Ajit and Bipin start travelling at the same time from the same point and in the same direction. Ajit travels u

Feb 05

sameerteacher01 asked:


Ajit and Bipin start travelling at the same time from the same point and in the same direction. Ajit travels uniformly at 10 km/hr, while Bipin starting at 8 km/hr increases his speed by 0.5 km/hr after each hour. In how many hours will they be together again?

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4 comments

1. 

   np_rt / February 8, 2009

   Ajit: d = 10t
   Bipin: d = 8t + 0.25t^2 (a=0.5 km/hr^2)

   10t = 8t+0.25t^2
   0 = 0.25t^2-2t
   0 = t^2 – 8t
   0 = t(t-8)

   t = 0 (start of race)
   t = 8 hours (correct answer)

2. 

   MaX / February 10, 2009

   va = 10
   vb = 8 + 0.5t
   da = 10t
   db = 8t + 0.25t^2
   da = db
   10t = 8t + 0.25t^2
   8t + 0.25t^2 -10t = 0
   0.25t^2 -2t = 0
   t^2 – 8t = 0
   t(t – = 0
   t = {0,8}
   In 8 hours…

3. 

   iyiogrenci / February 12, 2009

   10 t =8+8.5+9+9.5+10+10.5+11+11.5+12=90
   t=9

   hint:
   you may use the sum formula for aritmetic progression.
   difference is 0.5
   the first term is 8

4. 

   kalyanr / February 14, 2009

   In Bipin’s reference frame, Ajit is initially travelling at a velocity of 2 km/hr in a straight line. His velocity decreases by 0.5 km/hr as he moves forward, assuming that the velocity change occurs only after 1 hr interval of time.Then the direction of his relative speed changes.From symmetry Ajit will meet Bipin when his relative velocity is again 2km/hr in the opposite direction to his original motion.
   So, a difference of 2*2km/hr occurs in a time of (4km/hr)/(0.5km/hr/hr) = 8 hr.

   So they will meet again after 8 hr.