Question

If $s=\frac{t^3}{3}-\frac{1}{2}{t}^2-6t+5$, find the value of $\frac{d^{2}s}{d{t}^2}$ at t = 1.

Answer 1

Given $s=\frac{t^3}{3}-\frac{1}{2}{t}^2-6t+5$

Differentiate on both sides w.r.t 't'

$\frac{ds}{dt}=\frac{3t^2}{3}-\frac{2t}{2}-6+0$

$\frac{ds}{dt}=t^2-t-6$

Again differentiate on both sides w.r.t 't'

$\frac{d^{2}s}{d{t}^2}=2t-1$

$\frac{d^{2}s}{d{t}^2}{|}_{t=1}=2\left(1\right)-1=1$