Speed, distance, time and stopping distance

Average speed, distance and time with unit changes, converting between km/h and m/s, decimal hours, and how reaction distance and braking distance add up to a car's stopping distance.

A car at $60$ km/h covers about $17$ metres every second. In the second it takes you to notice a child stepping out, you have already travelled the length of four cars, and you haven't touched the brake yet. This section puts numbers on that.

Speed, distance and time

Speed, distance, time

$$S = \frac{D}{T} \qquad D = S \times T \qquad T = \frac{D}{S}$$

These are one formula rearranged three ways. The units must match: kilometres with hours gives km/h, metres with seconds gives m/s. Change the units before you substitute.

Example 1●

A train travels $23$ km in $25$ minutes. Find its average speed in km/h.

$T = \dfrac{25}{60}$ h

The answer is wanted in km/h, so turn minutes into hours: divide by 60.

$S = 23 \div \dfrac{25}{60} = 55.2$ km/h

Distance over time.

Calculators give time as decimal hours, like $2.7$ h. That is not $2$ h $70$ min. The decimal part is a fraction of an hour, so multiply it by $60$: $0.7 \times 60 = 42$, giving $2$ h $42$ min.

km/h and m/s

One km/h means $1000$ m in $3600$ s, which is $\dfrac{1}{3.6}$ m/s. So:

Converting speeds

$$\text{km/h} \xrightarrow{\ \div 3.6\ } \text{m/s} \qquad \text{m/s} \xrightarrow{\ \times 3.6\ } \text{km/h}$$

Example 2●

Convert $60$ km/h to m/s, then find how far a car at that speed travels in $1.5$ seconds.

$60 \div 3.6 \approx 16.67$ m/s

Divide by 3.6.

$D = \dfrac{60}{3.6} \times 1.5 = 25$ m

Distance is speed times time. Using the exact value 60 ÷ 3.6 avoids rounding error.

Stopping distance

When a driver sees a hazard, the car keeps going at full speed while the driver reacts. Only then do the brakes start to slow it down.

Stopping distance

$$\text{stopping distance} = \text{reaction distance} + \text{braking distance}$$

Reaction distance is speed times reaction time. An alert driver reacts in about $0.75$ to $1$ second. Tired, distracted or drink drivers often take $2$ seconds or more.
Braking distance depends on the square of the speed, and on the road, tyres and brakes. Wet or gravel roads make it much longer.

Exam questions give you the formula. A common one, for a car on a dry sealed road, is:

One common approximation

$$d = \frac{5Vt}{18} + \frac{V^2}{170}$$

Here $d$ is in metres, $V$ is the speed in km/h and $t$ is the reaction time in seconds. The first part is just $\dfrac{V}{3.6} \times t$, the reaction distance, because $\dfrac{5}{18} = \dfrac{1}{3.6}$. The second part is the braking distance.

Example 3●●

A driver is doing $60$ km/h with a reaction time of $1.5$ s. Use the formula to find the stopping distance, to the nearest metre.

$\dfrac{5 \times 60 \times 1.5}{18} = 25$ m

Reaction distance: same as Example 2.

$\dfrac{60^2}{170} = \dfrac{3600}{170} \approx 21.2$ m

Braking distance: 60 squared is 3600.

$d \approx 25 + 21.2 = 46.2 \approx 46$ m

Add them and round.

InteractiveStopping before the crossing

Try this

A child steps onto a crossing $40$ m ahead. At $50$ km/h on a dry road with an alert driver ($1$ s), does the car stop in time? Now change only the speed: try $60$ km/h, then $65$ km/h. Then go back to $50$ km/h and make the driver distracted ($2$ s). Which change is worse, and why does the impact speed jump so much?

Your turn

Use $d = \dfrac{5Vt}{18} + \dfrac{V^2}{170}$ to find the stopping distance at $80$ km/h with a reaction time of $1$ second, to the nearest metre.

Show the working

Reaction: $\dfrac{5 \times 80 \times 1}{18} \approx 22.22$ m. Braking: $\dfrac{6400}{170} \approx 37.65$ m. Total $\approx 59.87$, so about $60$ m.

Flashcards

km/h to m/s

Divide by $3.6$

Average speed

Total distance ÷ total time, never the average of the speeds

Stopping distance

Reaction distance + braking distance

Double the speed

About four times the braking distance

Practice

A car travels $150$ km in $2.5$ hours. Find its average speed in km/h.
Convert $72$ km/h to m/s.
How many hours does it take to drive $210$ km at $84$ km/h?
$3.4$ hours is the same as:
A car's reaction distance is $12$ m and its braking distance is $19$ m. What is the stopping distance?
A car's speed doubles. Its braking distance becomes about:
A car travels $40$ km at $80$ km/h on the motorway, then $40$ km at $40$ km/h through roadworks. Find the average speed, to one decimal place.
Use $d = \dfrac{5Vt}{18} + \dfrac{V^2}{170}$ to find the stopping distance, to the nearest metre, when $V = 80$ and $t = 1$.
Exam style: A driver at $100$ km/h is reading a text and takes $2$ seconds to react. Find the reaction distance in metres, to one decimal place.
Exam style: A driver at $70$ km/h with a reaction time of $1.2$ s sees a fallen tree $45$ m ahead. Using $d = \dfrac{5Vt}{18} + \dfrac{V^2}{170}$, which statement is correct?