Paper 2 · June 2024 · Direct variation

The extension $y$ of an elastic string varies directly as the magnitude of the force $F$ extending it. Given $y=0.45$ m when $F=6$ N, express $F$ as a function of $y$ .

Model answer

F=40y/3

Also accepted: 40y/3, F = 40y/3

Explanation

Direct variation gives

y=kF

, so

0.45=6k

and

k=\dfrac{3}{40}

. Then

y=\dfrac{3}{40}F

, and making

F

the subject,

F=\dfrac{40}{3}y

Derived from ZIMSEC Pure Mathematics Paper 2, June 2024, Q1

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