Time, Speed & Distance

Time, Speed & Distance

The one relationship behind everything here is

$\text{Distance}=\text{Speed}\times\text{Time}.$

Keep two of the three fixed and the third follows. TSD is among the most-tested arithmetic sub-topics and feeds train, boat-and-stream, and race questions.

Core results

• Constant distance: speed and time are inversely proportional — if speed rises by a factor $k$, time falls to $1/k$.
• Relative speed: moving towards each other, speeds add; moving in the same direction, speeds subtract.
• Average speed over two equal distances at $u$ and $v$ is the harmonic mean, not the simple average:

$\bar{v}=\frac{2uv}{u+v}$

• Boats & streams: downstream speed $=b+s$, upstream $=b-s$ (boat $b$, stream $s$).

A worked example

You drive to a town at 40 km/h and return along the same road at 60 km/h. What is your average speed for the whole trip?

It is not 50. The distances are equal, so use the harmonic mean:

$\bar v=\frac{2\cdot40\cdot60}{40+60}=\frac{4800}{100}=\mathbf{48\text{ km/h}}.$

The slower leg takes longer, so it weighs more — pulling the average below 50.

Common traps

• Averaging speeds directly. Equal distances → harmonic mean; equal times → simple mean. Read which is equal.
• Unit slips. Convert km/h ↔ m/s with the factor $\dfrac{5}{18}$ (km/h → m/s) before using train lengths in metres.
• Train crossing. Crossing a pole uses the train's length; crossing a platform uses length + platform.

Checklist

• Fix the constant quantity (distance? time?) before averaging
• Add/subtract speeds for relative motion
• Use $\dfrac{5}{18}$ for km/h → m/s
• For two equal legs, average speed $=\dfrac{2uv}{u+v}$