1.

A point P moves in counter-clockwise direction on a circular path as shown in the figure. The movement of 'P ' is such that it sweeps out a lengths=t ^ 3+5where s isinmetersandtisinseconds.Theradiusofthepathis20m.The acceleration of ' P ' when t = 2 s is nearly:

Answer»

` 13 m//s ^ 2`
`12 m//s ^ 2`
` 7.2 m//s^ 2`
` 14m//s ^ 2`

Solution :DISTANCE time relation of the particle, `s=t^(3)+5`
speed of the particle `V=(ds)/(dt)=(d)/(dt) (t^(3)+5)=3t^(2)`
Tangential acceleration, `a_(t)=(DV)/(dt)=(d)/(dt) (3t^(2))=6t`
At time t=2s, Speed of the particle `v=3(2)^(2)=12m//s`
Tangential acceleration, `a_(x)=6(2)=12m//s^(2)`
Centripetal acceleration, `a_(c)=v^(2)/R=((12)^(2))/(20)=(144)/(20)=7.2 m//s^(2)`
Net acceleration `a=sqrt((a_(c))^(2)+(a_(t))^(2))=sqrt((7.2)^(2)+(12)^(2))=14m//s^(2)`


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