Question 17
A particle moves in a straight line so that its acceleration (m s$^{-2}$) is given by $a=x+1$, where $x$ is its displacement (m) from the origin $O$ at time $t$ seconds.
Initially, the particle is at $O$ and has a velocity of 1 m s$^{-1}$.
Determine the time at which the particle has a speed of 5 m s$^{-1}$.
[6 marks]A particle that speeds up the further it goes
Its acceleration depends on where it is, not on the time. Watch velocity and displacement move together, then see when the speed reaches 5 m s−1.
Drawn to scale, in slow motion. The graph’s $x$-axis lines up with the number line above it, so the point on the graph sits directly below the particle.
Run the clock until the speed is 5
Move the time $t$. The particle moves along the number line and its velocity arrow grows. Find the moment its speed is 5 m s$^{-1}$. Each route shows the graph and readouts that route would use.
The speed is 5 m s$^{-1}$. Here $e^t=5$, so $t=\ln(5)\approx1.61$ seconds, and the particle is at $x=4$.
This is the start: the particle is at $O$ with velocity 1 m s$^{-1}$. These are the initial conditions that fix both constants and pick the branch $v=x+1$.
The speed is still under 5 m s$^{-1}$. Move later in time.
The speed is now over 5 m s$^{-1}$. Move back in time.
Very close. Use the nudge buttons.
Work through the solution one mark at a time
Try each step yourself before you reveal it.
Write the acceleration as $v\,\dfrac{dv}{dx}$, a form with only $v$ and $x$.
$$v\,\frac{dv}{dx}=x+1$$
The acceleration is given in terms of $x$, not $t$, so $a=\dfrac{dv}{dt}$ can’t be integrated directly. By the chain rule, $a=\dfrac{dv}{dx}\cdot\dfrac{dx}{dt}=v\,\dfrac{dv}{dx}$.
Marker: correctly uses an expression for acceleration in terms of displacement
Separate the variables and integrate each side, then use $x=0$ and $v=1$ at the start.
$$\begin{aligned}&\int v\,dv=\int(x+1)\,dx\\&\tfrac12v^2=\tfrac12x^2+x+c\\&x=0,\ v=1\Rightarrow c=\tfrac12\end{aligned}$$
Marker: correctly determines $\tfrac12v^2$ in terms of $x$, including the constant
Double both sides, factorise and take the positive root.
$$\begin{aligned}v^2&=x^2+2x+1=(x+1)^2\\v&=x+1\end{aligned}$$
Give the reason for the sign: $v=1>0$ when $x=0$. To change sign, $v$ would have to pass through $0$, which needs $x=-1$, but while $v>0$ the particle only moves right from $x=0$. So $v>0$ throughout.
A square root gives two branches, and the answer depends on the direction of motion, so you must justify the sign, not assume it.
Marker: correctly determines $v$ in terms of $x$, with reasoning for the sign
Write $v$ as $\dfrac{dx}{dt}$, separate the variables (no modulus is needed, since $x+1>0$) and use $x=0$ when $t=0$.
$$\begin{aligned}&\frac{dx}{dt}=x+1\Rightarrow\int\frac{1}{x+1}\,dx=\int1\,dt\\&\ln(x+1)=t+c_2,\quad c_2=0\\&x=e^t-1\end{aligned}$$
Marker: correctly determines a relationship between $t$ and $x$, or between $t$ and $v$
Speed is $\left\lvert v\right\rvert$, and $v=x+1=e^t$ is always positive, so set $e^t=5$.
$$\text{speed}=|v|=e^t=5\Rightarrow t=\ln(5)$$
Marker: correctly determines the time
Set out the acceleration form, both constants from the initial conditions and the reasoning for the sign in order, then finish with a concluding statement.
Marker: shows logical organisation, having fully attempted the question
The first three marks are gained exactly as in Method 1.
$$\begin{aligned}v\,\frac{dv}{dx}&=x+1\\\tfrac12v^2&=\tfrac12x^2+x+\tfrac12\\v&=x+1\quad(v>0\text{ throughout})\end{aligned}$$
Marker: correctly uses an expression for acceleration in terms of displacement
Marker: correctly determines $\tfrac12v^2$ in terms of $x$, including the constant
Marker: correctly determines $v$ in terms of $x$, with reasoning for the sign
Find where the speed is $5$, then write the time as an integral over displacement.
$$\begin{aligned}&x+1=5\Rightarrow x=4\\&t=\int_{0}^{4}\frac{1}{x+1}\,dx\end{aligned}$$
$x=-6$ is not reached, since $v>0$ and $x$ increases from $0$. Flipping $\dfrac{dx}{dt}=x+1$ gives $\dfrac{dt}{dx}=\dfrac{1}{x+1}$, so integrating from $x=0$ to $x=4$ gives the time taken.
Marker: correctly determines a relationship between $t$ and $x$, or between $t$ and $v$
Evaluate the integral, then finish with a concluding statement.
$$\begin{aligned}t&=\Big[\ln(x+1)\Big]_{0}^{4}\\&=\ln(5)-\ln(1)=\ln(5)\end{aligned}$$
Marker: correctly determines the time
Marker: shows logical organisation, having fully attempted the question
The first three marks are gained exactly as in Method 1, giving
$$v=x+1\quad(v>0\text{ throughout})$$
Marker: correctly uses an expression for acceleration in terms of displacement
Marker: correctly determines $\tfrac12v^2$ in terms of $x$, including the constant
Marker: correctly determines $v$ in terms of $x$, with reasoning for the sign
Notice that $a=v$, then separate the variables and use $v=1$ when $t=0$.
$$\begin{aligned}&a=x+1=v\Rightarrow\frac{dv}{dt}=v\\&\ln(v)=t+c_3,\quad c_3=0\\&v=e^t\end{aligned}$$
$a$ and $v$ are the same expression, $x+1$, so this differential equation involves only $v$ and $t$. Dividing by $v$ is allowed since $v>0$.
Marker: correctly determines a relationship between $t$ and $x$, or between $t$ and $v$
Set the speed equal to $5$ and finish with a concluding statement.
$$e^t=5\Rightarrow t=\ln(5)$$
Marker: correctly determines the time
Marker: shows logical organisation, having fully attempted the question
Each phrase of the question gave you something
If I had to guess where you might lose marks…
- Dropping the constant $c$ in Step 2. This gives $v^2=x^2+2x$, so $v^2=0$ at $x=0$, which contradicts the initial velocity of 1 m s$^{-1}$. Pushing on, $x^2+2x=25$ gives $x=-1+\sqrt{26}$ (rejecting $x=-1-\sqrt{26}$) instead of $x=4$, and every later value is wrong. You lose Mark 2 and everything that depends on it.
- Integrating $a$ with respect to $x$ to get $v$. Writing $v=\int(x+1)\,dx=\tfrac12x^2+x+1$ confuses $a=\dfrac{dv}{dt}$ with $\dfrac{dv}{dx}$. It leads to $\tfrac12x^2+x+1=5$, so $x^2+2x-8=0$, $(x+4)(x-2)=0$ and $x=2$ (rejecting $x=-4$).
- Choosing $v=-(x+1)$, or giving no reason for the sign. The negative branch gives $v=-1$ at $t=0$, contradicting the given initial velocity. Following it through gives $x=e^{-t}-1$, $v=-e^{-t}$, and $e^{-t}=5$ gives $t=-\ln(5)$, a negative time. Writing $v=\pm(x+1)$, or just $v=x+1$ with no justification, loses Mark 3.
- Stopping at $x=4$. In Method 2, $x=4$ is where the particle is when its speed is 5 m s$^{-1}$, not the time. The question asks for $t$.
- Confusing speed with displacement. Setting $x=5$ in $x=e^t-1$ gives $e^t=6$ and $t=\ln(6)$ instead of setting $v=e^t=5$.
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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