Displacement (y) of a body is represented by `y = pt + qt^(2) - rt^(3)` where p,q and r are constants of motion. The velocity of the body when its acceleration is 0 :

Options

A. `p + (q^(2))/(2r)`
B. `p + (q^(2))/(r)`
C. `p + (q^(2))/(4r)`
D. `p + (q^(2))/(3r)`

Answer

D

Explanation

Displacement y ` = pt + qt^(2) - rt^(3)`
Velocity ` v = (dy)/(dt) = p + 2qt - 3rt^(2) " ". . . (1)`
Acceleration is ` a = (d^(2)y)/(dr^(2) )= 2q - 6rt`
When acceleration = 0
2q - 6rt = 0
`t = (2q)/(6r) = (q)/(3r)`
Substituting the value of t in the equation (1) we get ,
`v = p + 2q ((q)/(3r)) - 3r xx ((q)/(3r))^(2)`
`:. v = p + (2q^(2))/(3r) - (3rq^(2))/(9r^(2))`
Velocity `v = p + (2q^(2))/(3r) - (q^(2))/(3r)`
`v = p + ((2q^(2) - q^(2))/(3r)) = p + (q^(2))/(3r)`