| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Distance from velocity-time graph |
| Difficulty | Moderate -0.8 This is a straightforward velocity-time graph question requiring only basic interpretation: finding areas under the graph (trapeziums/triangles) to calculate distances and recognizing that maximum distance occurs when velocity becomes zero. All techniques are standard M1 content with no problem-solving insight required, making it easier than average but not trivial due to the multi-part calculation. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area |
| Answer | Marks | Guidance |
|---|---|---|
| 2(i) | \(250 + \frac{1}{2}(290 - 250)\) | M1 |
| 2(i) | \(t = 270\) | A1 |
| 2(ii) | [2] | |
| 2(ii) | \(\frac{1}{2}x40(12+210)x12+210x12+\frac{1}{2}x20x12\) or \(\frac{1}{2}x40x(12+210)x12 - \frac{1}{2}x20x12\) or \(\frac{1}{2}x(210+x)x12etc\) | M1 |
| 2(ii) | Displacement is 2760m | M1 |
| 2(ii) | A1 | |
| 2(iii) | appropriate structure, ie triangle + rectangle + triangle + | triangle |
| 2(iii) | Distance is 3000m | A1 |
| 2(iii) | [2] |
2(i) | $250 + \frac{1}{2}(290 - 250)$ | M1 | Use of the ratio 12:12 (may be implied), or $v = u+at$
2(i) | $t = 270$ | A1 |
2(ii) | | [2] |
2(ii) | $\frac{1}{2}x40(12+210)x12+210x12+\frac{1}{2}x20x12$ or $\frac{1}{2}x40x(12+210)x12 - \frac{1}{2}x20x12$ or $\frac{1}{2}x(210+x)x12etc$ | M1 | The idea that area represents displacement
2(ii) | Displacement is 2760m | M1 | Correct structure, ie triangle1 + rectangle2 + triangle3 - [triangle4] with triangle3 = [triangle4], triangle1 + rectangle2, trapezium1&2, etc
2(ii) | | A1 |
2(iii) | appropriate structure, ie triangle + rectangle + triangle + |triangle|, triangle + rectangle + 2triangle, etc | M1 | All terms positive
2(iii) | Distance is 3000m | A1 | Treat candidate doing (ii) in (iii) and (iii) in (ii) as a mis-read.
2(iii) | | [2] |2\\
\includegraphics[max width=\textwidth, alt={}, center]{ae5d1e27-5853-48aa-9046-86ce1c1a154a-2_714_1048_1231_552}
A particle starts from the point A and travels in a straight line. The diagram shows the ( $\mathrm { t } , \mathrm { v }$ ) graph, consisting of three straight line segments, for the motion of the particle during the interval $0 \leqslant t \leqslant 290$.\\
(i) Find the value of ther which the distance of the particle from A is greatest.\\
(ii) Find the displacement of the particle from A when $\mathrm { t } = 290$.\\
(iii) Find the total distance travelled by the particle during the interval $0 \leqslant \mathrm { t } \leqslant 290$.
\hfill \mbox{\textit{OCR M1 2007 Q2 [7]}}