Time, Speed and Distance

1Distance is speed times time

The whole chapter is one identity: \text{Distance}=\text{Speed}\times\text{Time}. Rearrange it for whichever quantity is asked, and keep every quantity in one unit system before you multiply.

When the distance is fixed, speed and time are inversely proportional. If speed becomes \frac{3}{4} of usual, time becomes \frac{4}{3} of usual — the product is the same journey, so the factors must multiply to 1.

Figure. Length is speed and thickness is time, so each rectangle's area is the same 120 km journey. Shorten the speed bar to three-quarters and the time thickness must grow to four-thirds or the area — the distance — would change.

How it works

1. Name what is fixedA named route, a scheduled arrival, a race distance — that is the quantity that does not change when the speed does.
2. Write D = S × T onceFill the two knowns and solve for the third. Do not average speeds unless the next concept says you may.
3. Scale by the inverse factorIf speed is multiplied by k, time is multiplied by 1/k. Checking that the product of the two factors is 1 catches most slips.

2Convert before you divide

Train lengths arrive in metres and speeds in km/h. Dividing one by the other without converting produces a number with no meaning. Convert first: multiply km/h by \frac{5}{18} to get m/s, or multiply m/s by \frac{18}{5} to get km/h.

The direction of the conversion is a size check. A speed in m/s must be smaller than the same speed in km/h; if the number grew, the fraction was applied backwards.

Figure. Convert before you divide: km/h → m/s multiplies by 5/18, so the number must shrink (90 → 25). The reverse factor 18/5 grows the number — if your m/s reading came out larger than the km/h figure, the fraction ran backwards (the 324 trap on 90 km/h).

How it works

1. Pick the unit the lengths useMetres with metres. If the answer is wanted in km/h, convert back only at the end.
2. Multiply by 5/18 or 18/5km/h → m/s is × 5/18. The other way is × 18/5. The number must shrink going to m/s.
3. Then divideDistance over speed, now in matching units, is the time.

3Average speed is not an average of speeds

Average speed is total distance over total time, always. Over two equal distances at x and y that is the harmonic mean \frac{2xy}{x+y} — never \frac{x+y}{2}, because more time is spent on the slower leg.

The arithmetic mean is right when the times are equal instead. So the question is which quantity the problem holds equal — the distances or the durations — not which formula to memorise.

Figure. The outward leg is the shallower segment (50 km/h for three hours) and the return the steeper one (75 km/h for two). The dashed line is a car driven at a constant 60 km/h between the same endpoints — the definition of the average. More of the five hours sits under the shallow slope, which is why 60 lies below the midpoint of 50 and 75.

How it works

1. Read what is equal"There and back" or "each half of the journey" means equal distances. "For two hours, then for three" means times.
2. Total distance and total time separatelyAssume a convenient distance if none is given — the average does not depend on it.
3. Divide once, at the endTotal distance ÷ total time. Never average the two speeds and hope.

4Relative speed

Sit on one of the two moving objects and ask how fast the other seems to approach. Moving towards each other, the speeds add; moving the same way, they subtract. Every overtaking, meeting and crossing problem is that sentence plus a distance.

The two cases differ by a large factor. Two bodies at 60 and 40 km/h close at 100 km/h head-on and at only 20 km/h in the same direction — five times slower, so five times longer to pass, for the same pair.

Figure. Both bars are the relative speed of the same 60 km/h and 40 km/h pair. Opposite directions stack the speeds into 100; the same direction leaves only the 20 km/h difference — one-fifth as long, which is why an overtaking pass takes five times as long as a head-on one.

How it works

1. Add or subtract the speedsTowards each other: u + v. Same direction: |u − v|, faster minus slower.
2. Name the distance to closeFor objects with length, the sum of the lengths; for points, the gap between them.
3. Divide in matching unitsConvert the relative speed to m/s if the lengths are in metres, then divide.

5What distance does a train actually cover?

A train passing a pole covers its own length, because the pole has none. A train crossing a platform covers its own length plus the platform's, because the rear must clear the far end before the crossing is complete.

Give the pole time and the platform time and the difference between them is the platform's length divided by the speed — two facts, two unknowns, and almost no algebra.

Figure. The same train is drawn twice — at the start of the crossing and at the end. Between those positions it has moved its own 200 m plus the platform's 300 m, which is the long bracket. Shrink the platform to a point and only the train's length remains: the pole case.

How it works

1. Decide what has lengthA pole, a man, a signal: no length. A platform, a bridge, a tunnel, another train: length counts.
2. Add the lengths that countDistance = train + object. For a pole that is just the train.
3. Divide by the speed in m/sConvert first. A km/h figure divided into metres is meaningless.

6Boats and streams

Downstream the current adds to the boat; upstream it subtracts. If b is the boat's speed in still water and s the stream's, downstream is b+s and upstream is b-s. Solving the other way: b=\frac{\text{down}+\text{up}}{2} and s=\frac{\text{down}-\text{up}}{2}.

A round trip on a stream is an equal-distance average-speed problem in disguise. The harmonic mean of the downstream and upstream speeds is the average for the out-and-back, and it is always less than b.

Figure. Downstream and upstream sit equally far above and below the boat's still-water speed — 2.5 km/h either side of 12.5 — which is why averaging them recovers b and halving their gap recovers s. The stream bar is the common offset.

How it works

1. Read the two speeds off the timesDownstream speed = distance / downstream time; upstream likewise.
2. Average for the boat, halve the gap for the streamb = (down + up)/2 and s = (down − up)/2. Both come out in the same unit as the two speeds.
3. Refuse a negative streamIf upstream came out faster than downstream, the legs were swapped — fix the labels before trusting either number.

7Meeting on a circular track

On a circular track of length L, two runners first meet when together they have covered one full lap relative to each other. Opposite directions: meeting time =\frac{L}{v_1+v_2}. Same direction: meeting time =\frac{L}{|v_1-v_2|}.

They next meet at the starting point after the LCM of their individual lap times. First meeting anywhere and next meeting at the start are different clocks — the first uses relative speed, the second uses each runner's own lap.

How it works

1. Choose add or subtractOpposite directions add the speeds; the same direction subtracts them. That is the relative speed around the track.
2. Divide the lap length by itFirst meeting time = L / v_rel. Keep L and the speeds in matching units.
3. For a start-line reunion, take lap-time LCMEach runner's lap time is L/v. They are both at the start together at every common multiple of those times.

8Late or early at two speeds

When a journey at speed v_1 finishes late by t_1 and the same journey at v_2 finishes early by t_2, the two distance expressions share the scheduled time. Equating them solves for the distance in one step: \frac{D}{v_1}-\frac{D}{v_2}=t_1+t_2 when v_2>v_1.

The scheduled duration itself drops out. You never need to know what time the traveller was supposed to arrive — only how late and how early the two trials were.

Figure. Times in tenths of an hour for the 80 km worked example: 2.0 h is 12 min late, 1.6 h is 12 min early. Both offsets are measured from the schedule rule, so they add: 0.2+0.2=0.4 h is the gap that unlocks D. Bars are to a common time scale.

How it works

1. Write both trial timesD/v₁ is late by t₁; D/v₂ is early by t₂. Their difference is t₁ + t₂.
2. Factor D outD(1/v₁ − 1/v₂) = t₁ + t₂, with times converted to hours if the speeds are in km/h.
3. Solve for DOne multiplication. Check by recovering the scheduled time from either trial.