Question

A particle is moving in a straight line such that its distance at any time $t$ is given by $s=\frac{t^4}{4}-2t^3+4t^2-7$. The acceleration of the particle is minimum when $t=$

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (B)

$2$

$s=\frac{t^4}{4}-2t^3+4t^2-7$

$v=\frac{ds}{dt}=t^3-6t^2+8t$

$a=\frac{dv}{dt}=3t^2-12t+8$

For maxima or minima,

$\frac{da}{dt}=0$

$\Rightarrow 6t-12=0$

$\Rightarrow t=2sec$

Now, $\frac{d^{2}a}{d{t}^2}=6>0$

Hence, a is minimum at $t=2sec$