Skip to main content
### Number and algebra

### Geometry and measure

### Probability and statistics

### Working mathematically

### Advanced mathematics

### For younger learners

# Speedo

## You may also like

### Board and Spool

Or search by topic

Age 16 to 18

Challenge Level

- Problem
- Student Solutions

In this question, we need to make some assumptions about acceleration. A good model is that the power output of a car is a constant throughout the motion. However, this makes the calculations quite complicated. If instead we assume acceleration is constant throughout the motion, this makes the equations simpler and easier to use.

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 acceleratingÂ 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 $10\text{ km h}^{-1}$.

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 acceleratingÂ 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 $10\text{ km h}^{-1}$.

Think about this mechanical configuation and compute the time taken for the man to reach the spool