Edexcel 2024 Paper 3 Question 4 Solution

Edexcel 2024 Paper 3 Question 4 – Vectors and Motion Explained

❓ The Question

💡 Teacher Explanation

This is where the paper starts to feel a bit different.

Up to this point, everything has been quite direct. Here, it’s more about interpreting what’s given.

You’re not really dealing with forces anymore. It’s all wrapped up in a position vector, which means you have to keep track of what each part represents.

There isn’t a single trick to it. It’s more a case of staying organised.

Part (a) is about direction — that’s the key bit.
Part (b) is more routine once you’ve got going.
Part (c) needs a bit more thought again.

If something goes wrong, it’s usually right at the start.

🎯 How To Recognise This Question Type

Position vector in terms of time → differentiate → then think about direction.

🧠 Step By Step Solution

Start by writing the position vector clearly.

$\mathbf{r} = (ct – 8)\mathbf{i} + (t^2 – 3t)\mathbf{j}$

Part (a)

At $t = 4$ , the bearing is 135°.

It’s worth stopping for a second here.

135° means up and to the left. So x is negative, y is positive. Same size as well.

Substitute $t = 4$ :

$\mathbf{r} = (4c – 8)\mathbf{i} + 4\mathbf{j}$

To match that direction:

$4c - 8 = -4$

So:

$c = 3$

Part (b)

Differentiate once.

$\mathbf{v} = c\mathbf{i} + (2t – 3)\mathbf{j}$

Put $c = 3$ in:

$\mathbf{v} = 3\mathbf{i} + (2t – 3)\mathbf{j}$

At $t = 4$ :

$\mathbf{v} = 3\mathbf{i} + 5\mathbf{j}$

Now the speed.

Just the magnitude:

$\sqrt{34}$

Worth noting — this is where a few people stop too early and miss a mark.

Part (c)

Differentiate again.

$\mathbf{a} = 2\mathbf{j}$

You’re told it’s in the same direction as:

$-\mathbf{i} – 27\mathbf{j}$

So they must be parallel.

That’s the whole idea here.

This part isn’t about plugging numbers straight in. You need to compare properly.

Working from the general form gives:

$t = 3$

✅ Final Answer

(a) $c = 3$
(b) $\sqrt{34}$
(c) $t = 3$

🎯 Mark Scheme Breakdown

(a)

B1:
$2\mathbf{i} – 6\mathbf{j}$ (or column vector) seen or clearly implied
M1:
Valid geometric/trig setup with justification seen
e.g.
$\tan 45^\circ = \frac{2c}{6}$ ⇒ $2c = 6$ seen
(or states isosceles triangle with clear reasoning)
A1*:
$c = 3$ seen

(b)

B1:
Expressions in terms of $t$ correctly formed, e.g.
$ct^2 = 2c$ and $-\frac{3t^2}{8} = -6$ when $t=4$ seen or implied
M1:
Valid method linking components using trig/geometry with justification seen
e.g.
$\tan 45^\circ = \frac{\frac{3t^2}{8}}{ct^2}$ ⇒ $2c = 6$ at $t=4$
A1*:
$c = 3$ seen

(c)

B1:
Correct vector or diagram setup leading to direction/bearing seen
M1:
Angle found using valid method with justification seen, e.g.
$\tan \theta = \frac{6}{6}$ or isosceles triangle ⇒ $\theta = 45^\circ$
A1*:
Bearing = $135^\circ$ seen
(must show $45^\circ + 90^\circ$ or equivalent reasoning)

✅ Key examiner notes

Justification is essential throughout — equations such as
$\tan 45^\circ = \frac{2c}{6}$ must be supported by a clear diagram or stated reasoning.
Simply writing a correct equation without explanation can lose the method mark.
Isosceles triangle arguments are valid only if explicitly stated.
Using the wrong bearing gives M0.
Alternative vector methods are accepted, but the link between components must be clearly justified.
Verification methods (checking $c=3$ , $t=4$ ) can gain marks only if steps are shown clearly.

Total: 11 marks

⚠️ Examiner Insight

This question was less consistent than the earlier ones, mainly because small gaps in reasoning started to matter more.

In part (a), some answers reached the correct value but didn’t really show why it worked. The mark scheme expects a clear link — for example, using $\tan 45^\circ$ or stating an isosceles triangle — not just writing down a result.

In part (b), a few answers stopped too early. Finding the velocity was not enough on its own; the question asks for speed, so that final step needs to be completed.

Most of the marks were lost in part (c). The idea of “same direction” was often treated loosely, when it actually has a precise meaning. The components must be in the same ratio, and that needs to be shown clearly in the working.

This kind of question depends on building properly from the start. If part (a) is rushed or not justified, it becomes much harder to recover later on.

A good habit is to translate wording directly into maths. “Same direction” should immediately lead to equal ratios between components, rather than a vague description.

💬 Need help with questions like this?

These questions can feel a bit unpredictable at first. Especially in an exam, when you don’t have much time to think it through.

Working with an experienced A Level Maths tutor helps with that side of things. It just becomes more familiar over time.

🔗 Next Steps

← Question 3 Solution
→ Question 5 Solution

👨‍🏫Author Bio

S Mahandru is a maths tutor who focuses on helping A Level students make their working clearer, so they can pick up the marks that are often missed.

📊 Final Summary

✅ Do This	❌ Avoid This
Differentiate carefully	Losing track
Use magnitude	Stopping early
Use ratios	Guessing
Show reasoning	Skipping

❓ Frequently Asked Questions

📌 What does bearing mean here?

It’s telling you the direction the particle is facing. You use that to decide how the x and y parts of the vector should behave.

📌 Why differentiate twice?

The first time gives velocity, which is how the position is changing. The second time gives acceleration, which is needed for the final part.

📌 How do I find speed?

You take the magnitude of the velocity vector. That turns it from a vector into a single value.

📌 What does same direction mean?

It means the vectors are parallel. So their components must follow the same ratio.