In NSW, speeding, fatigue and alcohol are the three main causes of death and injury on our roads. Of these, speeding is the biggest factor, contributing to around $40%$40% of road fatalities each year.
Research has shown that even small reductions in speed, can help reduce the number of deaths and the severity of injury. If a car hits a pedestrian at $50$50 km/h, the pedestrian is twice as likely to die than if they'd been hit at $40$40 km/h.
The average speed of a vehicle is related to the total distance travelled and the total time taken by the formula:
$\text{average speed}=\frac{\text{total distance travelled}}{\text{total time taken}}$average speed=total distance travelledtotal time taken
$S=\frac{D}{T}$S=DT
We can easily rearrange this formula to make either $D$D or $T$T the subject:
Rearranging to make $D$D the subject:
$S$S | $=$= | $\frac{D}{T}$DT | ||
$S\times T$S×T | $=$= | $\frac{D}{T}\times T$DT×T | (Multiply both sides by $T$T) | |
$ST$ST | $=$= | $D$D | ||
$D$D | $=$= | $ST$ST |
Rearranging to make $T$T the subject:
$S$S | $=$= | $\frac{D}{T}$DT | ||
$S\times T$S×T | $=$= | $\frac{D}{T}\times T$DT×T | (Multiply both sides by $T$T) | |
$ST$ST | $=$= | $D$D | ||
$\frac{ST}{S}$STS | $=$= | $\frac{D}{S}$DS | (Divide both sides by $S$S) | |
$T$T | $=$= | $\frac{D}{S}$DS |
Jim drives his car from Sydney to Canberra, a distance of $279$279 km. The journey takes him $3$3 hours and $45$45 minutes, including lunch and fuel stops along the way.
(a) What is Jim's average speed for the journey, correct to the nearest km/h?
Think: Whenever we use a formula, we must keep units consistent. In this case, speed is expressed in kilometres per hour and distance is in kilometres, so time must be expressed in hours.
Our first step is to convert the time of $3$3 hours $45$45 minutes into hours. Because there are $60$60 minutes in an hour, $45$45 minutes is equivalent to $\frac{45}{60}$4560 or $0.75$0.75 hours. Therefore $3$3 hours $45$45 minutes is the same as $3.75$3.75 hours.
Do: Substituting into the speed-distance-time formula:
$S$S | $=$= | $\frac{D}{T}$DT | ||
$=$= | $\frac{279}{3.75}$2793.75 | (Substitute $D=279$D=279 and $T=3.75$T=3.75) | ||
$=$= | $74.4$74.4 | |||
$=$= | $74$74 km/h (nearest km/h) |
(b) How far could Jim travel at this same average speed in $40$40 minutes. Round your answer to the nearest kilometre.
Think: A time of $40$40 minutes is equivalent to $\frac{40}{60}$4060 hours, which simplifies to $\frac{2}{3}$23 of an hour. In this case it is easier to leave the time as a fraction because the decimal is recurring. It is also easier to substitute our values into the formula when $D$D is the subject.
Do:
$D$D | $=$= | $ST$ST | ||
$=$= | $74\times\frac{2}{3}$74×23 | (Substitute $S=74$S=74 and $T=\frac{2}{3}$T=23) | ||
$=$= | $\frac{148}{3}$1483 | |||
$=$= | $49.333\ldots$49.333… | |||
$=$= | $49$49 km (nearest km) |
(c) If Jim travelled at this same average speed, how long would it take him to drive from Sydney to Melbourne, a distance of $875$875 km. Give your answer in hours and minutes.
Think: This time, it will be easier to substitute our values into the formula when $T$T is the subject.
Do:
$T$T | $=$= | $\frac{D}{S}$DS | ||
$T$T | $=$= | $\frac{875}{74}$87574 | (Substitute $D=875$D=875 and $S=74$S=74) | |
$T$T | $=$= | $11.824\ldots$11.824… | ||
$T$T | $=$= | $11$11 hours $49$49 minutes (nearest minute) |
Reflect: In the final steps of the calculation, notice that our value for time was equal to $11.82432\ldots$11.82432… hours. To convert this time into hours and minutes, we simply multiplied the decimal portion by $60$60. To isolate the decimal portion, subtract $11$11 first, then multiply by $60$60. This gives $49.4594\ldots$49.4594… minutes, or $49$49 minutes, rounded to the nearest minute.
Speed limits are used to encourage safe driving on our roads. They are displayed on white signs with a red circle surrounding the speed limit in black text.
In NSW, the following speed limits are commonly used and enforced by law:
There are also advisory speed limits. These are yellow signs with black text and symbols. They indicate a recommended safe speed for certain road conditions like sharp bends and slippery surfaces.
Few people may realise just how much distance is required to stop a vehicle travelling at speed.
Let's say we are driving a car at $60$60 km/h and suddenly a dog runs onto the road. The time it takes for us to see the dog, recognise the danger, and apply the brakes is called our reaction-time. The distance our car travels during that time is called the reaction-time distance. Once the brakes are applied, the additional distance covered by the car before it comes to a complete stop, is called the braking distance.
If we add these distances together we get the total stopping distance:
$\text{stopping distance}=\text{reaction-time distance}+\text{braking distance}$stopping distance=reaction-time distance+braking distance
The reaction-times of most people will fall into a range between $0.7$0.7 and $3$3 seconds. Even for a driver who is fit, concentrating, alert and not affected by alcohol, drugs, fatigue or distraction, their reaction-time on average will still be about $1.5$1.5 seconds.
The following graphic shows the reaction-time distance and braking distance for an average family car travelling at different speeds on different road conditions.
Consider the following situation, where one car is travelling at $40$40 km/h and another car is travelling at $50$50 km/h. Each car is travelling on a dry road.
Both drivers see a pedestrian $27$27 metres ahead, recognise the danger, and apply their brakes.
The car travelling at $40$40 km/h will take $26$26 metres to stop, and will avoid hitting the pedestrian.
The car travelling at $50$50 km/h will take an extra nine metres to stop, and will still be travelling at $41$41 km/h when it hits the pedestrian.
If a driver's reaction-time is $1.5$1.5 seconds and their vehicle is travelling at $60$60 km/h, what is the reaction-time distance, in metres.
Solution
First, we convert $60$60 km/h into metres per second (m/s). To do this we multiply $60$60 by $1000$1000, to convert from kilometres to metres. We then divide by $3600$3600, the number of seconds in one hour. This gives us $\frac{50}{3}$503 m/s.
$D$D | $=$= | $ST$ST |
$=$= | $\frac{50}{3}\times1.5$503×1.5 | |
$=$= | $25$25 m |
So the reaction-time distance is $25$25 metres. This is the distance travelled by the car during the $1.5$1.5 seconds it takes the driver to react to a potential hazard.
A formula for calculating the stopping distance $D$D, in metres, of a vehicle travelling at speed $V$V, in metres per second, is given as follows:
$D=1.5V+\frac{V^2}{13.72}$D=1.5V+V213.72
Amanda is driving her car at $75$75 km/h, when she notices a hazard on the road ahead. She recognises the danger and applies the brakes to bring her car to a complete stop.
(a) Use the formula to calculate her stopping distance. Round your answer up to the next whole metre.
Think: First convert $75$75 km/h to m/s by multiplying by $1000$1000 and dividing by $3600$3600. So $75$75 km/h is equivalent to $\frac{125}{6}$1256 m/s. Notice that we keep the speed as a fraction to avoid rounding.
Do: We can now substitute our values into the formula:
$D$D | $=$= | $1.5V+\frac{V^2}{13.72}$1.5V+V213.72 | ||
$=$= | $1.5\times\frac{125}{6}+\frac{\left(\frac{125}{6}\right)^2}{13.72}$1.5×1256+(1256)213.72 | (Substitute $V=\frac{125}{6}$V=1256) | ||
$=$= | $31.25+31.634\ldots$31.25+31.634… | |||
$=$= | $62.884\ldots$62.884… | |||
$=$= | $63$63 m (next whole metre) |
(b) If the term $1.5V$1.5V in the formula is the reaction-time distance, what percentage of her total stopping distance is determined by her reaction-time? Round your answer to one decimal place.
Think: From our working above, we can see that the term $1.5V$1.5V is equal to $1.5\times\frac{125}{6}$1.5×1256 or $31.25$31.25 m. This is the reaction-time distance. To express this value as a percentage, we divide by the total stopping distance, and multiply by $100$100.
Do:
Percentage of total stopping distance | $=$= | $\frac{31.25}{62.884}\times100$31.2562.884×100 |
$=$= | $49.694\ldots$49.694… | |
$=$= | $49.7%$49.7% ($1$1 d.p.) |
Stopping distance is determined by a number of factors. The first four, listed below, affect the reaction-time distance, while the rest are more likely to affect the braking distance.
Drink driving is a factor in about one in every seven fatal car crashes in NSW. In the five year period from 2013 to 2017, $93%$93% of all those killed in alcohol-related car accidents, were men and $67%$67% were under the age of $40$40.
In 1980, there were $389$389 deaths in NSW due to alcohol-related car accidents, $30%$30% of the road toll. In 1982, Random Breath Testing (RBT) was introduced as a way of reducing these fatalities. RBT and advertising campaigns against drink driving have proven successful with a steady decline in fatalities in the years since their introduction. By 2017 the figure had dropped to $55$55 deaths, about $14%$14% of the road toll.
Even very small amounts of alcohol can affect a person's concentration, judgement and performance. Although a person may think they are fine to drive, their reaction-times will be slower than normal.
For the safety of everyone on the road, there are legal limits for the amount of alcohol a person can have in their system while driving. Known as blood alcohol content (BAC), it is a measure of the concentration of alcohol in the blood. A reading of $0.05$0.05 ('point o five') for example, means there are $0.05$0.05 grams ($50$50 mg) of alcohol in every $100$100 mL of blood.
In NSW, the following BAC limits apply:
There are several formulas that are used to determine a person's blood alcohol content (BAC), each based on the concept of a standard drink.
By describing an alcoholic drink in terms of the number of standard drinks it contains, we can compare its alcoholic content with that of other drinks. Therefore, a standard drink is a unit of measurement.
In Australia, a standard drink is any drink containing $10$10 grams of alcohol. One standard drink always contains the same amount of alcohol, regardless of the size of the container or the type of alcohol within it.
As a guide, one standard drink is equivalent to:
We can work out the number of standard drinks in a particular quantity of alcoholic drink using a formula. If $V$V is the volume, in litres (L), of the drink, and $A$A is the percentage of alcohol ($%$% alc/vol), then the number of standard drinks, $N$N, is given by:
$N=0.789\times V\times A$N=0.789×V×A
The number $0.789$0.789 in the formula is a constant known as the specific gravity of ethyl alcohol.
Calculate the number of standard drinks in each of the following. In each case, round your answer to one decimal place.
(a) A $375$375 mL bottle (stubbie) of beer, with an alcohol content of $4.8%$4.8% alc/vol.
Think: When we use this particular formula, there is no need to convert the percentage to a decimal, so $A=4.8$A=4.8. The volume, however, needs to be in litres. We know there are $1000$1000 mL in $1$1 litre, so we divide $375$375 by $1000$1000 to get $V=0.375$V=0.375 L.
Do: Substituting these values into the formula, we get the number of standard drinks:
$N$N | $=$= | $0.789\times V\times A$0.789×V×A | ||
$=$= | $0.789\times0.375\times4.8$0.789×0.375×4.8 | (Substitute $V=0.375$V=0.375 and $A=4.8$A=4.8) | ||
$=$= | $1.4202$1.4202 | |||
$=$= | $1.4$1.4 ($1$1 d.p.) |
So there are $1.4$1.4 standard drinks in a $375$375 mL bottle of beer ($4.8%$4.8% alc/vol).
(b) A $750$750 mL bottle of red wine, with an alcohol content of $13%$13% alc/vol.
Think: Once again, we convert the volume of the container to litres and use the percentage of alcohol as it is given.
Do:
$N$N | $=$= | $0.789\times V\times A$0.789×V×A | ||
$=$= | $0.789\times0.75\times13$0.789×0.75×13 | (Substitute $V=0.75$V=0.75 and $A=13$A=13) | ||
$=$= | $7.692\ldots$7.692… | |||
$=$= | $7.7$7.7 ($1$1 d.p.) |
So there are $7.7$7.7 standard drinks in a $750$750 mL bottle of red wine ($13%$13% alc/vol).
The following BAC formulas are used to predict a person's blood alcohol content after they consume a given number of standard drinks, $N$N, in a given number of hours, $H$H.
The formula is different for males and females and depends on the person's mass, $M$M, in kilograms.
$BAC_{male}=\frac{10N-7.5H}{6.8M}$BACmale=10N−7.5H6.8M | $BAC_{female}=\frac{10N-7.5H}{5.5M}$BACfemale=10N−7.5H5.5M |
A male weighing $78$78 kg consumes $4$4 standard drinks in $1.5$1.5 hours. What is his predicted blood alcohol content? Answer to three decimal places.
Solution
$BAC_{male}$BACmale | $=$= | $\frac{10N-7.5H}{6.8M}$10N−7.5H6.8M | ||
$=$= | $\frac{10\times4-7.5\times1.5}{6.8\times78}$10×4−7.5×1.56.8×78 | (Substitute $N=4$N=4, $H=1.5$H=1.5 and $M=78$M=78) | ||
$=$= | $\frac{28.75}{530.4}$28.75530.4 | |||
$=$= | $0.0542\ldots$0.0542… | |||
$=$= | $0.054$0.054 ($3$3 d.p.) |
After a person stops drinking alcohol, it will take time for their blood alcohol content to return to zero.
The body, but specifically the liver, processes alcohol in the blood over time. Although everyone is different, the amount of alcohol in the blood is reduced at a rate of $0.015$0.015 BAC per hour.
To determine the duration of time $t$t, in hours, it will take for a person's BAC to return to zero, we can use the following formula:
$t=\frac{BAC}{0.015}$t=BAC0.015
Let's look at an example of how all these formulas are used. There is no need to memorise the formulas, as they will be given in questions.
Sarah is $18$18 years old and holds a provisional P1 license, with a zero alcohol limit. She started drinking at $10$10 pm on Friday night and had $6$6 alcoholic mixer drinks in $4$4 hours. Sarah has a mass of $58$58 kg.
(a) Each alcoholic mixer drink that Sarah drank was a $330$330 mL bottle with an alcohol content of $5%$5% alc/vol. Calculate the equivalent number of standard drinks she consumed over the four hours, rounding your answer to one decimal place.
Think: Here we use the formula for standard drinks. We can use the alcohol percentage as is, but the volume of liquid consumed must be in litres. For six $330$330 mL drinks, the total volume of liquid consumed is $6\times330=1980$6×330=1980 mL. Divide this by $1000$1000 to get $1.98$1.98 L.
Do:
$N$N | $=$= | $0.789\times V\times A$0.789×V×A | ||
$=$= | $0.789\times1.98\times5$0.789×1.98×5 | (Substitute $V=1.98$V=1.98 and $A=5$A=5) | ||
$=$= | $7.8111$7.8111 | |||
$=$= | $7.8$7.8 ($1$1 d.p.) |
So Sarah consumed the equivalent of $7.8$7.8 standard drinks.
(b) Using your answer from part (a), calculate her BAC when she stopped drinking at $2$2 am on Saturday morning. Round your answer to two decimal places.
Think: We use the BAC formula for females to calculate her blood alcohol content after she finished drinking.
Do:
$BAC_{female}$BACfemale | $=$= | $\frac{10N-7.5H}{5.5M}$10N−7.5H5.5M | ||
$=$= | $\frac{10\times7.8-7.5\times4}{5.5\times58}$10×7.8−7.5×45.5×58 | (Substitute $N=7.8$N=7.8, $H=4$H=4 and $M=58$M=58) | ||
$=$= | $\frac{48}{319}$48319 | |||
$=$= | $0.1504\ldots$0.1504… | |||
$=$= | $0.15$0.15 ($2$2 d.p.) |
So Sarah's BAC after she finished drinking was $0.15$0.15.
(c) Using your answer from part (b), how many hours will it take for Sarah's BAC to return to zero?
Think: To work out the number of hours it will take for Sarah's BAC to return to zero, divide her BAC by $0.015$0.015.
Do:
$t$t | $=$= | $\frac{BAC}{0.015}$BAC0.015 | ||
$=$= | $\frac{0.15}{0.015}$0.150.015 | (Substitute $BAC=0.1508\ldots$BAC=0.1508…) | ||
$=$= | $10$10 hours |
So it will take at least $10$10 hours for Sarah's blood alcohol content to return to zero.
(d) A what time on Saturday would Sarah be fine to drive her car?
Solution: Sarah stopped drinking at $2$2 am on Saturday morning when her BAC was measured. If we add on $10$10 hours, we get a time of $12$12 pm. So it will be midday on Saturday, before Sarah can consider driving her car.
Alcohol affects people differently. Two people can drink exactly the same amount of alcohol but have different readings for their blood alcohol content.
The formulas used in this topic for BAC provide estimates only. Their accuracy will vary depending on a variety of factors:
To calculate the number of standard drinks we use the formula $N=0.789\times V\times A$N=0.789×V×A; where
$V$V is the volume of the drink in litres (L); and
$A$A is the percentage of alcohol (% alc/vol) in the drink.
Calculate the number of standard drinks in a $375$375 mL can of pre-mix drink that has an alcohol content of $7%$7% alc/vol to the nearest decimal place.
To calculate the blood alcohol content (BAC) of a person we use the formula $BAC_{male}=\frac{10N-7.5H}{6.8M}$BACmale=10N−7.5H6.8M for males and $BAC_{female}=\frac{10N-7.5H}{5.5M}$BACfemale=10N−7.5H5.5M for females; where
$N$N is the number of standard drinks consumed;
$H$H is the number of hours of drinking; and
$M$M is the person's mass in kilograms.
Bill is an adult male that weighs $87$87kg and has consumed $4$4 standard drinks in $4$4 hours. Calculate his blood alcohol content correct to three decimal places.
We want to calculate the BAC of a female with a mass of $53$53kg, who drinks three $425$425ml bottles of full strength beer with an alcohol content of $4.8%$4.8% alc/vol, over $3$3 hours.
First calculate the number of standard drinks correct to one decimal place. Remember we use the formula $N=0.789\times V\times A$N=0.789×V×A; where
$V$V is the volume of the drink(s) in litres (L); and
$A$A is the percentage of alcohol (% alc/vol) in the drink(s).
Using your answer to part (a), calculate her blood alcohol content (BAC) correct to three decimal places. Remember, we use the formula $BAC_{female}=\frac{10N-7.5H}{5.5M}$BACfemale=10N−7.5H5.5M where;
$N$N is the number of standard drinks consumed;
$H$H is the number of hours of drinking; and
$M$M is the person's mass in kilograms.
makes predictions about everyday situations based on simple mathematical models