Question

A particle is moving in a straight line such that its distance at any time $t$ is given by $s=\frac{1}{4}{t}^4-2t^3+4t^2-7.$ Find t when its velocity is maximum($t_v$) and acceleration minimum($t_a$).

• A. $t_v=2-2/\sqrt{3}$, $t_a=4$
• B. $t_v=2$, $t_a=1$
• C. $t_v=2-2/\sqrt{3}$, $t_a=2$
• D. $t_v=-2/\sqrt{3}$, $t_a=2$

Answer 1

The correct option is C $t_v=2-2/\sqrt{3}$, $t_a=2$

$s=\frac{1}{4}{t}^4-2t^3+4t^2-7$
$v=\frac{ds}{dt}=t^3-6t^2+8t$
$\frac{dv}{dt}=3t^2-12t+8$
For $v$ to be maximum or minimum,
$\frac{dv}{dt}=0$
$\Rightarrow 3t^2-12t+8=0$
$\Rightarrow t=\frac{12\pm \sqrt{48}}{6}$
$\Rightarrow t=2\pm \frac{2}{\sqrt{3}}$
Now, $\frac{d^{2}v}{d{t}^2}=6t-12$
At $t=2-\frac{2}{\sqrt{3}}$, $\frac{d^{2}V}{d{t}^2}<0$
Hence, $v$ is maximum at $t=2-\frac{2}{\sqrt{3}}$
Now, acceleration $a=dv/dt=3t^2-12t+8$
$\frac{da}{dt}=6t-12$
For acceleration to maximum or minimum,
$\frac{da}{dt}=0$
$\Rightarrow t=2$
Now, $\frac{d^{2}a}{d{t}^2}=6>0$
Hence, acceleration is minimum at $t=2$