Answer
$21$ m; No
Work Step by Step
The position-time relation is: $S(t)=\dfrac{at^2}{2}+v_0t+s_0$
Plug in the values.
$S(t)=\dfrac{2t^2}{2}+20t=t^2+20t$
$S(t)=t^2+20t \implies S(0)=0+20(0)=0$
$S''(t)=2t+20 \implies S''(0)=20$
$S'''(t)=2 \implies S'''(0)=2$
The second degree Taylor polynomial $T_2(x)$ for when $t=1$
$T_2(1)=0t^0+20 \dfrac{t^1}{1!}+2 \dfrac{t^2}{2!}0t^0+20 \dfrac{1^1}{1!}+2 \dfrac{1^2}{2!}=21 m$
Thus, the position of the car after 1 second is 21 m. We should not use this polynomial for the full minute because the acceleration of $2~m/s^2$ would not be accurate over that entire time period.
