The distance ' $\mathrm{s}$ ' in meters covered by a particle in $\mathrm{t}$ seconds is given by $s=2+27…
The distance ' $\mathrm{s}$ ' in meters covered by a particle in $\mathrm{t}$ seconds is given by $s=2+27 t-t^3$. The particle will stop after distance.
65 meters
80 meters
56 meters
60 meters
Solution
$\mathrm{s}=2+27 \mathrm{t}-\mathrm{t}^3$ and particle stops when its velocity is zero.
$\therefore \frac{\mathrm{ds}}{\mathrm{dt}}=27-3 \mathrm{t}^2=0 \Rightarrow \mathrm{t}^2=9 \Rightarrow \mathrm{t}=3 \mathrm{sec} .$
$\therefore$ Distance covered in $3 \mathrm{sec}$, is
$\mathrm{S}_{(\mathrm{t}=3)}=2+27(3)-(3)^3=56 \mathrm{~m}$
