Level 3 Integration Walkthrough

2022 NCEA Level 3 Integration Question 2(d)

2022 Paper

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Question

An object's acceleration can be modelled by

\[ a(t)=0.9e^{0.3t}, \]

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

The object had a velocity of \(10\text{ m s}^{-1}\) after \(2\) seconds.

How far did the object travel during the 5th second of its motion?

The 5th second means the interval from \(t=4\) to \(t=5\).

First walkthrough idea

Focus to try first

Integrate acceleration to get velocity, use the given velocity to fix the constant, then integrate velocity from \(t=4\) to \(t=5\).

Step 1

Build the velocity function

Acceleration is \(\frac{dv}{dt}\), so integrate \(a(t)\) with respect to time.

Show the first step’s working
\[ \frac{dv}{dt}=0.9e^{0.3t} \] \[ v(t)=\int 0.9e^{0.3t}\,dt \] \[ v(t)=3e^{0.3t}+C \]

Walkthrough overview

What this question practises

This 2022 walkthrough is part of AS91579 — Apply integration methods in solving problems.

Method: Turning acceleration into velocity, then distance over one second.

This is Question 2(d) from the 2022 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 turning acceleration into velocity, then distance over one second. Use the hints to plan the method, then repeat the question without hints and check each step.

Common mistake to avoid

Check each step against the original condition, preserve signs and restrictions, and confirm that the final result answers the question asked.

Continue practising