Question 19
Two objects, A and B, move on a horizontal surface. The velocity (m s−1) of Object A at time $t$ (s) is given by
$$\boldsymbol{v}_{\mathrm{A}}=0.7\,\hat{\boldsymbol\imath}+2.2\cos(0.4t)\,\hat{\boldsymbol\jmath},\quad t\geq0$$
At $t=0$, the position (m) of Object A relative to the origin, $O$, is $6.9\hat{\boldsymbol\imath}+2.3\hat{\boldsymbol\jmath}$.
At $t=0$, Object B is at $O$ and is moving directly towards Object A with a speed of 2.4 m s−1. Object B moves with a constant acceleration of magnitude 1.35 m s−2.
Given that the two objects collide, use vector calculus to determine the acceleration of Object B. Express your answer in Cartesian form, giving each component to two decimal places.
[6 marks]What the question doesn’t tell you
Have a go at the question cold first. Then open this to see the steps the question expects you to find for yourself.
- You have to build A’s position yourself. Integrate $\boldsymbol{v}_{\mathrm{A}}$, then use $\boldsymbol{r}_{\mathrm{A}}(0)=6.9\hat{\boldsymbol\imath}+2.3\hat{\boldsymbol\jmath}$ to find the constant.
- “Directly towards Object A” only fixes B’s starting direction. B is at $O$, so that direction is $6.9\hat{\boldsymbol\imath}+2.3\hat{\boldsymbol\jmath}$, and $\boldsymbol{v}_{\mathrm{B}}(0)$ is 2.4 times its unit vector.
- The direction of $\boldsymbol{a}_{\mathrm{B}}$ is unknown. “Magnitude 1.35” gives one equation, $p^2+q^2=1.35^2$, not the vector. B does not travel along the line $OA$.
- B’s position needs two integrations. The two constants come from $\boldsymbol{v}_{\mathrm{B}}(0)$ and $\boldsymbol{r}_{\mathrm{B}}(0)=\boldsymbol{0}$.
- “Collide” means the same position at the same time: $\boldsymbol{r}_{\mathrm{A}}(t)=\boldsymbol{r}_{\mathrm{B}}(t)$, with one value of $t$ in both components. Paths crossing is not enough.
- There are three unknowns, $p$, $q$ and $t$. The two component equations and the magnitude give three equations. Squaring and adding removes $p$ and $q$ and leaves an equation in $t$ alone.
- That equation mixes $t^4$ with $\sin(0.4t)$, so you can only solve it numerically, and only in radian mode.
- The acceleration only comes out once $t$ is known, by substituting back.
B is aimed at A, but its path has to curve to meet it
First watch a straight chase along the line of sight miss. Then watch both objects on their true paths, drawn to scale, arrive at the same point at the same moment.
The size of B’s acceleration is fixed and its direction is not. Only one direction puts B where A is at the moment A is there, and that direction is what the question wants.
Find the moment B can be where A is
Slide the time. The hollow dot is the ghost: where B would be with no acceleration. An acceleration of size 1.35 m s−2 moves B a distance of $0.675t^2$ away from the ghost, in a direction you don’t know yet. So at each time, B can only be somewhere on the dashed circle. Find the time when A sits exactly on it.
A is exactly on the circle, so B can be there at the same moment. That is the collision, at $t\approx2.893$ s. The orange arrow from the ghost to A is $\tfrac12\boldsymbol{a}_{\mathrm{B}} t^2$, so it points the same way as the acceleration, at about $65.5^\circ$ to $\hat{\boldsymbol\imath}$. Every route lands on $\boldsymbol{a}_{\mathrm{B}}\approx0.56\hat{\boldsymbol\imath}+1.23\hat{\boldsymbol\jmath}$.
A is outside the circle. The gap from the ghost to A is longer than $0.675t^2$, so no acceleration of size 1.35 can have put B where A is by this time.
A is inside the circle. The gap is now shorter than $0.675t^2$, so an acceleration of size 1.35 would carry B past where A is. Meeting this late would need a smaller acceleration.
Very close. Use the 0.01 s buttons.
Work through the solution one mark at a time
Try each step yourself before you reveal it.
Integrate each component of A’s velocity separately. For the $\hat{\boldsymbol\jmath}$ part, use $\displaystyle\int\cos(kt)\,dt=\frac1k\sin(kt)$ with $k=0.4$.
$$\begin{aligned}\boldsymbol{r}_{\mathrm{A}}&=\int\left(0.7\,\hat{\boldsymbol\imath}+2.2\cos(0.4t)\,\hat{\boldsymbol\jmath}\right)dt\\&=0.7t\,\hat{\boldsymbol\imath}+\frac{2.2}{0.4}\sin(0.4t)\,\hat{\boldsymbol\jmath}+\boldsymbol{c}\\&=0.7t\,\hat{\boldsymbol\imath}+5.5\sin(0.4t)\,\hat{\boldsymbol\jmath}+\boldsymbol{c}\end{aligned}$$
A collision is about positions, but the question gives you A’s velocity. Position is the antiderivative of velocity.
One vector constant covers both components. At $t=0$ both integrated terms are zero, so the constant is A’s starting position.
$$\begin{aligned}\boldsymbol{r}_{\mathrm{A}}(0)&=\boldsymbol{c}=6.9\hat{\boldsymbol\imath}+2.3\hat{\boldsymbol\jmath}\\\boldsymbol{r}_{\mathrm{A}}&=(6.9+0.7t)\,\hat{\boldsymbol\imath}\\&\quad+\left(2.3+5.5\sin(0.4t)\right)\hat{\boldsymbol\jmath}\end{aligned}$$
Check by differentiating: $5.5\times0.4\cos(0.4t)=2.2\cos(0.4t)$.
Marker: correctly determines the position vector of Object A
Find the direction from $O$ to A’s starting point, and its length.
$$\begin{aligned}\boldsymbol{r}_{\mathrm{A}}(0)-\boldsymbol{r}_{\mathrm{B}}(0)&=6.9\hat{\boldsymbol\imath}+2.3\hat{\boldsymbol\jmath}\\\sqrt{6.9^2+2.3^2}&=\sqrt{47.61+5.29}\\&=\sqrt{52.9}\approx7.27324\end{aligned}$$
“Directly towards Object A” at $t=0$ means B’s velocity points from $O$ to A’s starting point. A direction can only be scaled to a speed once it is a unit vector.
Make it a unit vector.
$$\begin{aligned}\frac{6.9\hat{\boldsymbol\imath}+2.3\hat{\boldsymbol\jmath}}{\sqrt{52.9}}&=\frac{3\hat{\boldsymbol\imath}+\hat{\boldsymbol\jmath}}{\sqrt{10}}\\&\approx0.948683\hat{\boldsymbol\imath}+0.316228\hat{\boldsymbol\jmath}\end{aligned}$$
Multiply by the speed.
$$\begin{aligned}\boldsymbol{v}_{\mathrm{B}}(0)&=2.4\times\frac{3\hat{\boldsymbol\imath}+\hat{\boldsymbol\jmath}}{\sqrt{10}}\\&=\frac{7.2}{\sqrt{10}}\hat{\boldsymbol\imath}+\frac{2.4}{\sqrt{10}}\hat{\boldsymbol\jmath}\\&\approx2.27684\hat{\boldsymbol\imath}+0.758947\hat{\boldsymbol\jmath}\end{aligned}$$
Velocity is speed times the unit vector of the direction of motion.
Store both components in your calculator rather than retyping rounded values.
Treating B as starting from rest does not earn this mark, although follow-through applies to the marks after it.
Marker: correctly determines the initial velocity of Object B
Give $\boldsymbol{a}_{\mathrm{B}}$ two unknown components and write its size down as an equation.
$$\begin{aligned}\boldsymbol{a}_{\mathrm{B}}&=p\,\hat{\boldsymbol\imath}+q\,\hat{\boldsymbol\jmath}\\p^2+q^2&=1.35^2=1.8225\end{aligned}$$
“Constant acceleration of magnitude 1.35” gives you the size of $\boldsymbol{a}_{\mathrm{B}}$, not its direction.
Do not assume it points along $3\hat{\boldsymbol\imath}+\hat{\boldsymbol\jmath}$. Nothing says so, and with that assumption B never gets closer to A than about 3.79 m. There is no collision time to find.
Integrate. The first constant of integration is $\boldsymbol{v}_{\mathrm{B}}(0)$.
$$\begin{aligned}\boldsymbol{v}_{\mathrm{B}}&=\int\left(p\,\hat{\boldsymbol\imath}+q\,\hat{\boldsymbol\jmath}\right)dt\\&=pt\,\hat{\boldsymbol\imath}+qt\,\hat{\boldsymbol\jmath}+\boldsymbol{c}_1\\\boldsymbol{c}_1&=\boldsymbol{v}_{\mathrm{B}}(0)\\\boldsymbol{v}_{\mathrm{B}}&=(2.27684+pt)\,\hat{\boldsymbol\imath}\\&\quad+(0.758947+qt)\,\hat{\boldsymbol\jmath}\end{aligned}$$
Here $p$ and $q$ are constants, so they integrate like numbers.
Integrate again. B starts at $O$, so the second constant is $\boldsymbol{0}$.
$$\begin{aligned}\boldsymbol{r}_{\mathrm{B}}&=\left(2.27684t+\tfrac12pt^2\right)\hat{\boldsymbol\imath}\\&\quad+\left(0.758947t+\tfrac12qt^2\right)\hat{\boldsymbol\jmath}+\boldsymbol{c}_2\\\boldsymbol{c}_2&=\boldsymbol{r}_{\mathrm{B}}(0)=\boldsymbol{0}\\\boldsymbol{r}_{\mathrm{B}}&=\left(2.27684t+\tfrac12pt^2\right)\hat{\boldsymbol\imath}\\&\quad+\left(0.758947t+\tfrac12qt^2\right)\hat{\boldsymbol\jmath}\end{aligned}$$
The $\tfrac12$ comes from $\displaystyle\int t\,dt=\tfrac12t^2$.
Assuming $\boldsymbol{a}_{\mathrm{B}}$ points along $3\hat{\boldsymbol\imath}+\hat{\boldsymbol\jmath}$ does not earn the fourth, fifth or sixth marks.
Leaving out the $\tfrac12$ gives $t\approx2.107$ and $0.81\hat{\boldsymbol\imath}+1.08\hat{\boldsymbol\jmath}$. That response does not earn this mark, but follow-through applies to the marks after it.
Marker: correctly determines the position vector of Object B in terms of the components of its acceleration
Set $\boldsymbol{r}_{\mathrm{A}}(t)=\boldsymbol{r}_{\mathrm{B}}(t)$ and match the components.
$$\begin{aligned}\hat{\boldsymbol\imath}&:\ 6.9+0.7t=2.27684t+\tfrac12pt^2\\\hat{\boldsymbol\jmath}&:\ 2.3+5.5\sin(0.4t)\\&\qquad=0.758947t+\tfrac12qt^2\end{aligned}$$
Collide means the same place at the same time. Two vectors are equal when their components are equal, so you get two equations with the same $t$ in both.
Get the acceleration terms on their own, using $0.7-2.27684=-1.57684$.
$$\begin{aligned}\tfrac12pt^2&=6.9-1.57684t\\\tfrac12qt^2&=2.3+5.5\sin(0.4t)\\&\quad-0.758947t\end{aligned}$$
That is three unknowns, $p$, $q$ and $t$, and three equations: these two and $p^2+q^2=1.8225$. You don’t know $p$ or $q$, but you do know $p^2+q^2$.
Square both equations and add them.
$$\begin{aligned}&\left(\tfrac12pt^2\right)^2+\left(\tfrac12qt^2\right)^2\\&\quad=\tfrac14t^4\left(p^2+q^2\right)\\&\quad=\tfrac14(1.8225)\,t^4\\&\quad=0.455625t^4\end{aligned}$$
$$\begin{aligned}&(6.9-1.57684t)^2\\&\quad+\left(2.3+5.5\sin(0.4t)-0.758947t\right)^2\\&\quad=0.455625t^4\end{aligned}$$
Squaring can’t add a false answer here. Once $t$ is known, the two equations above give exactly one $p$ and one $q$, and this equation says that they satisfy $p^2+q^2=1.8225$.
Marker: correctly uses the collision condition and the magnitude of the acceleration of Object B to form an equation in $t$
Solve the equation numerically, with your calculator in radian mode.
$$\begin{aligned}f(x)&=\left(6.9+\left(0.7-\frac{7.2}{\sqrt{10}}\right)x\right)^2\\&\quad+\left(2.3+5.5\sin(0.4x)-\frac{2.4}{\sqrt{10}}\,x\right)^2\\&\quad-0.455625x^4\end{aligned}$$
The equation mixes $t^4$ with $\sin(0.4t)$, so no algebra will isolate $t$. Radian mode matters because the integral you did in Step 1 only holds when the angle is in radians.
Find the zero, then write the time. Keep the stored value for the next step.
$$t\approx2.893\text{ s}$$
Graph it on $0\le x\le4$, $-100\le y\le60$ and use the zero tool, or solve $f(x)=0$ numerically from a starting value of 3. The calculator gives $x=2.8925546$.
Make sure there is no other root.
$$f(0)=52.9>0$$
The graph starts above the axis and crosses it once, at this root. So this is the only root, and there is no branch to reject.
In degree mode the calculator gives $t\approx2.251$ and $1.32\hat{\boldsymbol\imath}+0.27\hat{\boldsymbol\jmath}$. That response does not earn this mark, but follow-through applies to the sixth.
Marker: correctly determines the time at which the objects collide
Rearrange each component equation from Step 4 for its unknown. Multiply by 2 and divide by $t^2$.
$$\begin{aligned}p&=\frac{2(6.9-1.57684t)}{t^2}\\q&=\frac{2\left(2.3+5.5\sin(0.4t)-0.758947t\right)}{t^2}\end{aligned}$$
Now that $t$ is known, each component equation from Step 4 has one unknown. Dividing by $t^2$ is fine because $t\neq0$.
Substitute the stored $t$.
$$\begin{aligned}t^2&=8.36687\\6.9-1.57684t&=2.33890\\2.3+5.5\sin(0.4t)-0.758947t&=5.14056\end{aligned}$$
$$\begin{aligned}p&=\frac{2(2.33890)}{8.36687}=0.559087\\q&=\frac{2(5.14056)}{8.36687}=1.22879\end{aligned}$$
Carry $t$ to at least two decimal places, and use the stored value if you can. Rounding $t$ to 2.9 first gives $0.55\hat{\boldsymbol\imath}+1.22\hat{\boldsymbol\jmath}$, which does not earn this mark.
The size of the answer, $\sqrt{0.56^2+1.23^2}\approx1.35$, checks the components, but it does not replace the working.
Marker: correctly determines the acceleration of Object B
Integrate each component, dividing by 0.4 for the $\hat{\boldsymbol\jmath}$ part.
$$\begin{aligned}\boldsymbol{r}_{\mathrm{A}}&=\int\left(0.7\,\hat{\boldsymbol\imath}+2.2\cos(0.4t)\,\hat{\boldsymbol\jmath}\right)dt\\&=0.7t\,\hat{\boldsymbol\imath}+\frac{2.2}{0.4}\sin(0.4t)\,\hat{\boldsymbol\jmath}+\boldsymbol{c}\\&=0.7t\,\hat{\boldsymbol\imath}+5.5\sin(0.4t)\,\hat{\boldsymbol\jmath}+\boldsymbol{c}\end{aligned}$$
Collision is about positions, and $\boldsymbol{v}_{\mathrm{A}}=\dfrac{d\boldsymbol{r}_{\mathrm{A}}}{dt}$.
Find the constant from A’s starting position.
$$\begin{aligned}\boldsymbol{r}_{\mathrm{A}}(0)&=\boldsymbol{c}=6.9\hat{\boldsymbol\imath}+2.3\hat{\boldsymbol\jmath}\\\boldsymbol{r}_{\mathrm{A}}&=(6.9+0.7t)\,\hat{\boldsymbol\imath}\\&\quad+\left(2.3+5.5\sin(0.4t)\right)\hat{\boldsymbol\jmath}\end{aligned}$$
At $t=0$ both integrated terms are 0, so the constant is A’s starting position.
Marker: correctly determines the position vector of Object A
Make B’s first direction a unit vector.
$$\begin{aligned}\sqrt{6.9^2+2.3^2}&=\sqrt{52.9}\\&=2.3\sqrt{10}\approx7.27324\end{aligned}$$
$$\begin{aligned}\frac{6.9\hat{\boldsymbol\imath}+2.3\hat{\boldsymbol\jmath}}{\sqrt{52.9}}&=\frac{3\hat{\boldsymbol\imath}+\hat{\boldsymbol\jmath}}{\sqrt{10}}\\&\approx0.948683\hat{\boldsymbol\imath}+0.316228\hat{\boldsymbol\jmath}\end{aligned}$$
B starts at $O$ heading straight at A’s starting point, so its first direction is $6.9\hat{\boldsymbol\imath}+2.3\hat{\boldsymbol\jmath}$.
Scale it to the speed 2.4.
$$\begin{aligned}\boldsymbol{v}_{\mathrm{B}}(0)&=2.4\times\frac{3\hat{\boldsymbol\imath}+\hat{\boldsymbol\jmath}}{\sqrt{10}}\\&=\frac{7.2}{\sqrt{10}}\hat{\boldsymbol\imath}+\frac{2.4}{\sqrt{10}}\hat{\boldsymbol\jmath}\\&\approx2.27684\hat{\boldsymbol\imath}+0.758947\hat{\boldsymbol\jmath}\end{aligned}$$
Treating B as starting from rest does not earn this mark, although follow-through applies to the marks after it.
Marker: correctly determines the initial velocity of Object B
Write the acceleration in polar form, with $\theta$ measured anticlockwise from $\hat{\boldsymbol\imath}$.
$$\boldsymbol{a}_{\mathrm{B}}=1.35\cos(\theta)\,\hat{\boldsymbol\imath}+1.35\sin(\theta)\,\hat{\boldsymbol\jmath}$$
A vector of size 1.35 at angle $\theta$ to $\hat{\boldsymbol\imath}$ has components $1.35\cos(\theta)$ and $1.35\sin(\theta)$. The magnitude is built in straight away, and you are left with one unknown direction instead of $p$ and $q$.
Integrate, and use $\boldsymbol{v}_{\mathrm{B}}(0)$ for the first constant.
$$\begin{aligned}\boldsymbol{v}_{\mathrm{B}}&=1.35t\cos(\theta)\,\hat{\boldsymbol\imath}\\&\quad+1.35t\sin(\theta)\,\hat{\boldsymbol\jmath}+\boldsymbol{c}_1\\\boldsymbol{c}_1&=\boldsymbol{v}_{\mathrm{B}}(0)\\\boldsymbol{v}_{\mathrm{B}}&=\left(2.27684+1.35t\cos(\theta)\right)\hat{\boldsymbol\imath}\\&\quad+\left(0.758947+1.35t\sin(\theta)\right)\hat{\boldsymbol\jmath}\end{aligned}$$
The acceleration is constant, so $\theta$ is a constant too.
Integrate again. B starts at $O$, so $\boldsymbol{c}_2=\boldsymbol{0}$, and $\tfrac12\times1.35=0.675$.
$$\begin{aligned}\boldsymbol{r}_{\mathrm{B}}&=\left(2.27684t+0.675t^2\cos(\theta)\right)\hat{\boldsymbol\imath}\\&\quad+\left(0.758947t+0.675t^2\sin(\theta)\right)\hat{\boldsymbol\jmath}\end{aligned}$$
Leaving out the $\tfrac12$ puts $1.35t^2$ where $0.675t^2$ should be, and gives $t\approx2.107$ and $0.81\hat{\boldsymbol\imath}+1.08\hat{\boldsymbol\jmath}$. That response does not earn this mark, but follow-through applies to the marks after it.
Marker: correctly determines the position vector of Object B in terms of the components of its acceleration
Set $\boldsymbol{r}_{\mathrm{A}}=\boldsymbol{r}_{\mathrm{B}}$ at the same $t$. Match the components and get the acceleration terms on their own.
$$\begin{aligned}0.675t^2\cos(\theta)&=6.9-1.57684t\\0.675t^2\sin(\theta)&=2.3+5.5\sin(0.4t)\\&\quad-0.758947t\end{aligned}$$
Square both and add.
$$\begin{aligned}&0.675^2t^4\cos^2(\theta)\\&\quad+0.675^2t^4\sin^2(\theta)\\&\quad=0.455625t^4\left(\cos^2(\theta)+\sin^2(\theta)\right)\\&\quad=0.455625t^4\end{aligned}$$
$$\begin{aligned}&(6.9-1.57684t)^2\\&\quad+\left(2.3+5.5\sin(0.4t)-0.758947t\right)^2\\&\quad=0.455625t^4\end{aligned}$$
$\cos^2(\theta)+\sin^2(\theta)=1$ removes the unknown angle. This is the same equation as Method 1.
Marker: correctly uses the collision condition and the magnitude of the acceleration of Object B to form an equation in $t$
The equation cannot be solved algebraically. Solve it with the same calculator procedure as Method 1, Step 5.
$$x=2.8925546$$
Use radian mode, because the calculus of $\sin(0.4t)$ assumes radians. Keep the stored value.
Write the time. It is the only positive root (Method 1, Step 5).
$$t\approx2.893\text{ s}$$
In degree mode the calculator gives $t\approx2.251$, which does not earn this mark, although follow-through applies to the sixth.
Marker: correctly determines the time at which the objects collide
With the stored $t$, $0.675t^2=5.64764$. Divide each component equation by it.
$$\begin{aligned}\cos(\theta)&=\frac{2.33890}{5.64764}=0.414138>0\\\sin(\theta)&=\frac{5.14056}{5.64764}=0.910214>0\end{aligned}$$
Check the signs before you trust $\tan^{-1}$, because it only returns angles between $-90^\circ$ and $90^\circ$. Both are positive, so $\theta$ is in the first quadrant and the calculator’s value needs no adjustment.
Find $\theta$ from $\tan(\theta)$.
$$\begin{aligned}\tan(\theta)&=\frac{5.14056}{2.33890}=2.19785\\\theta&=\tan^{-1}(2.19785)\\&\approx65.5349^\circ\ \ (1.14380\text{ radians})\end{aligned}$$
Substitute $\theta$ back into the polar form.
$$\begin{aligned}\boldsymbol{a}_{\mathrm{B}}&=1.35\cos(65.5349^\circ)\,\hat{\boldsymbol\imath}\\&\quad+1.35\sin(65.5349^\circ)\,\hat{\boldsymbol\jmath}\\&=0.559087\hat{\boldsymbol\imath}+1.22879\hat{\boldsymbol\jmath}\end{aligned}$$
Either keep $\theta$ in radians with the calculator still in radian mode, or switch to degree mode for this step only.
Carry $\theta$ to at least one decimal place. $\theta=65.5^\circ$ still gives $0.56\hat{\boldsymbol\imath}+1.23\hat{\boldsymbol\jmath}$, but $\theta=66^\circ$ gives $0.55\hat{\boldsymbol\imath}+1.23\hat{\boldsymbol\jmath}$, which does not earn this mark.
The size of the answer, $\sqrt{0.56^2+1.23^2}\approx1.35$, checks the components, but it does not replace the working.
Marker: correctly determines the acceleration of Object B
Integrate each component, dividing by 0.4 for the $\hat{\boldsymbol\jmath}$ part.
$$\begin{aligned}\boldsymbol{r}_{\mathrm{A}}&=\int\left(0.7\,\hat{\boldsymbol\imath}+2.2\cos(0.4t)\,\hat{\boldsymbol\jmath}\right)dt\\&=0.7t\,\hat{\boldsymbol\imath}+\frac{2.2}{0.4}\sin(0.4t)\,\hat{\boldsymbol\jmath}+\boldsymbol{c}\\&=0.7t\,\hat{\boldsymbol\imath}+5.5\sin(0.4t)\,\hat{\boldsymbol\jmath}+\boldsymbol{c}\end{aligned}$$
Collision is about positions, and $\boldsymbol{v}_{\mathrm{A}}=\dfrac{d\boldsymbol{r}_{\mathrm{A}}}{dt}$.
Find the constant from A’s starting position.
$$\begin{aligned}\boldsymbol{r}_{\mathrm{A}}(0)&=\boldsymbol{c}=6.9\hat{\boldsymbol\imath}+2.3\hat{\boldsymbol\jmath}\\\boldsymbol{r}_{\mathrm{A}}&=(6.9+0.7t)\,\hat{\boldsymbol\imath}\\&\quad+\left(2.3+5.5\sin(0.4t)\right)\hat{\boldsymbol\jmath}\end{aligned}$$
At $t=0$ both integrated terms are 0, so the constant is A’s starting position.
Marker: correctly determines the position vector of Object A
Make B’s first direction a unit vector.
$$\begin{aligned}\sqrt{6.9^2+2.3^2}&=\sqrt{52.9}\\&=2.3\sqrt{10}\approx7.27324\end{aligned}$$
$$\begin{aligned}\frac{6.9\hat{\boldsymbol\imath}+2.3\hat{\boldsymbol\jmath}}{\sqrt{52.9}}&=\frac{3\hat{\boldsymbol\imath}+\hat{\boldsymbol\jmath}}{\sqrt{10}}\\&\approx0.948683\hat{\boldsymbol\imath}+0.316228\hat{\boldsymbol\jmath}\end{aligned}$$
B starts at $O$ heading straight at A’s starting point, so its first direction is $6.9\hat{\boldsymbol\imath}+2.3\hat{\boldsymbol\jmath}$.
Scale it to the speed 2.4.
$$\begin{aligned}\boldsymbol{v}_{\mathrm{B}}(0)&=2.4\times\frac{3\hat{\boldsymbol\imath}+\hat{\boldsymbol\jmath}}{\sqrt{10}}\\&=\frac{7.2}{\sqrt{10}}\hat{\boldsymbol\imath}+\frac{2.4}{\sqrt{10}}\hat{\boldsymbol\jmath}\\&\approx2.27684\hat{\boldsymbol\imath}+0.758947\hat{\boldsymbol\jmath}\end{aligned}$$
Treating B as starting from rest does not earn this mark, although follow-through applies to the marks after it.
Marker: correctly determines the initial velocity of Object B
Keep $\boldsymbol{a}_{\mathrm{B}}$ as a single unknown vector and integrate.
$$\begin{aligned}\boldsymbol{v}_{\mathrm{B}}&=\int\boldsymbol{a}_{\mathrm{B}}\,dt=\boldsymbol{a}_{\mathrm{B}} t+\boldsymbol{c}_1\\\boldsymbol{c}_1&=\boldsymbol{v}_{\mathrm{B}}(0)\end{aligned}$$
$\boldsymbol{a}_{\mathrm{B}}$ is constant, so it integrates like a constant.
Integrate again.
$$\begin{aligned}\boldsymbol{r}_{\mathrm{B}}&=\int\left(\boldsymbol{a}_{\mathrm{B}} t+\boldsymbol{v}_{\mathrm{B}}(0)\right)dt\\&=\tfrac12\boldsymbol{a}_{\mathrm{B}} t^2+\boldsymbol{v}_{\mathrm{B}}(0)\,t+\boldsymbol{c}_2\\\boldsymbol{c}_2&=\boldsymbol{r}_{\mathrm{B}}(0)=\boldsymbol{0}\\\boldsymbol{r}_{\mathrm{B}}&=\boldsymbol{v}_{\mathrm{B}}(0)\,t+\tfrac12\boldsymbol{a}_{\mathrm{B}} t^2\end{aligned}$$
The $\tfrac12$ is the part people drop. Without it you get $t\approx2.107$ and $0.81\hat{\boldsymbol\imath}+1.08\hat{\boldsymbol\jmath}$. That response does not earn this mark, but follow-through applies to the marks after it.
Marker: correctly determines the position vector of Object B in terms of the components of its acceleration
Set $\boldsymbol{r}_{\mathrm{A}}(t)=\boldsymbol{r}_{\mathrm{B}}(t)$ and get the acceleration term on its own.
$$\begin{aligned}\boldsymbol{r}_{\mathrm{A}}(t)&=\boldsymbol{v}_{\mathrm{B}}(0)\,t+\tfrac12\boldsymbol{a}_{\mathrm{B}} t^2\\\boldsymbol{r}_{\mathrm{A}}(t)-\boldsymbol{v}_{\mathrm{B}}(0)\,t&=\tfrac12\boldsymbol{a}_{\mathrm{B}} t^2\end{aligned}$$
$\boldsymbol{v}_{\mathrm{B}}(0)\,t$ is where B would be with no acceleration: the ghost. At the collision, $\tfrac12\boldsymbol{a}_{\mathrm{B}} t^2$ has to cover the gap from the ghost to A exactly.
Write the left side in components.
$$\begin{aligned}&\boldsymbol{r}_{\mathrm{A}}(t)-\boldsymbol{v}_{\mathrm{B}}(0)\,t\\&\quad=(6.9+0.7t-2.27684t)\,\hat{\boldsymbol\imath}\\&\qquad+\left(2.3+5.5\sin(0.4t)-0.758947t\right)\hat{\boldsymbol\jmath}\\&\quad=(6.9-1.57684t)\,\hat{\boldsymbol\imath}\\&\qquad+\left(2.3+5.5\sin(0.4t)-0.758947t\right)\hat{\boldsymbol\jmath}\end{aligned}$$
Take the magnitude of both sides.
$$\begin{aligned}&\sqrt{\begin{aligned}&(6.9-1.57684t)^2\\&+\left(2.3+5.5\sin(0.4t)-0.758947t\right)^2\end{aligned}}\\&\quad=0.675t^2\end{aligned}$$
You don’t know the direction of $\boldsymbol{a}_{\mathrm{B}}$, but you do know its size. Taking the magnitude of both sides removes the direction. Since $t^2\ge0$, $\left\lvert\tfrac12\boldsymbol{a}_{\mathrm{B}} t^2\right\rvert=\tfrac12t^2\lvert\boldsymbol{a}_{\mathrm{B}}\rvert=0.675t^2$.
Marker: correctly uses the collision condition and the magnitude of the acceleration of Object B to form an equation in $t$
Each side is easy to graph, and the collision time is where the curves meet. Graph both.
$$\begin{aligned}y_1&=\sqrt{\begin{aligned}&\left(6.9+\left(0.7-\frac{7.2}{\sqrt{10}}\right)x\right)^2\\&+\left(2.3+5.5\sin(0.4x)-\frac{2.4}{\sqrt{10}}\,x\right)^2\end{aligned}}\\y_2&=0.675x^2\end{aligned}$$
Use radian mode, for the same reason as in Method 1. Graph both on $0\le x\le4$, $0\le y\le12$ and use the intersect tool for $x>0$. The calculator gives $x=2.8925546$ and $y=5.64764$. Keep the stored value.
Write the time.
$$t\approx2.893\text{ s}$$
There is only one intersection. $y_1$ starts at $\sqrt{52.9}\approx7.27324$ and falls to 4.79861 at $x=4$, while $y_2$ rises from 0 to 10.8.
In degree mode the calculator gives $t\approx2.251$, which does not earn this mark, although follow-through applies to the sixth.
Marker: correctly determines the time at which the objects collide
Rearrange the Step 4 equation for $\boldsymbol{a}_{\mathrm{B}}$ and substitute the stored $t$.
$$\begin{aligned}\boldsymbol{a}_{\mathrm{B}}&=\frac{2}{t^2}\left(\boldsymbol{r}_{\mathrm{A}}(t)-\boldsymbol{v}_{\mathrm{B}}(0)\,t\right)\\&=\frac{2}{8.36687}\left(2.33890\hat{\boldsymbol\imath}+5.14056\hat{\boldsymbol\jmath}\right)\\&=0.559087\hat{\boldsymbol\imath}+1.22879\hat{\boldsymbol\jmath}\end{aligned}$$
Carry $t$ to at least two decimal places. Rounding $t$ to 2.9 first gives $0.55\hat{\boldsymbol\imath}+1.22\hat{\boldsymbol\jmath}$, which does not earn this mark.
The size of the answer, $\sqrt{0.56^2+1.23^2}\approx1.35$, checks the components, but it does not replace the working.
Marker: correctly determines the acceleration of Object B
Each phrase of the question gave you something
The engine here is inversion. The unknown is the acceleration inside $\boldsymbol{r}_{\mathrm{B}}$, and you only know its size. So “collide” has to become two component equations, which you square and add. That lets the hidden third fact, $\lvert\boldsymbol{a}_{\mathrm{B}}\rvert=1.35$, remove the direction. Then you solve for $t$ on a calculator and substitute back. The wording also invites a straight-line chase along $OA$, which never collides, and it quietly expects a unit vector step for “directly towards”.
If I had to guess where you might lose marks…
- Assuming B moves in a straight line. Taking the acceleration along $6.9\hat{\boldsymbol\imath}+2.3\hat{\boldsymbol\jmath}$ means the objects never meet. They are closest at $t\approx2.36$ s, still 3.79 m apart, so there is no collision time to find.
- Dropping the $\tfrac12$. Writing $\boldsymbol{r}_{\mathrm{B}}=\boldsymbol{v}_{\mathrm{B}}(0)\,t+\boldsymbol{a}_{\mathrm{B}} t^2$ gives $t\approx2.107$ s and $\boldsymbol{a}_{\mathrm{B}}\approx0.81\hat{\boldsymbol\imath}+1.08\hat{\boldsymbol\jmath}$.
- Solving in degree mode. This gives $t\approx2.251$ s and $\boldsymbol{a}_{\mathrm{B}}\approx1.32\hat{\boldsymbol\imath}+0.27\hat{\boldsymbol\jmath}$.
- Starting B from rest. Losing the 2.4 m s−1 gives $t\approx4.317$ s and $\boldsymbol{a}_{\mathrm{B}}\approx1.065\hat{\boldsymbol\imath}+0.830\hat{\boldsymbol\jmath}$.
- Rounding $t$ to 2.9 before you substitute back. This gives $0.55\hat{\boldsymbol\imath}+1.22\hat{\boldsymbol\jmath}$. Carry $t$ to at least two decimal places, and use the stored value if you can.
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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