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4. A particle moves along the x axis. Its position is given by the equation `x=2+3t -4t^2`,with `x` in meters and `t` in seconds. Determine
(a) its position when it changes direction and
(b) its velocity when it returns to the position it had at `t = 0`.
`v =(dx)/(dt)`
(a)`2.56 m`
(b)`-3m/s`
Given:
`x = 2 + 3t - 4t^2`
(a)
from given equation we get:
`v = (dx)/(dt) = 3 - 8t`
when `v = 0` it changes direction, so
` 0 = 3-8t`
`=> t =3/8 s`
`x = 2 + 3(3/8) -4(3/8)^2 = color(red)(2.56 m)`
(b)
when `t=0` `x = 2 m`
`2 = 2 + 3t - 4t^2`
`t(3-4t) = 0`
`t= 0` or `t = 3/4 s`
When`t = 3/4 s` it returns to the position is had at t = 0
`v = 3 - 8t`
`=>v = 3-8(3/4) = color(red)(-3m//s)`
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