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The positione of a particle is expressed...

The positione of a particle is expressed as ` vecr = ( 4t^(2) hati + 2thatj)` m, where t is time in second. Find the velocity o the particle at t = 3 s

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`vecr = ( 4t^(2)hati + 2thatt) m`
Velocity ` vecc = (dvecr)/(dt) = hati d/(dt) (4t^(2)) + hatj d/(dt) (2t)`
` vecv = ( 8t) hati + 2hatj`
At t = 3 s , velocity is given by
` vecv_(t=3) = (8 xx 3) hati + 2hatj`
` vecv_(t =3) = ( 24hati + 2hatj) m s^(-1)`
`= sqrt580`
` = 24.08 m s^(-1)`
Direction ` theta = tan ^(-1) 2/24`
` = tan^(-1) 1/12 = 4.76^(@)`
Thus the particle has velocity `24.08 m s^(-1)` at an angle ` 4.76^(@)` with x -axis.
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