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The position of a particle is expressed ...

The position 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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`bar( r)=(4t^(2)hati+2thatj)m`
Velocity `bar( v) = (dbarr)/(dt)=hati(d)/(dt)(4t^(2))+hatj(d)/(dt)(2t)`
`barv=(8t)hatt2hatj" "cdots(i)`
At t=3 s velocity is given by
`barv_(t=3)=(8xx3)hatj+2hatj`
`barv_(t+)=(24hatj+2hatj)m.s^(-1)`
Magnitude `|bar(v)_(t=3)|=sqrt((24)^(2)+2^(2))`
`= sqrt(580)`
`=24.08 m.s^(-1)`
Direction `theta = tan^(-1) (2)/(24)`
Thus the particle has velocity 24.08 `m.s^(-1)` at an angle `tan^(-1)((1)/(12))` with x-aix.
Note: For above example No 8 it can be seen that y - component of `barv ` i.e `v_(y)=2` remains constant with time its x - component `v_(x)` = 8t increases as the time increases.
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