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Position of a point object is given by x...

Position of a point object is given by `x = (2)/(3)t^(3) - 4t^(2)+6t+7` (x is in meter and t is in seconds). What would be the distance travelled by object from t = 0 to instant when particle its direction of velocity for last time.

A

`(16)/(3)m`

B

`(32)/(3)m`

C

`(8)/(3)m`

D

`0`

Text Solution

Verified by Experts

The correct Answer is:
A
Position of ……………
`x = (2)/(3)t^(3)-4t^(2)+6t+7`
`v = (dx)/(dt) = 2t^(2)-8t^(2)-8t+6`
`v = 2(t-1) (t-3)`
`v = 0 at t = 1.3`

`x(t = 0) = 7m`
`x(t = 1) = (2)/(3) - 4+6+7=(29)/(3)m`
`x(t=3) = (2)/(3).3^(3) - 4.3^(2)+6.3+7=7m`
distance travelled `= [x(t=1) - x (t=0)] + [x(t=1)- x (t=3)]`
`=((29)/(3) - 7) + ((29)/(3) - 7) = (16)/(3)m`
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