A formula for working out the braking distance \(d\), for a vehicle travelling at speed \(v\), is \(d=av^3+bv^2\), where \(a\) and \(b\) are constants. Values of \(d\) for different values of \(v\) have been collected. Explain how a straight line can be drawn to represent the formula, and state how the values of \(a\) and \(b\) could be obtained from the line.[4]
The value, \(\$V\), of a limited edition watch has been increasing each year from 2016 to 2025. A watch trader claims that the increase is exponential and so can be modelled by an equation in the form
\(V=V_0\mathrm e^{kt}\),
where \(V_0\) and \(k\) are constants and \(t\) is the time in years since 1st January 2016. The table below gives values of \(V\) and \(t\) for some of the years from 2016 to 2025.
Year
\(2016\)
\(2019\)
\(2022\)
\(2025\)
\(t\) years
\(0\)
\(3\)
\(6\)
\(9\)
\(\$V\)
\(12000\)
\(12950\)
\(13900\)
\(15000\)
On the grid provided, plot \(\ln V\) against \(t\) and draw a straight line graph to show that the model is valid for the years 2016 to 2025.[3]
Using your graph, estimate the value of \(V_0\) and \(k\).[4]
Explain the significance of the value of \(V_0\).[1]
Using your graph, estimate the value of the watch on 1st January 2023.[2]
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Answer:(a) Plot \(\dfrac{d}{v^2}\) against \(v\); gradient \(a\), vertical intercept \(b\). (b)(i) Approximately linear plot. (ii) \(V_0\approx12000\), \(k\approx0.025\). (iii) Initial value on 1st January 2016. (iv) Approximately \(\$14300\).