The space s described in time t by

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 $\mathrm{t}=2:$

$a=-260$