Estimating motion from video or sensor data

Video and sensors do not give you the motion itself but a set of numbers: frames separated by $\Delta t = 1/f$, or accelerometer samples taken at a rate $f_s$. The first step is to fix a scale from a known object (a ruler, a ball of known diameter) and convert pixels into metres, producing a table of coordinates $x(t)$. Velocity is estimated from differences between neighbouring frames: $v_{i+1/2} = (x_{i+1}-x_i)/\Delta t$, so each point of the $v(t)$ graph lies "between" frames. Acceleration is the slope of the $v(t)$ graph, or the second difference $a \approx (x_{i+1}-2x_i+x_{i-1})/\Delta t^{2}$. Real data are noisy and $\Delta t$ is finite, so it is useful to estimate the uncertainty $\delta v \approx \delta x/\Delta t$ and to check whether the second differences really are constant. The kinematics becomes a physical model only when the measured acceleration agrees with a free-body diagram and Newton's second law, $m\vec a = \sum \vec F$.

A video is recorded at 25 frames per second and the scale is set with a ruler: $1\ \text{m} = 250\ \text{px}$. A ball rolls along the floor, and in frames 0, 1, 2, 3 its coordinates are 0, 25, 50, 75 px. Find the ball's velocity and its uncertainty if the coordinate is measured to $\delta x = 1\ \text{px}$.

An object falls; the video was recorded at 20 frames per second, so $\Delta t = 0.05\ \text{s}$. The scaled coordinates (m) by frame are 0, 0.012, 0.049, 0.110, 0.196. Determine the type of motion, find the acceleration, and write the model $x(t)$. Does the result agree with a free-body diagram if air resistance is negligible?

A phone sensor fixed to a cart measures an acceleration $a = 2.6\ \text{m/s}^{2}$ along an inclined plane with angle $\theta = 20^{\circ}$. Estimate the coefficient of friction $\mu$ between the cart and the plane, and state what further checks are needed before the model can be trusted.

Self-study plan (safe, with nothing dropped from a height).

1. Recording. Put the phone on a tripod or a stack of books so that it cannot fall, and record slow-motion video (120–240 frames per second) of a ball rolling along the floor. Place a ruler in the frame. There is no need to throw objects or film a fall: free fall will be handled in a simulation.
2. Scale. In one frame measure the length of the ruler in pixels to obtain the factor $k$ (m per pixel).
3. Table and graph. Mark the ball's position in 6–10 successive frames, convert to metres, and plot $x(t)$ and $v(t)$ (velocity from differences).
4. Simulation. Open a motion simulation with graphs (for example, "Motion" or "Constant-acceleration motion"), set $a = 9.8\ \text{m/s}^{2}$ and compare the shape of the graph with your data.
5. Conclusion. Write down which model ($x = x_0 + v_0 t$ or $x = x_0 + v_0 t + \tfrac{1}{2}at^{2}$) fits your data, and what uncertainty $\delta v$ you expect at your $\Delta t$.

Show the transition from discrete frames to a continuous model: how position marks on frames become the graphs $x(t)$ and $v(t)$, and where the acceleration comes from.

Diagram + graphs. At the top, a row of frames (rectangles) with time labels $t_0, t_1, t_2, \dots$ and the ball's marked positions; below them, an $x$ axis with the same points and the intervals $\Delta x$ shaded. Bottom left: the graph $x(t)$ with points joined by a line and a tangent whose slope gives the velocity. Bottom right: the graph $v(t)$ with points lying between frames and the straight line $v = v_0 + at$ whose slope equals the acceleration $a$.

On a school track a runner passes the 60 m mark. A smartphone records at 240 frames/s; the frame interval is $\Delta t = 1/240\,\text{s} \approx 4.17\,\text{ms}$. In two neighbouring frames the runner’s hip moves $0.034\,\text{m}$ along the lane. The scale is set by a $1.00\,\text{m}$ lane marking that spans 100 pixels.

This uses the frame-to-position-time table step: calibrating pixels to metres, attaching timestamps, and taking a numerical derivative. The average speed over the interval is $v \approx \Delta x/\Delta t = 0.034\,\text{m}/0.00417\,\text{s} \approx 8.2\,\text{m/s}$, i.e. the slope of the $x(t)$ graph between two frames.

A phone is fixed to the dashboard of a $1400\,\text{kg}$ car. While braking from $50\,\text{km/h}=13.9\,\text{m/s}$ to zero in $3.2\,\text{s}$, the built-in accelerometer logs at $50\,\text{Hz}$, i.e. 160 samples. The mean acceleration along motion is $a \approx -13.9/3.2 \approx -4.3\,\text{m/s}^2$.

Estimating motion from sensor data requires numerical integration of acceleration: $\Delta v = a\,\Delta t$ for each of the 160 intervals, then summing. The car’s free-body diagram (weight, normal reaction, braking friction) shows that horizontal friction produces this acceleration; with $m=1400\,\text{kg}$ the mean braking force is about $F=ma \approx 6.0\times10^3\,\text{N}$.

On a footbridge an accelerometer measures vertical acceleration at $100\,\text{Hz}$ for $50\,\text{s}$ after a truck passes. The peak acceleration is $0.80\,\text{m/s}^2$, and the dominant frequency is $2.5\,\text{Hz}$. For a sinusoidal mode the displacement amplitude is $A = a_{\text{max}}/(2\pi f)^2 \approx 0.80/(2\pi\cdot2.5)^2\,\text{m} \approx 3.2\,\text{mm}$.

This is sensor-data processing: removing offset and drift, then converting acceleration to displacement by double numerical integration. The resulting $y(t)$ graph shows the bridge vibration; amplitude and period can be read from it.