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

$\frac{\mathrm{ds}}{\mathrm{dt}} = 6 t + 5$
This is the instantaneous velocity $v$;
$\frac{\mathrm{dv}}{\mathrm{dt}} = 6$ which is your instantaneous (and constant) acceleration.