Question

The position of a particle as a function of time $t,$ is given by $x\left(t\right)=at+b{t}^2-c{t}^3$

Where, $a,b$ and $c$ are constants. When the particle attains zero acceleration, then its velocity will be-

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

Answer 1

The correct option is C $a+\frac{b^2}{3c}$

Given, $x\left(t\right)=at+b{t}^2-c{t}^3$
$\Rightarrow v=\frac{dx}{dt}=\frac{d}{dt}(at+b{t}^2-c{t}^3)=a+2bt-3c{t}^2$
$\Rightarrow a=\frac{dv}{dt}=\frac{d}{dt}(a+2bt-3c{t}^2)=2b-3c\times 2t$
When $a=0\Rightarrow t=\left(\frac{b}{3c}\right)$
$\therefore v=a+2b\left(\frac{b}{3c}\right)-3c{\left(\frac{b}{3c}\right)}^2=a+\frac{b^2}{3c}$
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.