1.

The displacement of a particle is represented by following equation: s=3t^(3) +7t^(2) +5t +8 where is in metre and in second the acceleration of particle at s is

Answer»

18 `ms^(-2)`
14 `ms^(-2)`
32 `ms^(-2)`
zero

Solution :Here `s=3t^(3)+7t^(2)+5t+8`
Differentiating it TWICE w.r.t,.t. we get
`(DS)/(dt)=9T^(2)+14t+5`
`(d^(2)s)/(dt^(2))=18t+14`
This PUTTING t=1 and `(d^(2)s)/(dt^(2))=a`
a=(18+14)=32 `ms^(-2)`


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