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A particle is moving with uniform acc along a straight line abc . Its velocity at A n C are 5 n 25 . If bc= 2ab find tge ratio of the time of travel from A to b n fron B to c
Let the separation amongst An and B be x That is AB = x Speed is 5 m/sec At that point the normal speed =( (speed at point A)+(velocity at point C))/2 = (5+ 25)/2 = 15 m/sec Remove from A to B = x At that point t1 = remove/speed = x/15 Separate from B to C = 2x At that point, t2 = separate/speed t2 = 2x/15 Proportion amongst t1 and t2 = (x/15): (2x/15)
Take a line segment AC . Now take a point B in between AC such that BC= 2AB. Now, Speed=distance/time So, v1= AB/t1But v1= (v+v0)/2So, (v+v0)/2=AB/t15+25/2=AB/t115=AB/t1t1=AB/15In the same wayt2=2AB/15Now,t1/t2=AB/15÷2AB/15So, ans.--: t1:t2=2:1
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