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`        appointment with uniform acceleration and V1 V2 and V3 denote the average Lost Season 3 of the interval of time T1 T2 and T3 which of the following relation is correct`
2 years ago

Sanju
106 Points
```							Lets assume that the particles move distances AB, BC and CD during the time t1, t2 and t3 respectively..Velocity at B = u+at1 Vel. at C =  u+a(t1 + t2)Vel. at D =  u+a(t1 + t2 + t3)Avg velocity = [u+v] / 2Therefore, v1 = [u + (u+at1 )] / 2 = u+½ at1v2 = [(u+at1 ) + (u+a(t1 + t2))] / 2 = u+at1+½ at2 v3 = [ (u+a(t1 + t2) +  (u+a(t1 + t2 + t3)] / 2 = u+at1+at2+½ at3  Finally we arrive at:v1-v2 / v2-v3 = t1+t2 / t2+t3Ans: Option B
```
2 years ago
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