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Knowing a method and reaching for the right one are different skills. Every wrong answer here is built from one specific mistake, and we name it. All three subjects, untimed, free.
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Paper 1 · Physics · Section A2:47:12
A particle is projected vertically upward from the ground with speed u. Taking the upward direction as positive, which equation gives the displacement s from the point of projection at time t, while the particle is still in the air?
Question 9 of 20 · +4 / −1
Physics, Chemistry and Maths. Everything in here is clickable.
You're not short of questions
You've done thousands. You'll do thousands more before you sit the paper.
Buried in them is a short list of mistakes you make again and again. A sign convention you drop when you're rushing. A domain you never think to check. A quantity you conserve when it isn't conserved. Each one costs you four marks every time it surfaces, and it surfaces more often than you'd guess, because nothing in your routine is set up to notice it twice.
Doing more questions won't find them. The one thing that would show you the pattern — what you reached for, and what you reached for the last four times — is the one thing a worked solution doesn't contain. It tells you what was true. It never tells you what you did.
Thirty questions that teach you four things beat three hundred that teach you none.
What sitting Paper 1 actually looks like
| Mode | Computer-based |
|---|---|
| Duration | 3 hours |
| Subjects | Physics, Chemistry, Mathematics |
| Questions | 75 |
| Marks | 300 |
| Section A | 20 multiple choice per subject, +4 / −1 |
| Section B | 5 numerical per subject, all compulsory, +4 / −1 |
| Section B answers | Typed on screen, rounded to the nearest integer |
| Sessions | January and April; your better score is the one that counts |
There are separate B.Arch and B.Planning papers. We don't cover those.
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Practice · Physics
No clockA particle is projected vertically upward with speed u. Taking upward as positive, the displacement at time t is s = ut − ½gt², because g points down, so it is negative in this convention.
Which equation is consistent with the chosen sign convention?
As = ut + ½gt²
Bs = ut − ½gt²
Cs = −ut + ½gt²
Ds = −ut − ½gt²
The miss is dropping the sign on g, not forgetting the formula. That is the method you reached for.
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