Physics 9702/12 — May/June 2017

Background Concept

Average speed is defined as total distance travelled divided by the total time taken:

$$v_{\text{avg}} = \frac{\text{distance}}{\text{time}}$$

In an estimate question, you use a realistic, well-known reference value (here, a typical elite 100 m sprint time) and then calculate the speed.

Understanding the Question

You are asked for an approximate average speed of a winning female Olympic athlete over $100\ \text{m}$. The options given are $6, 9, 12,$ and $15\ \text{m s}^{-1}$.

So we need a reasonable estimate of the winning time and then compute $100/t$.

Approach

1. Recall that elite female 100 m winning times are a little above $10\ \text{s}$ (typically around $10.5$ to $11.5\ \text{s}$).
2. Use $v = 100/t$ to estimate the average speed.
3. Compare to the options and choose the nearest.

Step-by-Step Reasoning

Take a typical winning time as about $11\ \text{s}$.

Compute average speed:

$$v = \frac{100\ \text{m}}{11\ \text{s}} \approx 9.1\ \text{m s}^{-1}$$

The closest option to about $9\ \text{m s}^{-1}$ is B.

(If you used $10.5\ \text{s}$, you would get $\approx 9.5\ \text{m s}^{-1}$, still closest to $9\ \text{m s}^{-1}$.)

Key Takeaways

• Use $v = s/t$ for average speed.
• For “approximate” multiple-choice questions, a sensible estimate is sufficient.
• Elite 100 m sprint times are around $10$–$12\ \text{s}$, giving speeds around $8$–$10\ \text{m s}^{-1}$.

Common Mistakes

• Choosing an unrealistically small time (e.g. $7\ \text{s}$), which would imply an implausibly large speed.
• Confusing average speed with peak speed (sprinters’ top speeds are higher than their average over the whole race).
• Arithmetic slip: dividing $11/100$ instead of $100/11$.

Things to Be Careful About

• The question says average speed over $100\ \text{m}$, so you must use total distance/time, not acceleration data.
• Ensure the unit is $\text{m s}^{-1}$ (distance in metres, time in seconds).
• Since it is “approximate”, pick the closest option rather than overthinking exact world-record values.