Question

​​A particle moves along a straight line such that its displacement at any time t is given by
s = 3 t³ + 7t² + 14t + 5
find the acceleration of particle at t = 1 s. {explain briefly with formula used}
[ans.32m/s² ]

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{s}=3{\mathrm{t}}^3+7{\mathrm{t}}^2+14\mathrm{t}+5 \\ \mathrm{Use}\mathrm{differentiation}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{equation}, \\ \mathrm{v}=\frac{\mathrm{ds}}{\mathrm{dt}} \\ \mathrm{v}=3\left(3{\mathrm{t}}^2\right)+7\left(2\mathrm{t}\right)+14\left(1\right)+0 \\ \mathrm{v}=9{\mathrm{t}}^2+14\mathrm{t}+14 \\ \mathrm{Again}\mathrm{by}\mathrm{differentiation}, \\ \mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}} \\ \mathrm{a}=9\left(2\mathrm{t}\right)+14\left(1\right)+0 \\ \mathrm{a}=18\mathrm{t}+14 \\ \mathrm{accelration}\mathrm{at}\mathrm{t}=1\mathrm{s}, \\ \mathrm{a}=18\left(1\right)+14 \\ \mathrm{a}=18+14=32\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$