How do you find the instantaneous acceleration for the particle whose position at time $t$ is given by $s(t) = 3{t^2} + 5t$ ?

Hint:Double derivation of particle position expression which gives acceleration. The instantaneous velocity of a particle at time t is given by the gradient of its position-time graph at that time, we can now use the terminology of functions and derivatives to say that the velocity of the particle is given by the derivative of its position function. Similarly, we can say that the instantaneous acceleration of a particle is given by the derivative of the velocity function.

Complete step by step answer:
On deriving for the first time of $s(t) = 3{t^2} + 5t$, we get
$\dfrac{{ds}}{{dt}} = 6t + 5$
On first derivation, we obtain what is called as instantaneous velocity v and derivation of the velocity expression with respect to time we obtain acceleration or instantaneous acceleration. This emphasises that the acceleration of a particle is the rate of change of the rate of change of the position, or if you prefer, the derivative of the derivative of the position function.

Either of these formulations is a bit of a mouthful, so it is more conventional to refer to the acceleration as the second derivative of $x(t)$.So, for the instantaneous acceleration, we are going to derive again for the second time, we get;
$\dfrac{{dv}}{{dt}} = 6$

So, from above we can say that the instantaneous acceleration of the particle is 6.

Note:Remember that the derivative with respect to time tells you how your quantity changes with time. The first derivative of position then tells you how position changes with time (velocity). Deriving velocity tells you how velocity changes in time, which is here instantaneous acceleration.