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A particle starts from rest and has acceleration of 2 ms-2 for 10 s. After that it travels for 30 s. with constant speed and then undergoes a constant retardation of 4 ms and comes back to rest. The total distance covered by the particle is: |
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Answer» 650 m `v=u+at` `v_1=0+2xx10 =20 ms^(-1)` Also `s_1=(V_(1)^(2)-u_(1)^(2))/(2A)=(20xx20-0)/(2xx2)=100m` In SECOND case ,`v_(2)`be the final vel.Which remains constant `s_(2)=v_(2)xxt(2)` `=20xx30=600 m` In third case `v_(3)=0,u_(3)=20 ms^(-1)` `a=-4 ms^(-2)` Now `t=(v_(3)-u_(3))/(a)=(-20)/(-4)` TOTAL DISTANCE =`S_1+S_2+S_3`=100+600+50=750 m |
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