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A particle moves in a circle of radius 2...

A particle moves in a circle of radius 20 cm. Its linear speed is given by `v = (3t^(2) +5t)` where t is in second and v is in m/s. Find the resultant acceleration at t = 1s.

Text Solution

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Tangential acceleration `a_(t) =(dv)/(dt) =(d)/(dt) 3(2+5t) 6t+5`
At `ts: a_(t) =6xx1+5=11" m/s"^(2) and" velocity "v=3t^(2)+5t=3xx1+5xx1=8" m/s"`
? Radial acceleration `a_(r)=(v^(2))/(r ) =((8)^(2))/(0.2) =(64)/(0.2) =320" m/s"^(2)`
? Resultant acceleration `a= sqrt((a_(t))^(2)+(a_(t))^(2))=sqrt((11)^(2)+(a_(t))^(2))=320.189" m/s"^(2)`
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