Motion in One Dimension

Position and Velocity

how fast is it moving?

Velocity is how much position changes each second. We read two snapshots of a cart and divide the change in position by the elapsed time.

Example

Velocity is just how much your position changes each second. We read two snapshots of a cart and divide the change in position by the time. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Read the two snapshots

A cart is at 2 metres at the start and 8 metres after 3 seconds. How fast was it going?

xi=2 m,xf=8 mx_i = 2\ \text{m}, \quad x_f = 8\ \text{m}
A cart at its first and second positionsA cart shown at an earlier marked position and at its later position further along a level track.startcart

Find the change in position

Subtract the positions: 8 minus 2 is 6 metres.

Δx=8 m−2 m=6 m\Delta x = 8\ \text{m} - 2\ \text{m} = 6\ \text{m}

Find the elapsed time

The time taken is 3 minus 0, which is 3 seconds.

Δt=3 s−0 s=3 s\Delta t = 3\ \text{s} - 0\ \text{s} = 3\ \text{s}

Divide to get velocity

Velocity is change in position divided by time: 6 over 3 is 2 metres per second. The metres-over-seconds leaves metres per second. On a straight rightward track, this speed is just the size of the velocity.

v=ΔxΔt=6 m3 s=2 m/sv = \frac{\Delta x}{\Delta t} = \frac{6\ \text{m}}{3\ \text{s}} = \hl{2}\ \text{m}/\text{s}
The cart with its velocity arrowThe cart at its later position with an arrow showing how fast and which way it moves.cartv

More displacement gives more velocity

Hold the time at 3 seconds. A bigger displacement in the same time means a bigger velocity.

ΔxΔtv3 m3 s1 m/s6 m3 s2 m/s9 m3 s3 m/s\begin{array}{c|c|c}\Delta x & \Delta t & v \\ \hline 3\ \text{m} & 3\ \text{s} & 1\ \text{m}/\text{s} \\ 6\ \text{m} & 3\ \text{s} & 2\ \text{m}/\text{s} \\ 9\ \text{m} & 3\ \text{s} & 3\ \text{m}/\text{s}\end{array}

More time lowers the same-trip velocity

Hold the displacement at 12 metres. Taking more time for the same trip makes the velocity smaller.

ΔxΔtv12 m2 s6 m/s12 m3 s4 m/s12 m6 s2 m/s\begin{array}{c|c|c}\Delta x & \Delta t & v \\ \hline 12\ \text{m} & 2\ \text{s} & 6\ \text{m}/\text{s} \\ 12\ \text{m} & 3\ \text{s} & 4\ \text{m}/\text{s} \\ 12\ \text{m} & 6\ \text{s} & 2\ \text{m}/\text{s}\end{array}
mechanics Two clean positions and a whole number of seconds let us read the velocity off exactly, units and all.