Confusion regarding ($d = rt$) vs ($x_1 = x_0 + v_o t + 0.5 at^2$) usage.

Question

I'm following an online physics course and I can't understand why for the question below the equation distance = speed $\times$ time can't be used while the equation $x_1 = x_0 + v_0t + 1/2at^2$ can. Could somebody please clarify why this is the case?

"A plane is moving at $30 \ m/s$. It accelerates at $2 \ m/s^2$ for $11$ seconds. How far does it travel during this time?"

Here's how I solved it which is wrong according to my book which uses the other equation.

1. I started by calculating the final velocity using $v_1 = v_0 + at$ which is $52 \ m/s$.

2. Using the final velocity I calculated the average velocity using $(v_1 - v_0) / 2$ resulting in $ 11 \ m/s$.

3. Inserting it into the equation $d = rt$ ($d = 11 \ m/s \times 11 \ s$) the answer is $121 \ m$.

Answer 1

This is because you have to distinguish two different cases: (1) when you move at constant speed ($s=v t$) in a straight direction and when you have acceleration ($x_1=x_0+V_0 t+\frac{1}{2} a t^2$), i.e. a change of speed versus time.

Answer 2

The plane accelerates at $\ddot{x}=2\ \text{m}\cdot\text{s}^{-2}$ for $11\text{ s}$. Therefore, its velocity is given by: $$\dot{x}=\int_0^{11}\ddot{x}\ dt=\int_0^{11}2\ dt=2t+C_1$$ The integration constant $C_1$ is obtained by comparison with the boundary condition: $$\dot{x}(t=0)=v_0=30\ \text{m}\cdot\text{s}^{-1}=C_1$$ The velocity is then: $$\dot{x}=2t+30$$ The distance traveled in $11\text{ s}$ is, once again, obtained by integration: $$x=\int_0^{11}\dot{x}\ dt=\int_0^{11}(2t+30)\ dt=t^2+30t+C_2$$ And once again, we compare this result with the boundary condition for the position $x(t)$ to obtain the value of $C_2$: $$x(t=0)=x_0=C_2$$ We don't need to know the initial position of the plane, since we're only interested in the change of position: $$\Delta x=x(t=11)-x(t=0)=(11^2+30\times 11+x_0)-x_0=151$$ The distance traveled by the plane in $11\text{ s}$ is $\Delta x=451\text{ m}$.