Using the equations: stopping distance and the nth second

Why speed limits near schools are so low

A child runs out between parked cars. The driver sees her. What happens next takes two stages, and both cost road:

1. Thinking. For the driver's reaction time, about $0.7\ \text{s}$ to $1.5\ \text{s}$, the foot is still moving to the brake. The car keeps its full speed.
2. Braking. Then the brakes slow the car steadily to rest.

The braking part comes from $v^2 = u^2 + 2as$ with the final velocity $0\ \text{m/s}$ and $a = -b$.

A worked example. A car moves at $25\ \text{m/s}$ ($90\ \text{km/h}$). The driver reacts in $1.0\ \text{s}$, and the brakes give $5\ \text{m/s}^2$.

1. Thinking. $25\ \text{m/s} \times 1.0\ \text{s} = 25\ \text{m}$.
2. Braking. $\dfrac{(25\ \text{m/s})^2}{2 \times 5\ \text{m/s}^2} = 62.5\ \text{m}$.
3. Stopping distance. $87.5\ \text{m}$, close to the length of a football pitch.

The square law

Try the simulation: double the speed.

The thinking distance doubles, but the braking distance becomes four times as long, because it grows as $v^2$. At $50\ \text{km/h}$ instead of $30\ \text{km/h}$ near a school, the car needs almost three times the braking distance. That is the physics behind low speed limits.

Watch out. The reaction time adds to the stopping distance but not to the braking distance. A phone in the hand makes the reaction time longer, and at $25\ \text{m/s}$ every extra second is $25\ \text{m}$ more.

The distance in one second

From rest with a steady acceleration $a$, the distance covered in the first $n$ seconds is $\tfrac12 a n^2$. The distance in the $n$th second alone is the difference between $n$ seconds and $n - 1$ seconds:

A worked example. From rest at $3\ \text{m/s}^2$, the $4$th second covers $\dfrac{3\ \text{m/s}^2}{2} \times (8 - 1) \times (1\ \text{s})^2 = 10.5\ \text{m}$.

The distances in the $1$st, $2$nd, $3$rd, ... seconds go as $1 : 3 : 5 : 7 : \ldots$, the odd numbers. Galileo found this pattern by rolling balls down a gentle slope and timing them with a water clock.

Watch out. "In the $4$th second" and "in $4$ seconds" are different questions. One is a single second, the other all four.

Practise with new numbers

Try it in Python

The program lists the stopping distance at speeds from $0\ \text{m/s}$ to $35\ \text{m/s}$, and draws the curve.

Remember these

• Stopping distance $= v\,t_r + \dfrac{v^2}{2b}$. Braking distance grows as $v^2$.
• From rest: $s_n = \dfrac{a}{2}(2n - 1)$; successive seconds go $1 : 3 : 5 : \ldots$

Check yourself

1. A driver at $20\ \text{m/s}$ has a reaction time of $0.7\ \text{s}$. What is her thinking distance?

2. With the same brakes, a car's speed is tripled. What happens to its braking distance?

3. From rest at $6\ \text{m/s}^2$, how far does a body go in the $2$nd second?

4. Why does a longer reaction time add to the stopping distance but not to the braking distance?