Type on any line or blank, or switch Draw on to write with a finger, stylus or mouse. Your work is saved in this browser on this device only. Nothing is sent anywhere. The answer key after page 2 stays covered until you open it, and it never prints with your work: it has its own print button.

Practice sheet · Automobiles Path · Lens 1: Math

Rates & stopping

Practice for The Math of Automobiles: speed, fuel and trip cost, stopping distance and depreciation (pp.8–14), plus the Algebra Level Up (pp.17–18). Prices are made-up examples.

Warm-upCoreLevel Up

Name Date Time I spent

Remember

distance = speed × time. km/h ÷ 3.6 = m/s. L/100 km: litres ÷ km × 100 (lower is better).

Stopping = reaction + braking. Reaction = speed (m/s) × reaction time. Braking = v² ÷ (2 × μ × 9.81), with μ = 0.7 on dry pavement and about 0.1 on ice (p.11). Double the speed, four times the braking. These are model estimates for a dry road and an alert driver.

Warm-up

One step

One formula, used once.

Red Deer run

Edmonton to Red Deer is about 150 km. How long does it take at 100 km/h? Give hours and minutes.

Metres per second

Convert 80 km/h to metres per second, to one decimal place.

Pump detective

A car used 36 L of fuel to drive 480 km. What is its fuel use in L/100 km (p.9)?

Real average

A family drives 240 km in 2 h 40 min, breaks included. What was their average speed?

Core

Two or three steps

Write the formula, then put the numbers in.

Gas trip

A 520 km trip in a car that uses 7.5 L/100 km. Gas is $1.50/L (example). How many litres, and what does it cost?

Electric trip

The same 520 km in an EV that uses 18 kWh/100 km, charged at home for $0.20/kWh (example). How much energy, and what does it cost?

Rates Practice · Automobiles Path · Unizon1/2

Automobiles Path · Unizon · PracticeRates & stopping distance

Thinking distance

At 80 km/h, how far does a car travel during a 1.5 s reaction? During a 2.5 s reaction by a driver glancing at a phone?

Stopping at 80

Find the braking distance at 80 km/h on dry pavement (μ = 0.7) and the total with a 1.5 s reaction. Then the braking distance on glare ice (μ = 0.1).

Losing value

A $30,000 car loses 20% in year 1, then 15% a year (p.14). What is it worth after 1, 2 and 3 years?

Speed squared

In the book’s model, braking from 40 km/h takes about 9 m. Without the formula, predict the braking distance at 120 km/h. Then check with v² ÷ 178.

Level Up

Lines and curves

From A1–A2 (pp.17–18). Use a calculator for roots.

Solve for speed

Reaction distance with a 1.5 s reaction is R = v ÷ 2.4 metres, for v in km/h (p.17). What speed gives 25 m of reaction distance?

How far for $60?

The gas trip cost on p.17 is C = 0.116d dollars for d km. How far will $60 of gas go? Round to the nearest km.

Skid marks

Braking distance is B = v² ÷ 178 (p.18). A car braked for 36 m. Solve for its speed in km/h.

When does it pay off?

Gas: C = 0.116d. An EV at 20 kWh/100 km and $0.20/kWh costs $0.040 per km, plus a made-up $1,200 home charger: C = 1,200 + 0.040d. Solve for the distance where they cost the same.

Make your own problem

Your turn to be the road engineer

Pick a real trip or a speed from the book: a school zone, a highway, a winter road.
Write a speed, fuel or stopping question with one number missing.
Solve it, check it, then ask: what would make it safer?

My problem, and my worked answer:

Rates Practice · Automobiles Path · Unizon2/2

Automobiles Path · Unizon · PracticeRates & stopping distance

Try every problem first, then check here. Wrong answers are useful: find the step where your work and the key split apart.

The key never prints with Print my work or Print blank. Open it to get its own Print the answer key button.

Answer key

Worked solutions. Small differences from rounding are fine. Nothing here is a reason to drive faster: speed belongs on closed tracks, never on public roads.

Warm-up

1.5 h = 1 h 30 min time = distance ÷ speed = 150 ÷ 100 = 1.5 h.

≈ 22.2 m/s 80 ÷ 3.6 ≈ 22.2.

7.5 L/100 km 36 ÷ 480 × 100 = 7.5.

90 km/h 2 h 40 min = 2 + 40 ÷ 60 = 8/3 h. 240 ÷ (8/3) = 240 × 3 ÷ 8 = 90.

Core

39 L · $58.50 7.5 × 5.2 = 39 L (520 km is 5.2 lots of 100 km). 39 × $1.50 = $58.50.

93.6 kWh · $18.72 18 × 5.2 = 93.6 kWh. 93.6 × $0.20 = $18.72, about a third of the gas cost.

≈ 33.3 m · ≈ 55.6 m 80 ÷ 3.6 ≈ 22.2 m/s. × 1.5 ≈ 33.3 m; × 2.5 ≈ 55.6 m. That extra second costs about 22 m: 5 car lengths.

≈ 36.0 m · ≈ 69.3 m · ≈ 252 m 22.2² ÷ (2 × 0.7 × 9.81) ≈ 36.0 m; + 33.3 = 69.3 m (the table on p.18 agrees). On ice: 22.2² ÷ (2 × 0.1 × 9.81) ≈ 251.7 m, seven times as far.

$24,000 · $20,400 · $17,340 30,000 × 0.80 = 24,000; × 0.85 = 20,400; × 0.85 = 17,340.

≈ 81 m 120 is 3 × 40, so braking is 3² = 9 times as long: 9 × 9 = 81 m. Check: 120² ÷ 178 ≈ 80.9 m.

Level Up

v = 60 km/h 25 = v ÷ 2.4 → v = 25 × 2.4 = 60. Check: 60 ÷ 3.6 × 1.5 = 25 m.

≈ 517 km 60 = 0.116d → d = 60 ÷ 0.116 ≈ 517.2.

v ≈ 80 km/h 36 = v² ÷ 178 → v² = 6,408 → v = √6,408 ≈ 80.0. (Real investigators also measure the road surface.)

d ≈ 15,789 km 0.116d = 1,200 + 0.040d → 0.076d = 1,200 → d ≈ 15,789. Past that, the EV costs less to fuel (ignoring the price of the cars).

Make your own: use real-looking numbers (a trip from a map, a speed from a sign), say what the model assumes (dry road, alert driver), and check that the answer is possible.

Rates Practice · Automobiles Path · UnizonAnswer key