What is the difference between position, displacement, and distance traveled?
Position is a single point. Usually in space we indicate positions with coordinates like $(x,y,z)$ in Cartesian coordinates, $(r,\phi,\theta)$ in spherical coordinates, etc. We can also define the position as a vector, i.e. the position vector, that is a vector that points from the origin (subjectively defined) to the position of the particle in question. It could be $\mathbf r=x\hat x+y\hat y+z\hat z$ using Cartesian coordinates, $\mathbf r=r\hat r$ using spherical coordinates$^*$, etc. In 1D there really isn't anything different between the position coordinate and the position vector, so you don't need to worry about the distinction in the problem you have described in your question.
Displacement is the change in position. It is a vector quantity; it is the difference between two position vectors. So, for example, if you go around a circle exactly one time, your displacement over that time is $0$. You can get the displacement at some time $t$ by integrating the instantaneous velocity over time: $$\Delta\mathbf r=\mathbf r(t)-\mathbf r(t_0)=\int_{t_0}^t\mathbf v(\tau)\text d\tau$$ Notice that $\mathbf r(t)$ is the position at time $t$, and $\mathbf r(t_0)$ is the initial position you are calculating the displacement with respect to. Hence the integral gives displacement since it is the difference between the final and initial position vectors.
Distance is a little different, but it is easy to understand. You can think of it as what the odometer in your car gives to you. It just tells you how far you have traveled without reference to some starting point. Going back to the example of going around a circle one time, your displacement was $0$, but the distance you traveled is equal to the circumference of the circle. Distance is a scalar value. You can determine it by integrating the speed over time: $$D(t)=\int_0^t|\mathbf v(\tau)|\text d\tau$$ Note that this integral is the same thing as the curve length of the path the particle moves along (i.e. the distance you have traveled).
These explanations should help you in your own problem.
$^*$ Note that the position vector in spherical coordinates is not $\mathbf r=r\hat r+\phi\hat\phi+\theta\hat\theta$. This is because the position vector, by definition, points from the origin to the position (coordinate) of the particle. Therefore, the position vector can only have a radial component, i.e. $\mathbf r=r\hat r$
$\color{blue}{\text{Position}}$: $$\color{blue}{\vec p(t)}$$
$\color{red}{\text{Displacement}}$†: $$\color{red}{\vec p(t_2) - \vec p(t_1)}$$
$\color{green}{\text{Distance Traveled}}$: $$\color{green}{\int_{t_1}^{t_2} \left\|\frac{d\vec p(t)}{dt}\right\|dt}$$
†Displacement is sometimes defined as the scalar $\|\vec p(t_2) - \vec p(t_1)\|$.
What is the difference between position, displacement, and distance traveled?
Succinctly:
(1) the position of an object is a vector with tail at the origin of the coordinate system and head at the location of the object.
(2) the displacement of an object is the vector difference of the current position vector of the object and the position vector of the object at an earlier time. That is, the tail of this displacement vector is at the earlier position and the head is at the current position. The length of this vector is, generally, the shortest distance between the earlier and current positions.
(3) the distance traveled from the earlier position to the current position is not a vector but is, rather, a path length. There are an infinity of paths that an object may take from the earlier position to the current position and, in general, each has a different associated length.