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

Answer 1

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

Formula Used: $v=\frac{dx}{dt}$
$a=\frac{dv}{dt}$
Given, position of the particle
$x\left(t\right)=at+b{t}^2-c{t}^3$
Velocity, $v=\frac{dx}{dt}=a+2bt-3c{t}^2$
Acceleration, $a=\frac{dv}{dt}=2b-6ct$
According to question, $a=0$
$\therefore 2b-6ct=0$
$\Rightarrow t=\frac{b}{3c}$
So, velocity at $t=\frac{b}{3c}$
$v=a+2b\left(\frac{b}{3c}\right)-3c{\left(\frac{b}{3c}\right)}^2$
$v=a+\frac{b^2}{3c}$