Quick Revision

Speed, Distance, Time and Units

$\text{Speed} = \frac{\text{Distance}}{\text{Time}}$ , so distance = speed × time and time = distance ÷ speed.
km/h to m/s: multiply by $\frac{5}{18}$ . m/s to km/h: multiply by $\frac{18}{5}$ . So 108 km/h = 30 m/s and 12 m/s = 43.2 km/h.
1 m/s = 3.6 km/h, so a speed in m/s always looks smaller.
Turn minutes into hours before you divide: 2 hours 15 minutes = 2.25 hours, not 2.15.
Trap: mixing metres with km/h. Pick one unit at the start and keep it to the end.

Distance fixed: time goes the other way to speed. Speeds 3 : 4 give times 4 : 3.
Speed becomes $\frac{a}{b}$ of usual, so time becomes $\frac{b}{a}$ of usual. At $\frac{3}{4}$ speed and 20 minutes late: the extra $\frac{1}{3}$ of the usual time is 20 minutes, so usual = 60 minutes.
Time fixed: distance goes with speed.
25% faster cuts the time by 20%, not 25%; 20% slower adds 25% to the time.
Trap: check whether the question wants the usual time or the new time.

Always total distance ÷ total time. Never average the speeds blindly.
Equal distances at x and y: $\frac{2xy}{x + y}$ . At 30 km/h and 45 km/h: $\frac{2 \times 30 \times 45}{75}$ = 36 km/h, not 37.5.
Three equal distances: $\frac{3xyz}{xy + yz + zx}$ .
Equal times at x and y: the plain average $\frac{x + y}{2}$ .
Parts given as fractions of the distance: take the distance as an LCM and add up the times.
Trap: for equal distances the answer is always below the plain average.

Late by m minutes at x, early by n minutes at y: $\frac{d}{x} - \frac{d}{y}$ = (m + n) minutes. Both late, or both early: use m − n.
5 km/h: 7 minutes late; 6 km/h: 5 minutes early. $\frac{d}{30} = \frac{12}{60}$ , so d = 6 km.
k km/h faster saves t hours on D km: $\frac{D}{v} - \frac{D}{v + k} = t$ , so v(v + k) = $\frac{Dk}{t}$ . Testing the options is quickest.
Trap: forgetting to turn minutes into hours.

x km/h without stops and y km/h with stops: stop time per hour = $\frac{x - y}{x} \times 60$ minutes. 50 and 40 km/h give 12 minutes.
Walk one way and ride back in $T_1$ ; walk both ways in $T_2$ . Then ride both ways = $2T_1 - T_2$ .
A distance done partly at one speed and partly at another in a total time: call one part x and write one time equation.
Trap: in stoppages, divide by the speed without stops, never by the speed with stops.

Same direction: relative speed = difference. Opposite directions: sum.
Two people D apart walking towards each other at a and b meet after $\frac{D}{a + b}$ , and the first has covered $\frac{a}{a + b}$ of D. At 260 km apart, 30 and 35 km/h: 4 hours, 120 km from the first one's start.
One starts later: move the early starter forward first, then share the gap that is left.
Overtaking: time = lead ÷ difference of speeds.
Trap: a later start changes the clock time; count from the right moment.

Two start together towards each other. After meeting they need $t_1$ and $t_2$ more to finish. Then $\frac{S_1}{S_2} = \sqrt{\frac{t_2}{t_1}}$ .
They met $\sqrt{t_1 t_2}$ after the start.
After meeting, 16 hours and 9 hours: speeds $\sqrt{9} : \sqrt{16}$ = 3 : 4, and they met after $\sqrt{144}$ = 12 hours.
The one that still needs longer is the slower one; use that to check the order of the ratio.
Trap: forgetting the square root, or putting the times the same way round as the speeds.

The gap closes at the difference of the speeds: catch time = lead ÷ (police speed − thief speed).
A lead in time: lead in distance = thief's speed × head-start time.
Policeman's run = lead + thief's run.
250 m lead, thief 6 km/h and policeman 8 km/h: 0.25 ÷ 2 = $\frac{1}{8}$ hour = 7.5 minutes, and the policeman runs 1 km.
Steps type: speed = steps in a time × length of one step.
Trap: a lead in metres with speeds in km/h.

Passing a pole, a signal post or a standing man: distance = the train's own length.
Passing a platform, bridge or tunnel: distance = train + platform. A 195 m train at 54 km/h (15 m/s) crosses a 330 m platform in 525 ÷ 15 = 35 seconds.
Passing a man walking or a car: use the relative speed, minus for the same direction, plus for opposite.
Platform in $T_1$ seconds and a man in $T_2$ : speed = platform ÷ $(T_1 - T_2)$ .
Trap: forgetting the train's own length, or forgetting $\frac{5}{18}$ .

To cross each other completely they cover the sum of their lengths.
Opposite directions: add the speeds. Same direction: subtract them.
130 m and 170 m at 55 and 35 km/h: opposite, 90 km/h = 25 m/s, so 300 ÷ 25 = 12 seconds; same direction, 20 km/h = $\frac{50}{9}$ m/s, so 54 seconds.
They pass a standing man in $t_1$ and $t_2$ and each other in $t_3$ : speed ratio = $(t_3 - t_2) : (t_1 - t_3)$ .
Trap: using one train's length when the question says completely.

Downstream = b + s; upstream = b − s.
Still water b = half the sum; stream s = half the difference.
Round trip time = $\frac{d}{b + s} + \frac{d}{b - s}$ . Boat 12 km/h, stream 2 km/h, 70 km each way: 5 + 7 = 12 hours.
Upstream k times as long as downstream: $\frac{b + s}{b - s} = k$ .
Round-trip average speed = $\frac{(b + s)(b - s)}{b}$ , below b.
Trap: half the sum is the boat, half the difference the stream.

A beats B by x m over L m: A : B = L : (L − x).
Beats by x m or t seconds: B's speed = $\frac{x}{t}$ .
Start of x m, dead heat: A runs L while B runs L − x.
Three runners: multiply ratios, never add leads. 560 m, A beats B by 35 m, B beats C by 16 m: C runs 525 × $\frac{544}{560}$ = 510 m, so A wins by 50 m, not 51.
Trap: a lead in seconds is not a lead in metres.

Same point, same direction: first meeting after $\frac{L}{a - b}$ . Opposite directions: $\frac{L}{a + b}$ .
First meeting at the starting point: LCM of the lap times.
Three runners meeting anywhere: LCM of the pair meeting times.
Speeds a : b in lowest terms: a − b meeting points the same way, a + b opposite.
500 m track at 7 and 3 m/s: same direction 125 seconds, opposite 50 seconds; 4 meeting points the same way, 10 the opposite way.
Trap: meeting anywhere is not meeting at the start.