# Speedo

## Problem

I drove my car along a stretch of road $500\textrm{ m}$ long. My car can accelerate uniformly from $0$ to $60\textrm{ km h}^{-1}$ in $10$ seconds. Its maximum speed is $100\textrm{ km h}^{-1}$.

1. I looked at my speedometer three times on the journey and read the speeds $10\textrm{ km h}^{-1}$, then $50\textrm{ km h}^{-1}$, then $10\textrm{ km h}^{-1}$. What was the least possible time to travel along the stretch of road? What was the greatest possible time?

2. On the next $500\textrm{ m}$ of road, I looked at my speedometer twice: on one occasion it registered $50\textrm{ km h}^{-1}$, which was my maximum speed for the journey, and on another occasion is registered $10\textrm{ km h}^{-1}$, which was also my minimum speed for the journey.

What were the least and greatest possible times I spent on this section of road?

3. On the next $500\textrm{ m}$ section of road I alternately accelerate to $50\textrm{ km h}^{-1}$ and decelerate down to $10\textrm{ km h}^{-1}$.

What is the largest number of times I can record a speed of $10\textrm{ km h}^{-1}$?

4. On the final $500\textrm{ m}$ section of road, before I am arrested for dangerous driving, I alternately accelerate to $50\textrm{ km h}^{-1}$ and decelerate down to $10\textrm{ km h}^{-1}$.

What speed must I start the section of road to finish at $10\textrm{ km h}^{-1}$?

Don't forget that I start one section of road at the same speed that I finish the previous section.

## Student Solutions

1) The least possible time spent on this section corresponds to accelerating to $100\text{ km h}^{-1}$, maintaining this speed and then decelerating to $10\text{ km h}^{-1}$ at the end of the section.

It takes $10\text{s}$ for the car to accelerate constantly from $0$ to $60\text{ km h}^{-1}$, so it will take $\frac{100}{60} \times 10 = 16 \frac{2}{3}\text{s}$ to accelerate from $0$ to $100\text{ km h}^{-1}$. Note that $100\text{ km h}^{-1}$ is roughly $28\text{ m s}^{-1}$, since we need to work in metres per second.

The distance travelled in this time is $\text{average speed} \times \text{time} = \frac{0 + 28}{2} \times 16 \frac{2}{3} \approx 233\text{m}$.

Similarly, the time taken for the car to decelerate from $100\text{ km h}^{-1}$ to $10\text{ km h}^{-1}$ is $\frac{90}{60} \times 10 = 15\text{s}$.

The distance travelled in this time is $\frac{28 + 2.8}{2} \times 15 \approx 231\text{m}$.

Then the total distance travelled is $500\text{m}$, so the distance that the car travels at the constant speed of $100\text{ km h}^{-1}$ is $500 - 233 - 231 = 36\text{m}$. Then the time taken is $\frac{\text{distance}}{\text{speed}} = \frac{36}{28} \approx 1.3\text{s}$.

Then the total time taken is $16 \frac{2}{3} + 1.3 + 15 \approx 33\text{s}$.

The greatest time involves spending lots of time travelling at a very low speed, say $1\text{ km h}^{-1}$, then accelerating to $50\text{ km h}^{-1}$ and then decelerating to the very low speed again. The time taken is $2\times\frac{49}{6}s = 16.3\text{s}$ for the accelerating/decelerating, and $1380\text{s}$ for the long section at the low speed, taking $\approx 1400\text{s}$ in total!

2) Suppose we were starting this section at $10\text{ km h}^{-1}$. The shortest time comes from accelerating straight away to $50\text{ km h}^{-1}$ and then keeping the speed constant. This takes $38\text{s}$ in total. The greatest time involves remaining at $10\text{ km h}^{-1}$ and then accelerating to finish at $50\text{ km h}^{-1}$. This takes $167\text{s}$.

3) Given that the car is travelling at $50\text{ km h}^{-1}$ when we start this section, the maximum number of times we can record a speed of $10\text{ km h}^{-1}$ is $5$, as the car covers $\frac{500}{9}\text{m}$ when accelerating from $10\text{ km h}^{-1}$ to $50\text{ km h}^{-1}$ or decelerating from $50\text{ km h}^{-1}$ to $10\text{ km h}^{-1}$.

4) From the last question, we know that must have to start at $50\text{ km h}^{-1}$: after $9$ accelerations and decelerations to and from $10\text{ km h}^{-1}$, we'll have covered $500\text{m}$ with a finishing speed of $10\text{ km h}^{-1}$.