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Rate, Speed and Time

Secondary 1 Mathematics · G3 focus · Study guide

Connect distance, time and speed, convert units, combine rates and explain average speeds with journey graphs.

Concept → worked examples → practice20 min read · practice linked below
Converting seventy-two kilometres per hour to metres per secondSeventy-two kilometres in one hour is seventy-two thousand metres in three thousand six hundred seconds. Dividing both quantities by three thousand six hundred gives twenty metres in one second.Convert both distance and time72 km in 1 h = 72 000 m in 3 600 s20 m in 1 s → 20 m/s

A rate measures change per unit

A tap delivering 12 litres in 3 minutes has an average flow rate of 12÷3=412\div3=4 litres per minute. Speed is a rate: distance travelled per unit of time. A constant speed stays the same throughout the stated interval; an average speed summarises a journey whose speed may vary.

For the interval being measured, average speed=total distancetotal time.\text{average speed}=\frac{\text{total distance}}{\text{total time}}. At constant speed, d=vt,t=dv(v>0).d=vt,\qquad t=\frac dv\quad(v>0). Always make distance and time units match the speed unit before substituting.

“Per hour” means “for each hour”. If speed is in km/h, 45 minutes is 45/60=0.7545/60=0.75 hours. It is not 0.45 hours. A journey's total elapsed time includes its stops unless the question specifically asks for moving time only.

Converting seventy-two kilometres per hour to metres per secondSeventy-two kilometres in one hour is seventy-two thousand metres in three thousand six hundred seconds. Dividing both quantities by three thousand six hundred gives twenty metres in one second.Convert both distance and time72 km in 1 h = 72 000 m in 3 600 s20 m in 1 s → 20 m/s
1 km/h=10003600 m/s=518 m/s1\text{ km/h}=\frac{1000}{3600}\text{ m/s}=\frac5{18}\text{ m/s}. To go from m/s to km/h, multiply by 3.6.

Worked example 1: speed and flow-rate conversions

Convert 72 km/h to m/s. Then convert 3.6 litres per minute to millilitres per second.

  1. Convert both units in the speed: 72 km/h=72×10003600 m/s=20 m/s.72\text{ km/h}=\frac{72\times1000}{3600}\text{ m/s}=\boxed{20\text{ m/s}}.
  2. For the flow rate, 3.6 litres is 3600 ml and one minute is 60 s: 3.6 litres/min=360060 ml/s=60 ml/s.3.6\text{ litres/min}=\frac{3600}{60}\text{ ml/s}=\boxed{60\text{ ml/s}}.

Check: 20 m each second gives 20(3600)=7200020(3600)=72000 m in an hour; 60 ml each second gives 60(60)=360060(60)=3600 ml in a minute.

Average speed uses totals

Worked example 2: a journey with a rest

A cyclist covers 6 km in 20 minutes, rests for 10 minutes, then covers another 9 km in 30 minutes. Find the average speed for the whole outing.

  1. Total distance is 6+9=156+9=15 km. Resting adds no distance.
  2. Total time is 20+10+30=6020+10+30=60 minutes, which is 1 hour.
  3. Average speed is 151=15 km/h.\frac{15}{1}=\boxed{15\text{ km/h}}.

Check: each moving stage has speed 18 km/h, so including a rest must reduce the average below 18 km/h. Dividing by only 50 minutes would answer a different question: average speed while moving.

Worked example 3: equal distances do not mean equal times

A vehicle travels 60 km at 40 km/h and returns along the same route at 60 km/h. There are no stops. Find the average speed.

  1. The outward time is 60/40=1.560/40=1.5 hours; the return time is 60/60=160/60=1 hour.
  2. Total distance is 120 km and total time is 2.5 hours.
  3. Average speed is 1202.5=48 km/h.\frac{120}{2.5}=\boxed{48\text{ km/h}}.

Check: the result lies between 40 and 60 km/h and is closer to the slower speed because more time is spent at that speed. The arithmetic mean (40+60)/2=50(40+60)/2=50 would apply to equal time intervals, not equal distances.

Explore the effect of a stop

A traveller covers two equal 30 km stages. Choose the two speeds and a rest time. Watch how the elapsed time controls the average, even though the total distance stays 60 km.

First stage: 60 min. Stop: 15 min. Second stage: 30 min. Total: 105 min. Average speed: 34.29 km/h.

Cumulative distance travelled over a two-stage journeyThe traveller reaches thirty kilometres at sixty minutes, rests until seventy-five minutes and reaches sixty kilometres at one hundred and five minutes.Distance travelled (km)030600105Time (minutes)
The horizontal section represents the stop. The end-time label changes with the total time; compare the numerical speeds as well as the slopes.

Average speed = 60 ÷ (105/60) = 34.29 km/h. A stop adds time without adding distance.

Combine rates, then find the time

If two workers act independently at constant rates on the same job, add the amounts completed per unit time. Adding their completion times does not give the combined completion time. If one flow removes water while another adds it, subtract the outflow rate from the inflow rate instead.

Worked example 4: two taps fill a tank

Tap A fills 18 litres in 3 minutes and tap B fills 20 litres in 5 minutes. Both run together at those constant rates into an empty 120-litre tank. There is no leak. Find the filling time.

  1. Tap A's rate is 18/3=618/3=6 litres/min; tap B's rate is 20/5=420/5=4 litres/min.
  2. Combined rate is 6+4=106+4=10 litres/min.
  3. Time is amount divided by rate: 120÷10=12 minutes.120\div10=\boxed{12\text{ minutes}}.

Check: in 12 minutes A supplies 72 litres and B supplies 48 litres. Their sum is 120 litres. The conclusion relies on both rates staying constant and both taps starting together.

Use equal distances to form an equation

Worked example 5: an earlier arrival

A coach takes 3 hours to travel a route at a constant speed of vv km/h. At (v+12)(v+12) km/h, it would take 2.5 hours. Find the original speed and the route length.

  1. The route is unchanged. Its length is 3v3v km in the first case and 2.5(v+12)2.5(v+12) km in the second.
  2. Equate the distances: 3v=2.5(v+12)=2.5v+30.3v=2.5(v+12)=2.5v+30.
  3. Subtract 2.5v2.5v, giving 0.5v=300.5v=30. Therefore v=60 km/hv=\boxed{60\text{ km/h}}.
  4. The route is 3(60)=180 km3(60)=\boxed{180\text{ km}}.

Check: 180/72=2.5180/72=2.5 hours. The time saving is 0.50.5 hours, or 30 minutes. Adding 12 to the distance would mix incompatible units.

Read a distance–time graph

On a graph of cumulative distance travelled against time, gradient is speed. A straight rising section means constant speed; a horizontal section means no further distance is travelled. Cumulative distance cannot decrease. A graph of distance from a starting point can decrease on a return journey, so always read the vertical-axis label.

A fifteen-kilometre journey with a ten-minute stopThe plotted points are zero minutes zero kilometres, twenty minutes six kilometres, thirty minutes six kilometres, and sixty minutes fifteen kilometres. The section from twenty to thirty minutes is horizontal.Distance travelled (km)06150203060Time (minutes)Rest
The vertical change is distance; the horizontal change is elapsed time. Use their units when calculating a gradient.

Worked example 6: interpret each graph section

For the graph above, find the first-stage speed, the rest duration and the speed in the final stage.

  1. The first stage covers 6 km in 20 minutes: 620=0.3 km/min=18 km/h.\frac6{20}=0.3\text{ km/min}=\boxed{18\text{ km/h}}.
  2. Between 20 and 30 minutes the distance stays at 6 km, so the rest lasts 10 minutes\boxed{10\text{ minutes}}.
  3. The final stage covers 15−6=915-6=9 km in 60−30=3060-30=30 minutes: 930=0.3 km/min=18 km/h.\frac9{30}=0.3\text{ km/min}=\boxed{18\text{ km/h}}.

Check: the two rising segments are parallel because their speeds are equal. The whole-journey average is still 15 km/h, as the stop is included.

Common mistakes

  • Reading 1 h 30 min as 1.30 h: it is 1+30/60=1.51+30/60=1.5 h.
  • Averaging two speeds automatically: find total distance and total time.
  • Leaving out stops: include them in elapsed time for the whole journey.
  • Reversing a conversion: 1 m/s is 3.6 km/h; a running speed of 5 m/s is 18 km/h.
  • Adding times for two workers: first convert each worker's performance to a rate.
  • Dividing by zero: a zero speed cannot complete a positive distance in finite time.

Continue with the question bank

Use the rate, speed and time question bank for journeys, unit conversions and combined work. For plotting coordinates and understanding gradient, read Linear Graphs.