Question

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. $a + \frac{b^2}{4c}$
• B. $a + \frac{b^2}{c}$
• C. $a + \frac{b^2}{2c}$
• D. $a + \frac{b^2}{3c}$

Answer 1

Answer: (D)

$x =at +bt^{2} -ct^{3}$
$ v = \frac{dx}{dt}=a+2bt-3ct^{2} $
$ a= \frac{dv}{dt} =2b-6ct =0 \Rightarrow t = \frac{b}{3c} $
$ v_{\left(at t= \frac{b}{3c}\right)} = a +2b \left(\frac{b}{3c}\right)-3c\left(\frac{b}{3c}\right) $
$ =a +\frac{b^{2}}{3c} .$