Standard +0.8 This is a Further Maths question requiring integration of 1/√(9-t²) using trigonometric substitution (arcsin), then applying the mean value theorem. While the setup is clear, the integration technique is non-standard for typical A-level and requires recognizing the inverse trig form, placing it moderately above average difficulty.
In this question you must show detailed reasoning.
An ant starts from a fixed point \(O\) and walks in a straight line for \(1.5\) s. Its velocity, \(v\) cms\(^{-1}\), can be modelled by \(v = \frac{1}{\sqrt{9-t^2}}\).
By finding the mean value of \(v\) in \(0 \leq t \leq 1.5\), deduce the average velocity of the ant. [5]
So average velocity is \(0.35 \text{ cms}^{-1}\) along \(OA\)
A1 [5]
correct unit must be present
$\frac{1}{(1.5-0)} \int_0^{1.5} \frac{1}{\sqrt{9-t^2}} dt$ | B1 | Using the formula for mean value
$\left[\sin^{-1} \frac{t}{3}\right]_0^{1.5}$ | M1 | Ignore limits
$\sin^{-1} \frac{1.5}{3}$ | A1 | Correct use of limits.
$\frac{\pi}{6}$ | A1 | Soi by answer
So average velocity is $0.35 \text{ cms}^{-1}$ along $OA$ | A1 [5] | correct unit must be present | or accept awrt 0.349 or $\frac{\pi}{9}$
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In this question you must show detailed reasoning.
An ant starts from a fixed point $O$ and walks in a straight line for $1.5$ s. Its velocity, $v$ cms$^{-1}$, can be modelled by $v = \frac{1}{\sqrt{9-t^2}}$.
By finding the mean value of $v$ in $0 \leq t \leq 1.5$, deduce the average velocity of the ant. [5]
\hfill \mbox{\textit{OCR Further Pure Core 2 2018 Q5 [5]}}