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

Answer 1

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

Given,

$x=at+b{t}^2-c{t}^3$

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

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

When the particle attains zero acceleration,

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

$\therefore v_{(t=\frac{b}{3c})}=a+2b\left(\frac{b}{3c}\right)-3c{\left(\frac{b}{3c}\right)}^2=a+\frac{b^2}{3c}$

Hence, $\left(D\right)$ is the correct answer.