Answer
21 m and No
Work Step by Step
Consider the position-time relation as follows:
$s(t)=\dfrac{at^2}{2}+v_0t+s_0$
Then, we have $s(t)=\dfrac{2t^2}{2}+20t=t^2+20t$
$s(t)=t^2+20t; s(0)=0+20(0)=0\\s''(t)=2t+20;s''(0)=20\\s'''(t)=2 ;s'''(0)=2$
Calculate the second degree Taylor polynomial $T_2(x)$ when $t=1$
$T_2(1)=(0)(t^0)+ \dfrac{20t^1}{1!}+\dfrac{2t^2}{2!}(0)+ \dfrac{20(1^1)}{1!}+\dfrac{2(1^2)}{2!}=21 $m
Hence, the position of the car at $t=1$ seconds will be 21 m. We cannot calculate the distance traveled for a full minute using this polynomial (because the acceleration can not be maintained that long).