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S3 Science: Forces, Motion, Speed-Time Graphs and Net Force (Paper 2: intermediate)
The second paper connects formula, graph and friction, focusing on average speed and the direction of net force.
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 20 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: S3 Science: Forces, Motion, Speed-Time Graphs and Net Force |
| Level | S3 Science |
| # | Answer | One-line note |
|---|---|---|
| 1 | Total distance divided by total time | When the pace changes you cannot average the speeds, so use the whole distance over the whole time. |
| 2 | 6 m/s | Dividing 240 by 40 gives 6 in metres per second. |
| 3 | 32 N | Forces in the same direction add, so 20 plus 12 gives 32 newtons. |
| 4 | 8 N in the direction of the 20 N force | Opposite forces subtract and the net force follows the larger one. |
| 5 | 3 m/s² | Rearranging F = ma gives a as 6 divided by 2, which is 3. |
| 6 | 20 N | Using F = ma, 10 times 2 is 20 newtons. |
| 7 | 3 m/s² | The net force is 20 minus 8, and dividing 12 by the mass of 4 gives 3. |
| 8 | 15 N | Constant velocity means zero net force, so friction must equal the push and point the other way. |
| 9 | 4 m/s² | Acceleration is the change in speed over time, so 20 divided by 5 is 4. |
| 10 | 5 m/s² | In F = ma the force is divided by the mass, so 10 over 2 is 5. |
The intermediate step links formula and graph: average speed is total distance over total time, F = ma handles net force and acceleration, and friction always opposes relative motion.Tutor Registration →
Questions and answers are written originally by the TutorZone team and are not taken from any school, publisher or examination authority paper. Every value is checked by computer.

