Two boats. The obvious moment to check is t = 3.33, when one is level with where the other started, and the gap there is 3.33 km. The closest they ever actually get is 3.16 km, at t = 3.
The curve on the right is the gap against time. It has one lowest point, and that point is the answer.
Constant velocity motion is a vector equation of a line with the parameter renamed:
r = r0 + tv
and every word of 3.11 still applies. What is new is the names:
So "velocity" and "speed" are not interchangeable in a question. A boat travelling at (0, 3) has a speed of 3 and is going due north; a boat at (3, 4) has a speed of 5 and is heading on a bearing of 037°. Asking for speed and receiving a vector loses the mark, and so does the reverse.
The vector between the two objects is the whole method. A is at (3t, 4t) and B is at (10, 3t), so
AB = rB − rA = (10 − 3t, −t)
One vector, written once, and every question about the pair is a question about it. The 4t and the 3t cancel in the second component, which is why the gap closes so slowly in y.
Work with the square of the distance, because it has no square root in it and it is least exactly where the distance is:
|AB|² = (10 − 3t)² + t² = 100 − 60t + 9t² + t² = 10t² − 60t + 100
That is a quadratic in t, opening upwards, so its least value is at the vertex:
t = 60 / (2 × 10) = 3
and at t = 3 the squared distance is 90 − 180 + 100 = 10, so the distance itself is √10 = 3.16.
| t | AB | Distance apart |
|---|---|---|
| 0 | (10, 0) | 10.00 |
| 1 | (7, −1) | 7.07 |
| 2 | (4, −2) | 4.47 |
| 3 | (1, −3) | 3.16 |
| 3.33 | (0, −3.33) | 3.33 |
| 4 | (−2, −4) | 4.47 |
| 5 | (−5, −5) | 7.07 |
Read the symmetry: 4.47 at t = 2 and again at t = 4, 7.07 at t = 1 and again at t = 5. A table symmetric about t = 3 can only have its minimum at t = 3, and that check takes no algebra at all.
Two wrong times that feel right.
The only reliable route is to write AB as a function of t and minimise its square.
Do they ever collide? Only if AB = 0 for some single value of t, which needs both components zero at the same time. Here 10 − 3t = 0 gives t = 3.33 and −t = 0 gives t = 0. Different times, so no collision, and the least gap of 3.16 is how close they come. A question asking "do they collide" is asking you to show the two values of t disagree, not to find a minimum.
The method does not change, which is the reason for using vectors at all. Put the two boats in three dimensions, with B five units higher and neither of them changing height: A from the origin at (3, 4, 0) and B from (10, 0, 5) at (0, 3, 0). Then
AB = (10 − 3t, −t, 5)
and the squared distance picks up a constant:
|AB|² = 10t² − 60t + 125
The 25 raises the whole quadratic and leaves the vertex where it was, so the closest approach is still at t = 3, and the least distance is √35 = 5.92. The gap runs 11.18, 8.66, 6.71, 5.92, 6.71, still symmetric about t = 3.
A separation that neither object can close never moves the time of closest approach, only the distance. That is worth knowing, because it means a three-dimensional question with one constant component is a two-dimensional question with a constant added at the end.
The sub-topic does not stop at constant velocity. If the velocity is a function of time, the two operations are calculus, componentwise:
v = dr/dt and r = ∫v dt
and speed is still |v|, which now changes as well. Take v = (2t, 3 − t) with the object starting at the origin. Integrating each component:
r = (t², 3t − ½t²)
with no constants, because r(0) = 0. Then:
Notice that the method for the least speed is the one you just used for the least distance: write the square, find the vertex. The only thing that changed is which quantity is being minimised.
Distance travelled is not displacement. The displacement at t = 3 is (9, 4.5). The distance travelled is ∫|v| dt from 0 to 3, which is a different integral and is larger whenever the direction changes. A question asking "how far has it travelled" wants the second one, and on this course you would graph the speed and integrate it numerically.
The quadratic vertex is quick by hand, but questions with a non-constant velocity give a distance function that is not quadratic, and then graphing it and reading the minimum is the only practical route. Learn it on the easy case.
When you may use it. Applications. A calculator is allowed in every paper, and reading a minimum off a graph is accepted working provided you state what you graphed. Write down the distance function before you type it.
The mark people lose. Graphing one component instead of the distance. Typing 10-3X and finding where it is zero gives 3.33, the wrong answer, and the graph looks perfectly sensible. The function to minimise is the length of AB, with both components inside the square root. Write AB = (10 − 3t, −t) on the page first and type from that, not from the question.
1. Boat A has velocity (3, 4). Find its speed.
2. How far apart are the boats at t = 2? Give your answer to 2 decimal places.
3. At what time are they closest?
4. What is the least distance between them, to 2 decimal places?
5. The two paths cross at (10, 13.33). Why is that not where the boats are closest?
1 markAB as a function of t, both components.
1 markThe distance, or its square, as a function of t.
1 markThe minimising time.
1 markThe least distance, with units.
Four marks, and the first is the one the other three depend on. Write AB down before you do anything else, and write it as a column so neither component can go missing.
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