Question:
The space s described in time $t$ by a particle moving in a straight line is given by $S=t 5-40 t^{3}+30 t^{2}+80 t-250 .$ Find the minimum value of acceleration.
Solution:
Given : $s=t^{5}-40 t^{3}+30 t^{2}+80 t-250$
$\Rightarrow \frac{d s}{d t}=5 t^{4}-120 t^{2}+60 t+80$
Acceleration, $a=\frac{d^{2} s}{d t^{2}}=20 t^{3}-240 t+60$
$\Rightarrow \frac{d a}{d t}=60 t^{2}-240$
For maximum or minimum values of $a$, we must have
$\frac{d a}{d t}=0$
$\Rightarrow 60 t^{2}-240=0$
$\Rightarrow 60 t^{2}=240$
$\Rightarrow t=2$
Now,
$\frac{d^{2} a}{d t^{2}}=120 t$
$\Rightarrow \frac{d^{2} a}{d t^{2}}=240>0$
So, acceleration is minimum at $t=2$.
$\Rightarrow a_{\min =} 20(2)^{3}-240(2)+60=160-480+60=-260$
$\therefore$ At $\mathrm{t}=2:$
$a=-260$