A particle moves along a straight line such that its displacement at any time $$t$$ is given by $$s = 3{t^3} + 7{t^2} + 14t + 5.$$     The acceleration of the particle at $$t = 1\,s$$  is

Solution :
On double differentiation of displacement equation gives acceleration of body i.e. $$a = \frac{{{d^2}x}}{{d{t^2}}}$$
The displacement of a particle along a straight line is
$$s = 3{t^3} + 7{t^2} + 14t + 5\,......\left( {\text{i}} \right)$$
Differentiating Eq. (i) w.r.t. time, which gives the velocity
$$\eqalign{ & v = \frac{{ds}}{{dt}} = \frac{d}{{dt}}\left( {3{t^3} + 7{t^2} + 14t + 5} \right) \cr & = \frac{d}{{dt}}\left( {3{t^3}} \right) + \frac{d}{{dt}}\left( {7{t^2}} \right) + \frac{d}{{dt}}\left( {14t} \right) + \frac{d}{{dt}}\left( 5 \right) \cr & v = 3\frac{d}{{dt}}\left( {{t^3}} \right) + 7\frac{d}{{dt}}\left( {{t^2}} \right) + 14\frac{d}{{dt}}\left( t \right) + 0\,......\left( {{\text{ii}}} \right) \cr} $$
(as differentiation of a constant is zero)
$$\eqalign{ & {\text{Now}}\,{\text{use}}\,\frac{d}{{dt}}\left( {{x^n}} \right) = n{x^{n - 1}} \cr & {\text{So}}\,\,v = 3\left( 3 \right){t^{3 - 1}} + 7\left( 2 \right)\left( {{t^{2 - 1}}} \right) + 14\left( {{t^{1 - 1}}} \right) \cr & \Rightarrow v = 9{t^2} + 14t + 14\,......\left( {{\text{iii}}} \right) \cr & \left( {\because {t^0} = 1} \right) \cr} $$
Again differentiating Eq. (iii) w.r.t. time, which gives the acceleration
$$\eqalign{ & a = \frac{{dv}}{{dt}} = \frac{d}{{dt}}\left( {9{t^2} + 14t + 14} \right) \cr & = 18t + 14 + 0 = 18t + 14 \cr & {\text{At}}\,t = 1\,s, \cr & a = 18\left( 1 \right) + 14 = 18 + 14 = 32\,m/{s^2} \cr} $$