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A particle moving with constant acceleration of 2m/s² due west has an initial velocity of 9m/s due east . Find the distance covered in the fifth second of its motion |
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Answer» Here the PARTICLE velocity will become ZERO in 4.5 seconds, asv=u+at0=9−2×tt=4.5sec Displacement in the same PERIOD will bev 2 −u 2 =2as−81=2ss 4.5 =20.25m Let us call the point of velocity being zero as Point A.now body will start moving BACKWARDS,displacement in NEXT 0.5 sec from point A,s=ut+ 21 at 2 s 0.5 = 21 (−2)(0.25) 2 =−0.25m hence distance covered is 0.25 m(-ve sign indicates body moved in backward direction. It is to be noted that displacement and distance are not to be confused to be same)Distance traveled in 4 secondss 4 =9×4+ 21 (−2)(4) 2 s 4 =36−16=20 Thus,distance covered in 5th second isD=(s 4.5 +∣s 0.5 ∣−s 4 )=20.5+0.25−20=0.5 |
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