Motion in One Dimension

Constant Acceleration

speeding up from rest

When a cart speeds up steadily from rest, its speed grows with time and its distance grows with time squared, so equal seconds make growing gaps.

mechanics Starting from rest with a clean acceleration keeps every step exact, and the growing gaps between equal-time snapshots are the whole point.

Example

When a cart speeds up steadily from rest, its speed grows with time and its distance grows with time squared, so equal seconds make growing gaps. 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 motion

A cart starts from rest and speeds up at 2 metres per second squared for 4 seconds. Starting from rest means the first speed is zero.

a=2 m/s2,t=4 sa = 2\ \text{m}/\text{s}^{2}, \quad t = 4\ \text{s}
A cart at rest, about to accelerateA cart at the starting line with an acceleration arrow pointing in the direction it will speed up.carta

Speed after the time

With no starting speed, velocity is acceleration times time: 2 times 4 is 8 metres per second. The seconds-squared underneath cancels one second, leaving metres per second.

v=a t=2 m/s2 ⋅ 4 s=8 m/sv = a\,t = 2\ \text{m}/\text{s}^{2} \,\cdot\, 4\ \text{s} = \hl{8}\ \text{m}/\text{s}

Speed grows evenly with time

Hold the acceleration at 2 metres per second squared. Each longer time gives a larger speed by the same multiplication.

atv2 m/s21 s2 m/s2 m/s22 s4 m/s2 m/s24 s8 m/s\begin{array}{c|c|c}a & t & v \\ \hline 2\ \text{m}/\text{s}^{2} & 1\ \text{s} & 2\ \text{m}/\text{s} \\ 2\ \text{m}/\text{s}^{2} & 2\ \text{s} & 4\ \text{m}/\text{s} \\ 2\ \text{m}/\text{s}^{2} & 4\ \text{s} & 8\ \text{m}/\text{s}\end{array}

The distance formula from rest

Distance from rest is one half the acceleration times the time squared. The one half is there because the speed builds up evenly from zero.

x=12 a t2x = \tfrac{1}{2}\,a\,t^{2}

Substitute the numbers

Put in 2 for the acceleration and 4 for the time. The time squared is 16.

x=12 ⋅ 2 m/s2 ⋅ 42=12 ⋅ 2 ⋅ 16 mx = \tfrac{1}{2} \,\cdot\, 2\ \text{m}/\text{s}^{2} \,\cdot\, 4^{2} = \tfrac{1}{2} \,\cdot\, 2 \,\cdot\, 16\ \text{m}

Compute the distance

That gives 16 metres. Notice the snapshots at each second spread further apart: equal times, growing gaps.

x=16 mx = \hl{16}\ \text{m}
Equal time steps land at growing gapsSnapshots at each whole second spread further and further apart and the velocity arrows grow, because the cart keeps speeding up.t=1t=2t=3cartv

Fixed time makes distance track acceleration

Hold the time at 4 seconds. Stronger acceleration makes a proportionally larger distance from rest.

atx1 m/s24 s8 m2 m/s24 s16 m3 m/s24 s24 m\begin{array}{c|c|c}a & t & x \\ \hline 1\ \text{m}/\text{s}^{2} & 4\ \text{s} & 8\ \text{m} \\ 2\ \text{m}/\text{s}^{2} & 4\ \text{s} & 16\ \text{m} \\ 3\ \text{m}/\text{s}^{2} & 4\ \text{s} & 24\ \text{m}\end{array}

Time changes distance by a square

Hold the acceleration at 2 metres per second squared. Doubling time does more than double the distance because the time is squared.

atx2 m/s21 s1 m2 m/s22 s4 m2 m/s24 s16 m\begin{array}{c|c|c}a & t & x \\ \hline 2\ \text{m}/\text{s}^{2} & 1\ \text{s} & 1\ \text{m} \\ 2\ \text{m}/\text{s}^{2} & 2\ \text{s} & 4\ \text{m} \\ 2\ \text{m}/\text{s}^{2} & 4\ \text{s} & 16\ \text{m}\end{array}