radius vector displacement velocity and acceleration

The position of a material point in a chosen reference frame is given by the radius vector $\vec r(t)$, drawn from the origin to the point. In Cartesian coordinates it is expanded in the unit vectors of the axes:

$$\vec r = x\,\hat i + y\,\hat j + z\,\hat k.$$

The displacement over a time interval $\Delta t$ is the change of the radius vector

$$\Delta \vec r = \vec r(t+\Delta t) - \vec r(t),$$

that is, a vector directed from the initial position to the final one; its magnitude is generally not equal to the path length $s$ travelled. The average velocity over the interval is $\langle \vec v\rangle = \Delta \vec r/\Delta t$, while the instantaneous velocity is the time derivative of the radius vector:

$$\vec v = \frac{d\vec r}{dt},$$

and this vector is directed along the tangent to the trajectory. Acceleration characterises how fast the velocity changes:

$$\vec a = \frac{d\vec v}{dt} = \frac{d^2\vec r}{dt^2},$$

and $\vec a$ points in the direction of the change of the velocity vector, not necessarily along the trajectory itself.

The radius vector of a point is $\vec r(t) = (3t)\,\hat i + (4t^2)\,\hat j$ (SI). Find the displacement of the point over the interval from $t_1 = 0$ s to $t_2 = 2$ s and the magnitude of this displacement.

A point moves so that its radius vector is $\vec r(t) = (2t^2)\,\hat i + (3t)\,\hat j$ (SI). Find the velocity and acceleration vectors at $t = 1$ s and their magnitudes. Check whether the velocity is directed along the tangent to the trajectory.

A car travels a quarter of a circle of radius $R = 20$ m at a constant speed $v = 10$ m/s. Initially it is at the point $(20; 0)$ m and moves in the $+y$ direction, and at the end it is at the point $(0; 20)$ m. Find the travel time, the average acceleration $\vec a_{\text{av}} = \Delta \vec v/\Delta t$, and compare its magnitude with the instantaneous centripetal acceleration $v^2/R$.

Self-check with a safe simulation. Use a two-dimensional motion model that displays vectors (for example, the PhET "Projectile Motion" simulation): observation and measurement are done on the screen, with no throwing and no real-world measurements.

1. Set an initial velocity at an angle to the horizontal and switch on the display of velocity and acceleration vectors.
2. Pause the motion at equal intervals $\Delta t$ and record the coordinates $x$, $y$; draw the radius vectors $\vec r_i$ from the origin.
3. Connect neighbouring positions with segments: these are the displacements $\Delta \vec r_i$. Dividing each by $\Delta t$ gives the average velocities over the intervals.
4. Check two statements: as $\Delta t$ decreases, the average velocity vectors approach the direction of the tangent to the trajectory; the acceleration vector stays constant in magnitude and direction and points along the change $\Delta \vec v$.

Write your conclusion in two sentences: did the direction of the acceleration coincide with the direction of the change of velocity, and how did the magnitudes of displacement and path differ?

Show the geometric relation between the radius vector, displacement, velocity and acceleration on a curved trajectory: displacement is the chord between two radius vectors, velocity is tangent, and acceleration is decomposed into tangential and normal components; the normal component points toward the inside of the curvature, while the tangential component is along or opposite the velocity (for uniform circular motion the tangential component is zero).

Sketch (diagram). A curved trajectory of a point is drawn in the $xy$ plane. The origin $O$ and two nearby positions of the point, $A$ and $B$, are marked. Radius vectors are drawn from $O$: $\vec r(t)$ to $A$ and $\vec r(t+\Delta t)$ to $B$. The segment from $A$ to $B$ is labelled $\Delta \vec r$. At point $A$, the vector $\vec v$ is drawn along the tangent to the trajectory, and next to it the vector $\vec a$, shown as the sum of the tangential component $\vec a_{\tau}$ (along $\vec v$ or opposite to it) and the normal component $\vec a_n$, directed toward the inside of the bend of the trajectory. A dashed arc of the trajectory between $A$ and $B$ shows the path length $s$.

A cyclist rides a circular velodrome track of radius $R = 25$ m and covers exactly half a lap in $t = 10$ s. Along the track he travels $\pi R \approx 78.5$ m, so his average speed is $78.5/10 = 7.85$ m/s. A GPS unit on the handlebars logs coordinates and at the finish subtracts the starting ones: start and finish lie at diametrically opposite points of the circle.

This uses the definition of average velocity $\langle \vec v \rangle = \Delta \vec r / \Delta t$. Over half a lap the displacement vector $\Delta \vec r$ is the diameter chord of length $2R = 50$ m through the centre, so $|\langle \vec v \rangle| = 50/10 = 5.0$ m/s, not $7.85$ m/s. The example separates two quantities: the magnitude of the displacement vector and the length of the travelled path.

A car drives $3.0$ km east and then $4.0$ km north in $10$ min ($600$ s). The navigator saves a coordinate pair $(x, y)$ every second. Subtracting the coordinates gives the displacement vector $|\Delta \vec r| = \sqrt{(3.0)^2 + (4.0)^2} = 5.0$ km, directed at $\arctan(4/3) \approx 53^\circ$ to the east. The average velocity is $5000/600 = 8.3$ m/s, the average speed $7000/600 = 11.7$ m/s. Then the driver brakes from $15$ m/s to rest in $5.0$ s.

Both definitions are at work: $\vec v = d\vec r/dt$ and $\vec a = d\vec v/dt$. Subtracting coordinates in the navigator is a component-by-component computation of $\Delta \vec r$; during braking the velocity vector shrinks in magnitude from $15$ m/s to $0$, so $\vec a$ points opposite to $\vec v$ and equals $|\Delta v|/\Delta t = 15/5.0 = 3.0$ m/s$^2$. Acceleration changes the velocity vector, not only its magnitude.

A carousel seat moves on a circle of radius $R = 8.0$ m; one full turn takes $T = 12$ s. The speed of the seat is constant and equals $v = 2\pi R/T = 2\pi \cdot 8.0/12 \approx 4.2$ m/s. The passenger feels pressed into the back of the seat.

Acceleration is the rate of change of the velocity vector itself: $\vec a = d\vec v/dt$. Even with a constant magnitude $v$, the direction of $\vec v$ turns, and the vector $\Delta \vec v$ over a short interval points towards the centre of the circle. Hence the acceleration magnitude $a = v^2/R = (4.2)^2/8.0 \approx 2.2$ m/s$^2$, and the vector $\vec a$ is perpendicular to $\vec v$.