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Home » Forum » Mechanics » help neededGoutham 1995 GummA particle has an initial velocity of 9m/s due east and a constant accelaration of 2m/sec ^2 due west. the distance covered by the particle in the 5th second of its motion is?a) 0b)0.5c)2d) none of the aboveHINT- answer is 0.5 but i keep getting 0. help
since there is retardation,check velcoit ehen it becomes zero.hence v = 9 – 2*t = 0 at t = 4.5 seconds the particle stops hence the distnace would be b/w the 4th and 4.5th second for that mini 0.5 seconds u need to find the trvaelled distance.hence,at first 4 seconds it traversed,x = 9*4 – 0.5*2*16 = 36 – 16 = 20 meter.v = 9 – 2*4 = 9 – 8 = 1 m/sthus,in last 4.5 seconds,x = 9*4.5 – 0.5*2*4.5^2 = 81/2 – 81/4 = >81/4 = > 20.25 meterhence total distance20.25 – 20 = 0.25 meter is the answer. for first half secondsfor full 1th second => .25*2 = 0.50 meter.
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