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A particle moves along a straight line according to the law `s=16-2t+3t^(3)`, where `s` metres is the distance of the particle from a fixed point at the end of `t` second. The acceleration of the particle at the end of `2s` isA. `3.6m//s^(2)`B. `36m//s^(2)`C. `36km//s^(2)`D. `360 m//s^(2)` |
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Answer» Correct Answer - B Given, `s=16-2t+3t^(3)` `implies(ds)/(dt)=-2 +9t^(2)` `implies(d^(2)S)/(dt^(2))=18t` Now, the acceleratio of the particle at the end of `t=2` s is `f=(d^(2)s)/(dt^(2))=18xx2=36m//s^(2)` |
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