Question 1
A particle moves in a straight line. When its displacement from the origin is $x$ m, its velocity is $v=\sqrt{x^{2}+9}$ m s$^{-1}$.
Determine the acceleration (m s$^{-2}$) of the particle when $x=4$.
That’s it. Step through the hints below to see why.
Not quite. You took $a=\frac{dv}{dx}=\frac{x}{\sqrt{x^2+9}}=\frac{4}{5}=0.8$.
Not quite. You gave the velocity, $v=\sqrt{16+9}=5$.
Not quite. You used $a=\frac{d}{dx}\left(v^2\right)=2x=8$, dropping the $\tfrac12$.
Use $a=\dfrac{d}{dx}\left(\tfrac12v^2\right)$.
$$a=\frac{d}{dx}\left(\tfrac12\left(x^2+9\right)\right)=x=4$$
$v$ is given as a function of $x$.
$$a=\frac{d}{dx}\left(\tfrac12\left(x^2+9\right)\right)=x=4$$
The acceleration is $4$ m s$^{-2}$.
Only mark this done when you could do it without help. Reading the solution does not count.
A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.
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