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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. `5"ms"^(-2)`B. `2"ms"^(-2)`C. `3.2"ms"^(-2)`D. `4.3"ms"^(-2)` |
Answer» Correct Answer - C `"Here", r=10m, v=5ms^(-1), a_(t)=2m^(-2)` `a_(r)=(v^(2))/(r)=(5xx5)/(10)=2.5ms^(-2)` The net acceleration is `a=sqrt(a_(r)^(2)+a_(t)^(2))+sqrt((2.5)^(2)+2^(2))=sqrt(10.25)=3.2ms^(-2)` |
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