Level 3 Integration Walkthrough

2024 NCEA Level 3 Integration Question 1(c)

2024 Paper

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Question

An object's velocity can be modelled by the equation \[ v(t)=26.4t^{1/3}, \] where \(v\) is the velocity of the object in \(\text{m s}^{-1}\), and \(t\) is the time in seconds since the start of timing.

Initially, the object was \(360\) metres from a point \(P\).

Calculate the distance of the object from point \(P\) when it has reached a velocity of \(264\text{ m s}^{-1}\).

You must use calculus and show the results of any integration needed to solve the problem.

First walkthrough idea

Hint to try first

Velocity is the derivative of position, so integrate \(v(t)\) to get a position model.

Step 1

Build the position function

Integrating \(t^{1/3}\) gives \(\frac{t^{4/3}}{4/3}\).

Show the first step’s working

Integrating \(t^{1/3}\) gives \(\frac{t^{4/3}}{4/3}\).

Key result

\[ s(t)=19.8t^{4/3}+C \]

Walkthrough overview

What this question practises

This 2024 walkthrough is part of AS91579 — Apply integration methods in solving problems.

Method: Integrating velocity and using a position condition.

This is Question 1(c) from the 2024 NCEA Level 3 Integration paper for AS91579 — Apply integration methods in solving problems. Use the guided hints to practise the method before revealing the full working.

Calc.nz is an independent learning resource. Compare questions, diagrams, and assessment information with the official NZQA resources for AS91579.

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Learning summary

Review the method, not only the answer

This walkthrough helps you practise integrating velocity and using a position condition. Use the hints to plan the method, then repeat the question without hints and check each step.

Common mistake to avoid

Check the antiderivative by differentiating it, and handle constants and bounds explicitly.

Continue practising

← Question 1(b) Question 1(d) →