Question 9
A mechanism extends an initially stationary rod 15 mm at a final velocity of 2.5 m/s.
The acceleration of the rod is
Not quite. $2.5^2\div0.015=417$ leaves out the 2 in $v^2=u^2+2as$. With that acceleration the rod would reach 2.5 m/s after only 7.5 mm, half the extension. In 2022, 7.9% of Queensland students chose (A).
That is it. Step through the hints below to see why. In 2022, 63.2% of Queensland students chose (B).
Not quite. One way to land on 38 is $2.5\times15=37.5$, which multiplies the velocity by the extension left in millimetres. A velocity times a distance is not an acceleration at all. You need the velocity squared divided by twice the distance in metres. In 2022, 16.4% of Queensland students chose (C).
Not quite. $2.5^2\div(2\times0.15)=20.8$ turns 15 mm into 0.15 m. A millimetre is a thousandth of a metre, so 15 mm is 0.015 m and this answer is ten times too small. In 2022, 11.7% of Queensland students chose (D).
“Initially stationary” means the rod starts from rest, so $u=0$. You also know the final velocity and the distance. You want the acceleration, and the question says nothing about time.
$$u=0,\quad v=2.5\text{ m/s},\quad s=15\text{ mm},\quad a=\,?$$
The time is left over, so use the motion equation without $t$ in it, $v^2=u^2+2as$.
Convert the distance before you substitute. A millimetre is a thousandth of a metre, so 15 mm is 0.015 m.
$$a=\frac{v^2-u^2}{2s}=\frac{2.5^2-0^2}{2\times0.015}$$
$$a=208.3\text{ m/s}^2$$
Keep the 2. Leaving it out gives (A), and 0.15 m gives (D).
Check it on the velocity-time graph. From rest at a steady acceleration the graph is a straight line, and the triangle under it is the 15 mm.
$$\tfrac12\times2.5\times t=0.015$$
$$t=0.012\text{ s}$$
$$a=\frac{v-u}{t}=\frac{2.5}{0.012}=208.3\text{ m/s}^2$$
The slope of the line is the acceleration, and it rounds to 208 m/s².
$$a=\frac{v^2-u^2}{2s}=\frac{2.5^2-0^2}{2\times0.015}$$
$$a=208.3\text{ m/s}^2\approx208\text{ m/s}^2$$
The rod starts from rest, so $u=0$. No time is given, so $v^2=u^2+2as$ is the equation, and the 15 mm goes in as 0.015 m.
The most common wrong answer was (C). One route to it is $2.5\times15$, the velocity times the extension still in millimetres. Convert every length to metres first, then let the missing quantity choose the equation.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording and any figures are from the 2022 QCAA Engineering external assessment, © State of Queensland (QCAA) 2022, licensed under CC BY 4.0, and have been adapted, except where a third-party source is named. Tangent Tuition is not affiliated with the QCAA.
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