The position of a moving car at time t is given by f ( t ) = a t ^...

Given: position of the moving car at time $t\text{is} = f\left(t\right) = a{t}^{2} + bt + c$

So,

${v}_{\text{avg}} = \dfrac{f\left({t}_{2}\right) - f\left({t}_{1}\right)}{{t}_{2} - {t}_{1}}$

$⇒ {v}_{\text{avg}} = \dfrac{a\left({t}_{2}^{2} - {t}_{1}^{2}\right) + b\left({t}_{2} - {t}_{1}\right)}{{t}_{2} - {t}_{1}}$

$⇒ {v}_{\text{avg}} = a\left({t}_{1} + {t}_{2}\right) + b$

The instantaneous speed is given by:

${f}^{'}\left(t\right) = 2at + b$

So, to get the point where average speed is equal to car's actual speed,

${f}^{'}\left(t\right) = {v}_{\text{avg}}$

$⇒ a\left({t}_{1} + {t}_{2}\right) + b = at + b$

$⇒ t = \dfrac{{t}_{1} + {t}_{2}}{2}$