Find the constant of variation and complete the table

A 180 km journey is modelled at constant speed with no stops. At 60 km/h it takes three hours. Predict the time at 90 km/h

Vary the constant speed for a fixed 180 km journey. Compare 40 with 80 km/h and 60 with 90 km/h. Time is shown to two decimal places where necessary

Which relationship stays true throughout the investigation?

Use the fixed product

An inverse model makes one positive quantity smaller when the other grows, with their product fixed

1. 1Find the constant product Multiply a known input and output. Attach the product of their units \(\displaystyle k=xy\)
2. 2Find either unknown Divide the constant by the known quantity \(\displaystyle \begin{aligned}y&=k/x\\x&=k/y\end{aligned}\)
3. 3Check positivity and assumptions Positive physical quantities exclude zero. Fixed distance and constant speed are assumptions, not guarantees about a real journey
4. 4Test and revise Compare a new observation with the prediction. Investigate fixed delays or changed conditions before choosing a revised model

Data or rule

\(\displaystyle v=72\text{ km/h},\quad t=2.5\text{ h}\)

A journey takes 2.5 hours at a constant 72 km/h with no stops. Find the distance, which is the inverse constant for speed and time

Data or rule

\(\displaystyle vt=180\)

For this 180 km journey, find the time at 80 km/h.

Data or rule

\(\displaystyle vt=180\)

At what constant speed is the 180 km journey completed in 1.5 hours?

Data or rule

\(\displaystyle t=2.25\text{ h}\)

Express 2.25 hours as a number of minutes

Data or rule

\(\displaystyle xy=k,\quad x=2.4,\quad y=15\)

The positive variables have a constant product. Use the known pair shown below to find k

Data or rule

\(\displaystyle xy=36,\quad x=8\)

With xy=36, find y when x=8.

Data or rule

\(\displaystyle xy=36,\quad y=7.5\)

With xy=36, find x when y=7.5

A pupil uses the same product model but adds two to y whenever x increases by two. What is the error?

Could zero speed give a finite travel time for the nonzero 180 km distance?

Data or rule

\(\displaystyle \begin{gathered}t_{\mathrm{predicted}}=3\text{ h}\\t_{\mathrm{observed}}=3.5\text{ h}\end{gathered}\)

At 60 km/h the no-stop model predicts three hours. The observed total journey takes 3.5 hours. Find the extra time in minutes

For a 180 km journey, the observed total times are 3.5 hours at 60 km/h and 2.5 hours at 90 km/h. Which revised model fits both observations if each journey includes the same fixed delay?

Data or rule

\(\displaystyle t=\frac{180}{v}+0.5\)

Use the revised fixed-delay model to predict total time at 80 km/h.

For a 180 km journey, observed total times are 3.5 hours at 60 km/h and 2.5 hours at 90 km/h. Which conclusion about the model is justified?