The position vector of a body changes with time as vec r(t)…

The position vector of a body changes with time as $\vec r(t) = (1 - bt)\,kt\,\vec r_0$, where $\vec r_0$ is a unit vector and $b$, $k$ are positive constants. Find: (a) the velocity and acceleration vectors as functions of time; (b) the time $\Delta t$ after which the body returns to its starting position; (c) the distance travelled in that time.

Explanation of Method

A body that goes out and comes back.

Step 1: (a) Velocity and acceleration

Write $\vec r = (kt - bkt^2)\,\vec r_0$. The body moves along the fixed direction $\vec r_0$, so this is motion along a straight line.

\[ \vec v = \frac{d\vec r}{dt} = k\,(1 - 2bt)\,\vec r_0, \]

\[ \vec a = \frac{d\vec v}{dt} = -2bk\,\vec r_0. \]

\[\vec v = k(1 - 2bt)\,\vec r_0, \qquad \vec a = -2bk\,\vec r_0\]

The acceleration is constant and points opposite to $\vec r_0$: it is like a ball thrown upward.

Step 2: (b) Return to the start

At $t = 0$ the body is at $\vec r = 0$. It is back there when $(1 - bt)\,kt = 0$, i.e. at $t = 0$ or $t = 1/b$: $\Delta t = \frac{1}{b}$.

$r(t)$ $t$ $0$ $\tfrac{1}{2b}$ $\Delta t = \tfrac{1}{b}$ $\tfrac{k}{4b}$ $\text{turning point } (v = 0)$ $\text{out: } \tfrac{k}{4b}$ $\text{back: } \tfrac{k}{4b}$

Position along the line of $\vec r_0$ versus time. The body goes out, stops at $t = \tfrac{1}{2b}$ at the largest distance $\tfrac{k}{4b}$, and comes back to the start at $t = \tfrac{1}{b}$. The path length is out plus back: $\tfrac{k}{2b}$.

Step 3: (c) Distance travelled

The body goes out, stops, and comes back. It stops where $v = 0$: $1 - 2bt = 0 \;\Longrightarrow\; t = \frac{1}{2b}$.

\[ r_{\max} = \left(1 - \tfrac12\right)k\cdot\frac{1}{2b} = \frac{k}{4b}. \]

The path is $\tfrac{k}{4b}$ out plus $\tfrac{k}{4b}$ back: $s = 2 \cdot \frac{k}{4b} = \frac{k}{2b}$.

Result

\[\boxed{\,\vec v = k(1 - 2bt)\,\vec r_0, \qquad \vec a = -2bk\,\vec r_0\,}\]

\[\boxed{\,\Delta t = \frac{1}{b}\,}\]

\[\boxed{\,s = 2 \cdot \frac{k}{4b} = \frac{k}{2b}\,}\]

Remark

Distance vs displacement. After $\Delta t$ the displacement is zero, but the distance travelled is $\tfrac{k}{2b}$. The original solution got the same result by integrating $|v|$ in two parts; splitting the motion at the turning point gives it without integrals.