The position of a particle as a function of t | vedclass

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(D) Given the position function: $x(t) = at + bt^2 - ct^3$.The velocity $v(t)$ is the first derivative of position with respect to time:$v(t) = \frac{

Vedclass · Accepted Answer

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

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(D) Given the position function: $x(t) = at + bt^2 - ct^3$.The velocity $v(t)$ is the first derivative of position with respect to time:$v(t) = \frac{dx}{dt} = a + 2bt - 3ct^2$.The acceleration $a(t)$ is the derivative of velocity with respect to time:$a(t) = \frac{dv}{dt} = 2b - 6ct$.Set the acceleration to zero to find the time $t$ when this occurs:$0 = 2b - 6ct \implies 6ct = 2b \implies t = \frac{b}{3c}$.Now,substitute $t = \frac{b}{3c}$ into the velocity equation:$v = a + 2b\left(\frac{b}{3c}\right) - 3c\left(\frac{b}{3c}\right)^2$$v = a + \frac{2b^2}{3c} - 3c\left(\frac{b^2}{9c^2}\right)$$v = a + \frac{2b^2}{3c} - \frac{b^2}{3c}$$v = a + \frac{b^2}{3c}$.

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,its velocity will be:

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