A Thoroughbred gallops at 65 km/h. A migrating bird flies 800 km non-stop. Mia's family drives across Germany in a day. In every case, three quantities are linked: speed, distance, and time. Know any two and you can always find the third.
The triangle formula:
$$d = s \times t \qquad s = \frac{d}{t} \qquad t = \frac{d}{s}$$
Units: If speed is in km/h and time in hours, distance comes out in km. Always check your units match.
Tip: Cover the quantity you want — the triangle shows you how to compute it from the other two: $d$ on top, $s$ and $t$ on the bottom.
Math connection — Lessons 16 & 22: A distance–time graph is a straight line through the origin: $d = s \cdot t$. That is exactly $y = mx$ with slope $m = s$. A steeper line means a faster horse. The common difference in an arithmetic sequence of distances is the speed!
Worked Example
Worked Example — How far?
A horse trots at 12 km/h for 2.5 hours. How far does it travel?
$$d = s \times t = 12 \times 2.5 = 30 \text{ km}$$
If you then need the time to return: $t = d \div s = 30 \div 12 = 2.5$ h. (Same, obviously — a useful check!)
Warm-Up — the formula is given
-
Problem 1
A Thoroughbred gallops at 65 km/h. How far does it travel in 1.5 hours?
$d = 65 \times 1.5$
multiply →
-
Problem 2
Mia's family drove 420 km and it took 3.5 hours. What was their average speed?
$s = d \div t = 420 \div 3.5$
divide →
-
Problem 3
A migrating swallow flies 800 km at 60 km/h. How many hours does the journey take? Give your answer in hours and minutes.
$t = d \div s = 800 \div 60$
divide → convert remainder to minutes →
Core Problems — choose your own formula
-
Problem 4
A Quarter Horse sprints at 88 km/h. How long (in seconds) does it take to run exactly 1 km?
Hint: convert km/h to km/s first, or find time in hours then convert.
write formula → solve → convert to seconds →
-
Problem 5
Two horses leave the stable at the same time going in the same direction. Horse A trots at 18 km/h; Horse B walks at 7 km/h. After how many hours is Horse A exactly 33 km ahead of Horse B?
Gap per hour $= 18 - 7$ · Time $= \text{gap} \div \text{rate of separation}$
rate of separation → time →
-
Problem 6
Plot a distance–time graph for a horse trotting at 15 km/h. Mark the points at $t = 0, 1, 2, 3, 4$ hours. What is the slope of the line? What does that slope represent?
fill in table of (t, d) → sketch graph → slope = rise ÷ run →
-
Problem 7 Challenge
Mia walks to the stables at 4 km/h. Her friend cycles there at 16 km/h but starts 15 minutes after Mia. How long after Mia sets off does the cyclist catch up? How far from the start are they at that moment?
set up equation: Mia's distance = Cyclist's distance → solve for t → find distance →
-
Problem 8 Open
The distance from Munich to Berlin is approximately 584 km. Make a table showing travel time for: a car at 120 km/h, a train at 250 km/h, and a plane at 850 km/h. Then sketch all three lines on one distance–time graph. Which line is steepest, and why?
time = distance ÷ speed → fill table → sketch → explain slope →
Show answers
Problem 1
$d = 65 \times 1.5 = 97.5$ km
Problem 2
$s = 420 \div 3.5 = 120$ km/h
Problem 3
$t = 800 \div 60 = 13.\overline{3}$ h $= 13$ h $20$ min
Problem 4
$t = 1 \div 88$ h $= 3600 \div 88 \approx 40.9$ seconds
Problem 5
Rate of separation $= 18 - 7 = 11$ km/h. Time $= 33 \div 11 = 3$ hours.
Problem 6
Points: $(0,0),(1,15),(2,30),(3,45),(4,60)$. Slope $= 15$ km per hour $=$ the speed. The slope of a distance–time graph is always the speed.
Problem 7
Let $t$ = hours after Mia leaves. Mia's distance $= 4t$. Cyclist's distance $= 16(t - \tfrac{1}{4})$. Set equal: $4t = 16t - 4$ → $12t = 4$ → $t = \tfrac{1}{3}$ h $= 20$ min. Distance $= 4 \times \tfrac{1}{3} \approx 1.33$ km.
Problem 8
Car: $584 \div 120 \approx 4.87$ h. Train: $584 \div 250 \approx 2.34$ h. Plane: $584 \div 850 \approx 0.69$ h. The plane's line is steepest — highest speed = highest slope on a distance–time graph.
Coming up next → s06: Forces and Pressure
Why does a stiletto heel damage a wooden floor more than an elephant's foot? Spoiler: it's all about area. You'll learn $P = F \div A$ and discover why horse shoes matter more than you'd think.