Question

The position $x$ of a particle with respect to time $t$ along $x-$axis is given by $x=9t^2-t^3$ where $x$ is in metres and $t$ in seconds. What will be the position of this particle when it achieves maximum speed along the $+x$direction?

• A. $24m$
• B. $32m$
• C. $54m$
• D. $81m$

Answer 1

Answer: (C)

$54m$

$X=9t^2-t^3$
$\therefore V=18t-3t^2$
$\Rightarrow \frac{dv}{dt}=18-6t$

for maximum speed

$\frac{dv}{dt}=0,and\frac{d^{2}v}{d{t}^2}$ should be negative
$so18-6t=0\Rightarrow t=3s$
$att=3s,X=9(3{)}^2-(3{)}^3=81-27=54m$