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A particle is moving on a circular path of 10 m radius. At any instant of time, its speed is `5ms^(-1)` and the speed is increasing at a rate of `2ms^(-2)`. At this instant, the magnitude of the net acceleration will beA. `3.2ms^(-2)`B. `2ms^(-2)`C. `1.2ms^(-2)`D. `4.3ms^(-2)` |
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Answer» Correct Answer - A `a=sqrt(a_(t)^(2)+a_(n)^(2))` `a_(t)` = rate of change of speed =`2 ms^(-2)` `a_(n)=(v^(2))/(R)=((5)^(2))/(10)=2.5ms^(-2)` `therefore" "a=sqrt(a_(t)^(2)+a_(n)^(2))=sqrt((2)^(2)+(2.5)^(2))=3.2ms^(-2)` |
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