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The position of a particle as a functio...

The position of a particle as a function of time t, is given by `x(t)=at+bt^(2)-ct^(3)` where a,b and c are constants. When the particle attains zero acceleration, then its velocity will be :

A

`a + (b^(2))/(c )`

B

`a + (b^(2))/(3c)`

C

`a + (b^(2))/(4c)`

D

`a + (b^(2))/(2c)`

Text Solution

Verified by Experts

The correct Answer is:
B
`x(t) = at + bt^(2) - ct^(3), v = (dx(t))/(dt) = a + 2bt - ct^(2)`
`a = (d^(2) x (t))/(dt^(2)) = 2b - 6ct, a = 0]`
`g = (b)/(3c), v = a + 2b ((b)/(3c)) - 3c ((b)/(3c))^(2), v = a + (b^(2))/(3c)`
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