The position of a moving car at time t is given by f ( t ) = a t 2 + b t + c ,   t > 0 , where a ,…

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

So,

${\mathrm{v}}_{\text{avg}}=\frac{f\left(t_2\right)-f\left(t_1\right)}{t_2-t_1}$

$\Rightarrow {\mathrm{v}}_{\text{avg}}=\frac{a\left(t_2^2-t_1^2\right)+b\left(t_2-t_1\right)}{t_2-t_1}$

$\Rightarrow {\mathrm{v}}_{\text{avg}}=a\left(t_1+t_2\right)+b$

The instantaneous speed is given by:

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

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

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

$\Rightarrow a\left(t_1+t_2\right)+b=at+b$

$\Rightarrow t=\frac{t_1+t_2}{2}$