Question 4
An object moves in a straight line such that its velocity $\left(\text{m s}^{-1}\right)$ is given by $v(t)=t^2-e^{-t}$ for $0\le t\le5$, where $t$ represents time (s).
Determine the time, $t$, when the acceleration of the object is $2\ \text{m s}^{-2}$.
That’s it. Step through the hints below to see why. In 2025, 56.8% of Queensland students chose (A).
Not quite. $1.157$ solves $2t-e^{-t}=2$. The derivative of $-e^{-t}$ is $+e^{-t}$, because the chain rule brings down a factor of $-1$. In 2025, 12.0% of Queensland students chose (B).
Not quite. $3.865$ is $2(2)-e^{-2}$: the derivative with a sign slip, evaluated at $t=2$. The 2 in the question is an acceleration, so set $a(t)=2$ and solve for $t$. In 2025, 7.5% of Queensland students chose (C).
Not quite. $4.135$ is $a(2)=4+e^{-2}$, the acceleration at $t=2$. The question gives you the acceleration and asks for the time, so solve $a(t)=2$. In 2025, 23.3% of Queensland students chose (D).
Here $a(t)=2$. It happens only once, because $a(t)$ is increasing for every $t\ge0$.
At $t=2$, $a(2)=4+e^{-2}\approx4.135$. That’s the number in option (D), but it’s an acceleration, not a time.
The dashed curve is what you get if you differentiate $-e^{-t}$ as $-e^{-t}$. It crosses 2 at $t\approx1.157$, option (B), but it isn’t the acceleration. The chain rule gives $+e^{-t}$.
Here $a(t)$ is 2. Move the orange point until it sits on the dashed line $a=2$.
Acceleration is the rate of change of velocity, so differentiate $v(t)$. Watch the sign on the exponential term: the chain rule brings down a factor of $-1$.
$\displaystyle \frac{d}{dt}\left(-e^{-t}\right)=-e^{-t}\times(-1)$$\displaystyle =e^{-t}$
$$a(t)=v'(t)=2t+e^{-t}$$
The question gives you the acceleration and asks for the time. Set $a(t)$ equal to 2 and solve for $t$.
$$2t+e^{-t}=2$$
This can’t be solved by hand, so use technology. Graph both sides and find where they meet on $0\le t\le5$, or use the numerical solver.
$$t\approx0.768$$
Substitute back to check.
$\displaystyle a(0.768)=2(0.768)+e^{-0.768}$$\displaystyle \approx1.536+0.464$$\displaystyle =2.000$
There is only one answer. $a'(t)=2-e^{-t}$ is positive for every $t\ge0$, so $a(t)$ keeps increasing and passes 2 just once.
$$a(t)=v'(t)=2t+e^{-t}$$
$$2t+e^{-t}=2\ \Rightarrow\ t\approx0.768$$
Solve with technology on $0\le t\le5$. The acceleration is $2\ \text{m s}^{-2}$ at about 0.768 s.
How Queensland students answered in 2025
From the QCAA Mathematical Methods subject report for 2025. The percentages can add to less than 100 because some students didn’t answer.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2025 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2025, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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