Displacement (x) of a particle is related to | vedclass

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(C) The displacement is given by $x = at + bt^2 - ct^3$. Velocity $(v)$ is the first derivative of displacement with respect to time: $v = \frac{dx}{d

Vedclass · Accepted Answer

$a + \frac{b^2}{3c}$

Vedclass · Answer

(C) The displacement is given by $x = at + bt^2 - ct^3$. Velocity $(v)$ is the first derivative of displacement with respect to time: $v = \frac{dx}{dt} = a + 2bt - 3ct^2$. Acceleration $(a_{acc})$ is the derivative of velocity with respect to time: $a_{acc} = \frac{dv}{dt} = 2b - 6ct$. Setting acceleration to zero: $2b - 6ct = 0 \implies t = \frac{2b}{6c} = \frac{b}{3c}$. Now,substitute $t = \frac{b}{3c}$ into the velocity equation: $v = a + 2b(\frac{b}{3c}) - 3c(\frac{b}{3c})^2$ $v = a + \frac{2b^2}{3c} - 3c(\frac{b^2}{9c^2})$ $v = a + \frac{2b^2}{3c} - \frac{b^2}{3c}$ $v = a + \frac{b^2}{3c}$.

Displacement $(x)$ of a particle is related to time $(t)$ as: $x = at + bt^2 - ct^3$,where $a, b$,and $c$ are constants of the motion. The velocity of the particle when its acceleration is zero is given by:

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