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Networks, STD1 2026 HSC 27

The network shows the lengths of the paths between some exhibits at a zoo.
 

Nikolai is visiting the penguins and wants to visit the lions. The shortest path Nikolai can take is Penguins—Giraffes—Snakes—Lions.

  1. The paths around the zoo are not all flat. The network shows the same paths at the zoo with the gradients of each.
     
  3. Nikolai uses a wheelchair to get around the zoo. The wheelchair cannot access paths with a gradient greater than 0.083.
  4. Explain why the shortest path is not necessarily the best path for Nikolai to take to get to the Lions.   (1 mark)
  5. What is the best path for Nikolai to take from the Penguins to the Lions?   (1 mark)
  6. The speeds, in metres per second, that Nikolai can travel in the wheelchair are shown on some of the paths.
     
  8. Feeding time at the Lions is 12:15 pm. Nikolai plans to travel from the Otters to the Lions, passing the Koalas without stopping.
  9. What is the latest time Nikolai can leave the Otters to arrive for feeding time at the Lions?   (3 marks)
Show Answers Only

a.    \(\text{The shortest path uses Penguins}-\text{Giraffes (gradient 0.11) and}\)

\(\text{Giraffes}-\text{Snakes (gradient 0.09).}\)

\(\text{Both gradients are greater than 0.083, so Nikolai’s wheelchair}\)

\(\text{cannot access these paths.}\)

b.    \(\text{Penguins}-\text{Bears}-\text{Otters}-\text{Koalas}-\text{Lions}\)

c.    \(\text{11:55 am}\)

Show Worked Solution

a.    \(\text{The shortest path uses Penguins}-\text{Giraffes (gradient 0.11) and}\)

\(\text{Giraffes}-\text{Snakes (gradient 0.09).}\)

\(\text{Both gradients are greater than 0.083, so Nikolai’s wheelchair}\)

\(\text{cannot access these paths.}\)
 

b.    \(\text{Penguins}-\text{Bears}-\text{Otters}-\text{Koalas}-\text{Lions}\)

\(\text{(all gradients on this path are less than 0.083)}\)
 

c.    \(\text{Using}\ \ \text{Time}=\dfrac{\text{distance}}{\text{speed}}:\)

\(\text{Otters to Koalas}\) \(=\dfrac{400}{0.8}=500\ \text{seconds}\)
\(\text{Koalas to Lions}\) \(=\dfrac{700}{1}=700\ \text{seconds}\)

 

\(\text{Total time}\) \(=500+700\)
  \(=1200\ \text{seconds}\)
  \(=20\ \text{minutes}\)

 
\(\text{Latest time to leave}=\text{12:15 pm}-20\ \text{minutes}=\text{11:55 am}\)