Level 3 Integration Walkthrough

2023 NCEA Level 3 Integration Question 3(d)

2023 Paper

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Question

An object’s acceleration can be modelled by

\[ a=\frac{e^{2t}}{4e^{2t}-3}, \]

where \(t\ge 0\), \(a\) is measured in m s\(^{-2}\), and \(t\) is the time in seconds from the start of timing.

At \(t=0\) seconds, the object had velocity \(5\) m s\(^{-1}\).

Find the object’s velocity when \(t=4\) seconds.

First walkthrough idea

Focus to try first

Treat acceleration as \(\frac{dv}{dt}\), integrate carefully, then use \(v(0)=5\) before substituting \(t=4\).

Step 1

Integrate the acceleration

The derivative of \(4e^{2t}-3\) is \(8e^{2t}\).

Show the first step’s working

The derivative of \(4e^{2t}-3\) is \(8e^{2t}\).

Key result

\[ \frac{1}{8}\ln|4e^{2t}-3|+C \]

Walkthrough overview

What this question practises

This 2023 walkthrough is part of AS91579 — Apply integration methods in solving problems.

Method: Integrating acceleration to velocity with an initial condition.

This is Question 3(d) from the 2023 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 acceleration to velocity with an initial 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 3(c) Question 3(e) →