - 1)(b) Putting the values of u = 2 m/s, a = 1 m/s2 and t = 5 sec, we getS = 2 + 1/2(2 x 5 - 1) = 2 + 1/2 x 9= 2 + 4.5 = 6.5 m