# The position of a particles 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 : Option 1) $a+\frac{b^{2}}{4c}$ Option 2) $a+\frac{b^{2}}{3c}$ Option 3) $a+\frac{b^{2}}{c}$ Option 4) $a+\frac{b^{2}}{2c}$

S solutionqc

Given

$x(t)=at+bt^{2}-ct^{3}$

$v=\frac{dx}{dt}=a+2bt-3ct^{2}$

$a=\frac{dv}{dt}=2b-6ct$

$a=0 \Rightarrow t =\frac{2b}{6c}= \frac{b}{3c}$

$V \left ( t=\frac{b}{3c} \right )=a+2b\times \frac{b}{3c}-3c\times \frac{b^{2}}{9c^{2}}$

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

Option 1)

$a+\frac{b^{2}}{4c}$

Option 2)

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

Option 3)

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

Option 4)

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

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