Speedo
Investigate the relationship between speeds recorded and the distance travelled in this kinematic scenario.
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
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, we can assume that 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 $16 \frac{2}{3} + 1.3 + 15 \approx 33\text{s}$.
Note, this time would be $31.5\text{s}$ if we assume that we start this section of road at $10\text{ km h}^{-1}$ rather than starting from rest.
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 $38\frac{2}{3}\text{s}$ 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 $166\frac{2}{3}\text{s}$.
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 $10\text{ km h}^{-1}$.
Teachers' Resources
Why do this problem?
This problem provides practice in using both equations of constant acceleration and the principles behind velocity-time graphs to make calculations within a kinematics context. Students have the opportunity to explore upper and lower limits on speeds, times and maneouvres. They also gain practice in moving between realistic units ($\text{ km h}^{-1}$ and time taken from $0$ to $60\text{ km h}^{-1}$ in seconds) and SI units ($\text{m s}^{-1}$ and $\text{m s}^{-2}$).
Possible approach
Discuss the premise of the problem with students before diving in to any calculations. How does $100\textrm{ km h}^{-1}$ as a maximum speed compare to cars they are familiar with? (Probably a bit slower.) What about $0$ to $60\textrm{ km h}^{-1}$ in $10$ seconds? (Also not especially quick.) Can they foresee any difficulties in working with the information they have been given ($500\textrm{ m}$ stretch of road, speeds in $\text{ km h}^{-1}$ and acceleration as $0$ to $60\textrm{ km h}^{-1}$ in $10$ seconds)? How do they plan to deal with these difficulties?
(You may want to be prepared for the fact that working entirely in SI units simplifies the thinking but complicates the arithmetic, while working in a mixture of units can simplify some arithmetic but requires great care. It would be sensible to decide whether you will steer in one direction or another or leave students free choice.)
Look at the first question and ask students to sketch possible velocity-time graphs that meet the requirement of the speedometer registering $10\text{ km h}^{-1}$ at least twice and $50\text{ km h}^{-1}$ at least once. How could they adapt their graphs to maximise or minimise the time taken? How will their graphs reflect the maximum acceleration of the car and the distance being travelled? Students should sketch separate graphs for the maximum and minimum time taken, labelling as many values as they can and using these to calculate total time.
For each subsequent question, students should work in the same way, first sketching a graph which meets the stated requirements, then separate graphs for minimum and maximum times. At this stage, they could be working in pairs to provide ample opportunity to articulate their reasoning. Results could be compared with another pair initially. Having labelled sketches on individual whiteboards which could be passed aound or on large whiteboards around the room could facilitate work within and between pairs.
Possible extension
Students could go on to the problem Dangerous Driver?.