If $s=4t^3-2t^2+3t+7$, find velocity and acceleration at $t=2s$

The correct answer is $velocity = 43$ $unit/s$, $acceleration = 44$ $unit/s^2$

Given $s=4t^3-2t^2+3t+7$

Differentiating w.r.t $t$, $\dfrac{ds}{dt}$ = velocity = $12t^2 - 4t + 3$

∴ $ \left( \dfrac{ds}{dt} \right)_{t=2} $ = $v$ = $12(2)^2 - 4(2) + 3$ = $43$ $unit/s$



Since $v = 12t^2 - 4t + 3$

∴ $\dfrac{dv}{dt}$ = acceleration = $24t -4$


∴ $ \left( \dfrac{dv}{dt} \right)_{t=2} $ = $24 (2) - 4$ = $44$ $unit/s^2$