Time, Speed and Distance

Very often the speed is given in one unit and the answer is wanted in another. Keep the bridge ready: 1 km/h = 518 m/s and 1 m/s = 185 km/h.

Which fraction to multiply by? Think about the size of the number. The same speed always has a smaller number in m/s than in km/h (5 m/s = 18 km/h). So from km/h to m/s the number must shrink, and you multiply by 518. From m/s to km/h the number must grow, and you multiply by 185.

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Many questions hide the conversion inside a story. "A boy runs 300 m in 25 seconds; find his speed in km/h." Speed = 30025 = 12 m/s, and 12 × 185 = 43.2 km/h. Or "A car runs at 45 km/h; how many metres does it cover in 16 seconds?" 45 × 518 = 12.5 m/s, and 12.5 × 16 = 200 m.

When the distance is in metres and the time in minutes, first find metres per minute. Multiplying by 60 gives metres per hour, and dividing by 1000 gives km/h. Put together, metres per minute × 350 = km/h. For instance, 150 m per minute = 150 × 350 = 9 km/h.

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Radhika's school bus runs at 63 km/h. In m/s that is 63 × 518 = 17.5 m/s, so in one minute the bus covers 17.5 × 60 = 1050 m.

This is the direct use of D = S × T: the question gives two quantities and you find the third. Three habits make these questions easy.

Habit 1: keep time as a fraction. 40 minutes = 23 hour, 2 hours 30 minutes = 52 hours, 1 hour 45 minutes = 74 hours. Then speed = distance × 1time is a single multiplication. For example, 15 km in 45 minutes gives 15 × 43 = 20 km/h.

Habit 2: same speed means the unitary method. If 24 km take 3 hours, 40 km take 4024 × 3 = 5 hours. You do not even need the speed, because distance and time grow together.

Habit 3: break the journey into parts. If a journey has two or more parts, write time = distance ÷ speed for each part and add. Add the distances and add the times; never add the speeds.

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The two-mode journey is a standard question. "A man covers 26 km in 5 hours, partly walking at 4 km/h and partly cycling at 8 km/h. How far did he walk?" Let the walking time be t. Then 4t + 8(5 − t) = 26, so 40 − 4t = 26 and t = 3.5 h. He walked 14 km. The same answer comes without an equation. Had he cycled all 5 hours, he would have covered 40 km. Each hour of walking instead costs him 4 km, and he is 14 km short, so he walked for 144 = 3.5 h.

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Some questions have a speed that increases every hour (a progression), and some say what fraction of the journey is complete. In both cases write down how much is covered and how much is left before you calculate.

Radhika covers 25 of a 25 km route in 50 minutes. That is 10 km in 56 h, so her speed is 10 × 65 = 12 km/h. The remaining 15 km at this speed take 1512 = 114 h.

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Here two people or vehicles are compared, or one vehicle runs the same route at two speeds. The three ratio rules from the foundation do all the work.

Same distance: the time ratio is the speed ratio turned upside down. Speeds 5 : 3 give times 3 : 5. If a bus takes 10 hours for a route and a car's speed is to the bus's as 5 : 4, the car takes 10 × 45 = 8 hours. This also gives a handy product rule: for the same distance, S₁ × T₁ = S₂ × T₂. At 60 km/h a trip takes 6 h; at 45 km/h, 60 × 6 = 45 × T gives T = 8 h.

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Same time: the distance ratio equals the speed ratio. If speeds are 4 : 5 and both move for the same time, distances are 4 : 5.

Three vehicles, same distance. Times 3 : 4 : 6 give speeds 13 : 14 : 16. Multiply by the LCM 12 and you get 4 : 3 : 2. With three numbers, do not just reverse the order: 3 : 4 : 6 reversed is 6 : 4 : 3, which is wrong. Take reciprocals.

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Speed hidden in steps or jumps. Suppose A takes 5 jumps while B takes 4, and 3 jumps of A equal 2 jumps of B. Speed = (jumps per unit time) × (length of one jump). Let one jump of B be 3 units; then one jump of A is 2 units. Speeds = 5 × 2 : 4 × 3 = 10 : 12 = 5 : 6.

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Radhika walks at 4 km/h and her brother cycles at 12 km/h on the same 6 km road. Speeds are 1 : 3, so times are 3 : 1: Radhika takes 112 h and her brother 12 h.

Now the distance stays the same but the speed changes. The person walks slower than usual, the train speeds up, the car slows down. Because the distance is fixed, time changes in the inverse ratio of speed.

Speed becomes a fraction of the usual speed. If the new speed is ab of the usual speed, the new time is ba of the usual time. So the change in time is ba − 1 = b − aa of the usual time.

Say Radhika walks at 34 of her usual speed and reaches 12 minutes late. Her new time is 43 of the usual time, so she is late by 13 of the usual time. 13 × usual time = 12, which gives usual time = 36 minutes.

In parts it is even shorter. Speeds usual : new = 4 : 3, so times are 3 : 4. The 1-part gap is 12 minutes, so the usual time is 3 parts = 36 minutes.

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Speed changes by a percentage. Write the percentage as a fraction first. Speed up by 20% means speed × 65, so time × 56, and the time saved is 16 of the usual time. Speed down by 20% means speed × 45, so time × 54, and the delay is 14 of the usual time. Look carefully: a 20% rise in speed saves only 16 (16.67%) of the time, not 20%.

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Speed changes by a fixed amount (km/h) and the time is known. Find the distance first (usual speed × usual time). The new time is then distance ÷ new speed.

The train that is faster but stops. A train is 50% faster than a car but stops on the way, and both arrive together. Without the stop, the train would need 23 of the car's time, so the stop must be the remaining 13 of the car's time. That links the stop directly to the car's time.

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In these questions a bus has two speeds. One is the speed while it is actually moving, called the speed excluding stoppages. The other is the overall speed for the whole hour with the stops counted in, the speed including stoppages. The overall speed is smaller because a few minutes of every hour are spent standing still.

Take one hour. Including stops, suppose the bus covers 45 km. At its moving speed of 60 km/h, 45 km needs only 4560 h = 45 minutes, so the bus stood still for the other 15 minutes. That gives the rule:

Stoppage time per hour = Speed without stops − Speed with stopsSpeed without stops × 60 minutes.

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Other questions give the halts directly: "stops for 5 minutes after every 10 km" or "halts 3 times for 10 minutes each". Find the running time (distance ÷ speed) and add the halts. Read after every 10 km carefully. On a 45 km journey the stops come after 10, 20, 30 and 40 km, which is 4 stops, not 4510. Nobody stops at the very end of the journey.

Radhika's school van runs at 40 km/h but stops for 3 minutes after every 8 km. On a 30 km route, running time = 3040 h = 45 minutes, and the stops after 8, 16 and 24 km add 3 × 3 = 9 minutes. Total = 54 minutes.

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The story goes like this: at one speed a person is late, at another he is early. Or two people cover the same road and one takes some minutes more. In every version the distance is the same and the difference between two times is known.

If the distance is D and the two speeds are S₁ < S₂, the difference of times is
DS₁ − DS₂ = D × S₂ − S₁S₁S₂
So D = S₁ × S₂S₂ − S₁ × (difference of times).

What is the difference of times? If one trip is 10 minutes late and the other 5 minutes early, the times differ by 10 + 5 = 15 minutes. If both are late (12 minutes and 4 minutes), the difference is 12 − 4 = 8 minutes. When the speeds are in km/h, change the minutes into hours.

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Radhika walks to school at 3 km/h and is 10 minutes late; at 5 km/h she is 6 minutes early. The difference is 16 minutes = 415 h, so D = 3 × 55 − 3 × 415 = 152 × 415 = 2 km.

The ratio method gives the same answer. Speeds 3 : 5 mean times 5 : 3. The 2-part gap is 16 minutes, so 1 part = 8 minutes. The time at 3 km/h is 40 minutes, and D = 3 × 23 = 2 km.

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Speed up by x, time less by t, distance known. "A car covers 240 km; if its speed increases by 20 km/h, it takes 2 hours less." Then 240s − 240s + 20 = 2, which gives s(s + 20) = 240 × 202 = 2400 and s = 40. In general, s(s + x) = D × xt. Look for two numbers that differ by x and multiply to that value, or try the options.

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Times known, speed up by x. If the usual time is T and the time at x km/h more is T' over the same distance, then sT = (s + x)T', so s = xT'T − T'.

Average speed has only one definition, and every formula in this type comes out of it:

It is not the average of the speeds. Suppose Radhika goes to her aunt's house 90 km away at 45 km/h and comes back at 30 km/h. Going takes 2 hours and coming back 3 hours, so she covers 180 km in 5 hours. Her average speed is 36 km/h, not 45 + 302 = 37.5. She spends more time at the slower speed, so the average leans towards the slower speed.

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Case 1: equal times. If she drives 2 hours at 45 and 2 hours at 30, the distances are 90 and 60, a total of 150 km in 4 hours, which is 37.5 km/h. For equal times the average speed is the plain average, a + b2. In general, for times t₁, t₂, …, average = a t₁ + b t₂ + …t₁ + t₂ + ….

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Case 2: equal distances (going and returning, or a journey in equal halves or thirds). Take each distance as d. Total time = da + db, so average = 2dda + db = 2aba + b. For three equal distances it is 3abcab + bc + ca. The actual distance does not matter at all.

Case 3: fractions of the distance. If 14 of the way is at a, 14 at b and the rest at c, take the total distance as 1 and find each time: average = 114a + 14b + 12c. Taking the LCM of the speeds as the distance keeps fractions away.

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Rest and lunch count as time. If a driver stops for an hour, that hour joins the total time with zero distance. Trips with rest periods are set to check exactly this.

One more example. Radhika jogs 6 km at 6 km/h and walks 4 km at 4 km/h. Each part takes 1 hour, so her average is 102 = 5 km/h. The distances are unequal, but the times happen to be equal, and that is why the plain average works here.

This type runs the average-speed idea backwards. Instead of finding the average, you are given the total time or the average, and asked for a distance, a missing speed, or the speed needed for the rest of the journey.

Go at a, return at b, total time T. The same distance D is covered both ways, so Da + Db = T, which gives D = T × aba + b. Here aba + b is half of the round-trip average speed, so the round-trip distance 2D = average speed × T.

Radhika cycles to the library at 12 km/h and walks back at 4 km/h, taking 2 hours in all. D = 2 × 12 × 416 = 6 km.

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Missing speed from the average. Total time = Total distanceAverage speed. Subtract the times you already know; the time left and the distance left give the missing speed. For example, 120 km at an average of 40 km/h means 3 h in all. If the first 80 km took 2.5 h, the last 40 km must take 0.5 h, so the speed there is 80 km/h.

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Speed needed to finish on time = remaining distance ÷ remaining time. Someone has to cover 30 km in 2 h and has done 10 km in the first 50 minutes. He now needs 20 km in 70 minutes, that is 20 × 6070 ≈ 17.1 km/h.

Walking and riding. If walking both ways takes W hours and riding both ways takes R hours, one way on foot takes W2 and one way riding takes R2. So walking one way and riding back takes W + R2 hours. Any two modes of travel work the same way.

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When two bodies move at the same time, the useful question is how fast the gap between them is changing, not how fast each one goes. That rate is the relative speed.

Towards each other (from the two ends of a road): every hour the gap shrinks by a + b, so time to meet = Da + b. Both have moved for the same time, so the meeting point divides the road in the ratio of their speeds, a : b.

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Away from each other (from the same point in opposite directions): the gap grows by a + b every hour.

Same direction from the same point: the gap grows by a − b every hour, as the faster one pulls ahead.

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At right angles: after time t they are at and bt from the start along two perpendicular roads, so the distance between them is √((at)² + (bt)²) = t√(a² + b²). Look out for triplets like 3-4-5.

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"When they meet, one has travelled x km more than the other": in time t the extra distance is (a − b)t = x, so t = xa − b and the total distance = (a + b) × xa − b.

Radhika and her friend start 21 km apart and walk towards each other at 3 km/h and 4 km/h. They meet after 217 = 3 hours. By then Radhika has walked 9 km and her friend 12 km, in the ratio 3 : 4.

A person walking on a moving escalator works the same way. Walking with the escalator, the speeds add; walking against it, they subtract.

In the previous type both started together. Exam questions are rarely that kind. One starts at 8 a.m. and the other at 9:30 a.m., or you only know when each one leaves and arrives. Four tools handle all of these.

Tool 1: for a late start, move the early one first. On a 380 km road, A starts at 8 a.m. at 40 km/h and B starts from the other end at 10 a.m. at 60 km/h. By 10 a.m. A has already covered 80 km. The remaining 300 km gap closes at 40 + 60 = 100 km/h, so they meet 3 hours after 10 a.m., at 1 p.m. First bring the question to the point where both are moving together; then gap ÷ relative speed gives the time.

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Tool 2: only the start and arrival times are known. Take the distance as the LCM of the two journey times so that the speeds become whole numbers. If A takes 2 h and B takes 3 h, starting together from opposite ends, take D = 6 units. A's speed is 3 and B's is 2, so they meet after 65 h = 1 h 12 min. In general they meet after T_A × T_BT_A + T_B hours. If they start at different times, move the early one first (Tool 1) in these same units.

Tool 3: times taken after meeting (the square-root rule). If after crossing each other A needs t₁ hours and B needs t₂ hours to finish, then S_AS_B = √t₂t₁, and if they started together, they met √t₁ t₂ hours after starting. The reason: what A covered before the meeting is what B covers after it, so S_A × t = S_B × t₂. In the same way, S_B × t = S_A × t₁. Dividing one by the other, S_AS_B = S_B t₂S_A t₁, so (S_AS_B)² = t₂t₁.

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Tool 4: going to the end and coming back. If both start from A towards B and the faster one turns back at B, then by the time they meet they have together covered 2D, one full road each way. So the meeting time is 2Da + b, and the slower one is 2D × ba + b from A.

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A related question: buses leave a depot every T minutes at speed v, and a man walks towards them at u. He meets a bus every vTv + u minutes, because two buses in a row are vT apart and that gap closes at v + u.

Radhika leaves home at 7:00 a.m. for her aunt's town and reaches at 10:00 a.m. At the same moment her aunt leaves the town for Radhika's house and reaches at 1:00 p.m. They cross each other after 3 × 63 + 6 = 2 hours, at 9:00 a.m.

In a chase both move in the same direction, the one behind is faster, and the gap shrinks every hour by the difference of the speeds. So:

Time to catch = GapDifference of speeds

Once you have the time, the rest follows. Distance run by the thief = thief's speed × time. Distance run by the policeman = policeman's speed × time, which is also the gap + the thief's distance.

The ratio of speeds gives a neat shortcut. Both run for the same time, so their distances are in the ratio of their speeds. If the policeman runs at 9 and the thief at 7 with a gap of 300 m, the distances are 9 : 7, and the 2-part difference is the 300 m gap. So 1 part = 150 m: the thief runs 1050 m and the policeman 1350 m.

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Late start. If the thief starts at 9:00 and the police start from the same point at 9:20, the gap at 9:20 is the thief's speed × 2060 h. From there it is an ordinary chase. Give the clock time if that is what is asked.

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Gap after some time. If the chase lasts only a few minutes, gap left = starting gap − (difference of speeds × time). If the thief is faster, the gap grows instead.

Two unknown speeds. "In the same direction, caught in 10 minutes; in opposite directions, they meet in 2 minutes." This gives the difference and the sum of the speeds directly: difference = gap10, sum = gap2.

Radhika's younger brother runs off with her pen at 6 km/h. Radhika starts 50 m behind and chases him at 8 km/h. She catches him in 0.052 h = 1.5 minutes, after running 8 × 1.560 km = 200 m.

A race question is a TSD question written in race language. Learn the language once and the maths is only ratio.

• "A beats B by x metres" in an L-metre race: when A finishes L, B has run only L − x. The time is the same, so S_A : S_B = L : (L − x).
• "A beats B by t seconds": B finishes t seconds after A. B needs t seconds more for the full race, and in those t seconds B runs the remaining gap.
• "A gives B a start of x metres": B starts x metres ahead, so B runs only L − x.
• "A gives B a start of t seconds": B starts running t seconds earlier.
• Dead heat: both reach the finish at the same moment.

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If A beats B by x metres or t seconds, then B runs x metres in t seconds, so B's speed = xt. That single line settles many questions.

Chains. In a 100 m race, A beats B by 20 m and B beats C by 10 m. So A : B = 100 : 80 and B : C = 100 : 90. When A runs 100, B runs 80, and C runs 80 × 90100 = 72. A beats C by 28 m, not 20 + 10 = 30. In percentage terms, a 20% cut followed by a 10% cut is a 28% cut.

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Start with a speed ratio. When the speeds are 2 : 3 and the slower runner gets a big start, compare the distance each still has to run. In a 600 m race Radhika (2 parts) starts 300 m ahead, so she needs 300 m while her rival (3 parts) needs 600 m. By the time Radhika finishes her 300 m, the rival has run 450 m, so Radhika wins by 150 m.

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Two races with different starts. "With a start of 40 m, A wins by 10 s; with a start of 70 m, it is a dead heat." Moving the start from 40 m to 70 m wiped out exactly B's 10-second gap, so B runs 30 m in 10 s. See which change cancels which gap, and the question becomes one step.

On a circular track of length C, two runners who start together from the same point meet whenever one has gained exactly one full round on the other (same direction), or whenever together they have covered one full round (opposite directions).

First meeting anywhere on the track.

• Same direction: time = Ca − b, because the faster one has to gain one lap.
• Opposite directions: time = Ca + b, because together they complete one lap.

After that they keep meeting at the same interval. The n-th meeting comes at n × that time, and the number of meetings in a time T is the whole-number part of Tinterval.

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First meeting at the starting point. Each runner is back at the start after every lap time: Ca, Cb, … They are all at the start together at the LCM of their lap times, whatever the directions. For fractions, LCM = LCM of numeratorsHCF of denominators.

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Three runners meeting anywhere. When all three start together from one point, find the meeting interval of each pair. All three meet at the LCM of those pair intervals.

Number of distinct meeting points. Write the speeds in their lowest ratio a : b. In the same direction they meet at |a − b| distinct points; in opposite directions at a + b points. Speeds 4 : 1 give 3 points in the same direction and 5 points in opposite directions.

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Head start on a circle. If one runner starts some distance ahead in the direction of motion, the faster one catches up after gapa − b, just like a chase on a straight road.

Speeds may also be given in rounds per hour. Then take C = 1 round and the same formulas work: 7 and 3 rounds per hour in the same direction meet every 14 hour.

Radhika and her friend jog on a 600 m park track at 5 m/s and 3 m/s. Going the same way, they meet every 6002 = 300 s; going opposite ways, every 6008 = 75 s. Speeds 5 : 3 give 2 distinct meeting points in the same direction and 8 in opposite directions.

Buses leave a depot every T minutes and all run at the same speed v. Two consecutive buses are therefore always vT apart. A cyclist or walker on the road, with speed u, meets or is overtaken by a bus each time that gap is "used up".

• Buses coming from behind (same direction): the gap closes at v − u, so a bus overtakes you every a = vTv − u.
• Buses coming from the front (opposite direction): the gap closes at v + u, so you meet a bus every b = vTv + u.

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Radhika cycles on a road where buses overtake her every 15 minutes and she meets a bus coming from the front every 5 minutes. Then v − u = vT15 and v + u = vT5. Adding the two gives 2v = vT × 415, so T = 7.5 minutes. Subtracting gives 2u = vT × 215, so u : v = 1 : 2.

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This works every time, so remember the two results directly: T = 2aba + b and v : u = (a + b) : (a − b), where a is the longer interval (same direction) and b is the shorter (opposite direction).

Counting vehicles on the way. If a bus or train leaves each end every h hours and the journey takes x hours, the vehicles you meet from the other end are those that left it in the 2x-hour window from x before your start to x after it. Those whose departure lies at the very ends of this window are the ones standing at the terminals when you leave and when you arrive. So, excluding the two at the terminals, the number you meet on the road is 2xh − 1 (when xh is a whole number). If the two terminal vehicles are also counted, add 2.

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In chase questions the speed is not given directly. It is given through the steps: speed = (number of steps in a given time) × (length of one step). Before comparing, bring both runners to the same unit of length and the same unit of time.

Radhika's dog takes 4 leaps a second of 1.5 m each, so it runs 6 m every second. The hare it chases takes 7 leaps a second of 0.8 m each, which is 5.6 m a second. The dog gains 0.4 m every second, so a head start of 8 m is wiped out in 80.4 = 20 seconds.

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Steps given as a comparison. Many questions say "the thief takes 5 steps while the policeman takes 4 steps, and 3 steps of the policeman equal 5 steps of the thief". Take one thief step as 1 unit. Then one police step is 53 units, and in one unit of time the policeman covers 4 × 53 = 203 units against the thief's 5. The gap closes by 53 units for every unit of time.

What to ask for at the end. First find the number of time units the chase lasts: head start ÷ gain per unit of time. Then multiply by the steps (or leaps) per unit of time of whoever is asked about. If the question asks for distance, multiply the time by that runner's distance per unit of time.