Question

The position 'x' of a particle varies with time 't' as x=at^​2-bt³. Find the time 't' when the acceleration is equal to zero.​

Answer 1

$$\begin{gathered} \mathrm{Position}\mathrm{of}\mathrm{the}\mathrm{particle}: \\ \mathrm{x}\left(\mathrm{t}\right)={\mathrm{at}}^2-{\mathrm{bt}}^3 \\ \mathrm{Velocity}: \\ \mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}} \\ =2\mathrm{at}-3{\mathrm{bt}}^2 \\ \mathrm{Acceleration}: \\ \mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}} \\ =2\mathrm{a}-6\mathrm{bt} \\ \mathrm{Equate}\mathrm{the}\mathrm{acceleration}\mathrm{to}\mathrm{zero}: \\ 2\mathrm{a}-6\mathrm{bt}=0 \\ \mathrm{t}=\frac{2\mathrm{a}}{6\mathrm{b}} \\ =\frac{\mathbf{a}}{\mathbf{3}\mathbf{b}} \end{gathered}$$