Velocity and displacement

Question

My problem concerns displacement and velocity. Let $u$ denote displacement vector and $u'$ the velocity vector. Assume that $u(0)=0$ and $u'=0$ all the time. Then

$$u(t)=\int_{0}^{t}u'(s)ds+u(0)=0.$$

So, if velocity is constant and equals $0$, then there is no displacement. Now, let $u(0)=u_{0}$ while $u'$ is still $0$. Then

$$u(t)=\int_{0}^{t}u'(s)ds+u_{0}=u_{0}.$$

So, we observe constant displacement, although there is no velocity. This argues with my intuition. How can we observe any displacement if there is no velocity?

Does that mean that the particle was in a reference position some time ago and was displaced by $u_{0}$ and when it reached the point it stoped at time $t=0$ and so $u_{0}$ carries only the information about the change from reference position?

Answer 1

$u(t)=u_0$ means that at any time $t$ the position is the constant value $u_0$. This means that there is no displacement from $u_0$.

Answer 2

You need to distinguish these concepts clearly:

• $u(t)$ is the position at time $t$.
• Displacement means the change in position between two different times. For example, $u(t_1) - u(t_0)$ is the displacement from position at $t_0$ to position to $t_1$. This value has nothing to do with the actual distance you have traveled.
• Distance is the actual path length of travel between two different times.

Consider an example of walking along a straight line. Suppose at $t=0$ you start walking $10 m$ to the right, then stop and start walking $8 m$ back to the left.

• Your final position is just $2m$ to the right of the initial position. Thus, as far as positions are concerned, your displacement is just $2m$.
• But you have walked a total distance of $18m$.