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A particle is moving with velocity `v=4t^(3)+3 t^(2)-1 m//s`. The displacement of particle in time `t=1 s` to `t=2 s` will beA. 21 mB. 17 mC. 13 mD. 9 m |
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Answer» Correct Answer - A `v=(ds)/(dt)=4t^(3)+3t^(2)-1` `int_(0)^(s) ds=int_(1)^(2)(4t^(3)+3t^(2)-1)dt` `s=|t^(4)+t^(3)-t|_(1)^(2)` =`{(2)^(4)+(2)^(3)-(2)}-{(1)^(4)+(1)^(3)-(1)}` `=22-1=21 m` |
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