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

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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1. 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. 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 - 8) = 0 t = {0,8} In 8 hours...
3. 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. 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.
5. 8hrs.