Question

The position-time equation for a certain train is xf=2.8m+(8.1m/s)t+(2.9m/s^2)t^2.

Answer 1

The formula without numbers is
$xf = xi + vxi(t) + \frac{1}{2} (a) {t}^{2}$
do the initial velocity would be
8.1 m/s
and the acceleration would be
2.9 m/s^2