| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Sketch velocity-time graph |
| Difficulty | Easy -1.2 This is a straightforward three-stage SUVAT problem with all parameters explicitly given. Students simply apply v=u+at for each stage and calculate area under the velocity-time graph. No problem-solving insight required—purely routine application of standard mechanics formulas taught early in M1. |
| Spec | 3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Trapezium seen | B1 | \(v\)–\(t\) graph with three straight lines, with positive, zero and negative gradients, continuous |
| 0, 3, 9, 13 shown on the \(t\) axis | B1 | |
| \(v = 2.7\) soi in either part | B1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([0.5 \times (6 + 13) \times 2.7]\) | M1 | Using area of trapezium |
| Total distance \(= 25.65\) m | A1 | [2] Allow Distance \(= 513/20\) m |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Stage 1: \(s_1 = 0.5 \times 0.9 \times 3^2 = 4.05\); Stage 2: \(s_2 = 2.7 \times 6 = 16.2\); Stage 3: \(s_3 = 0.5 \times (2.7 + 0) \times 4 = 5.4\) | M1 | Complete method to find total distance using constant acceleration equations for all three stages |
| Total distance \(= 25.65\) m | A1 | [2] |
## Question 1:
### Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| Trapezium seen | B1 | $v$–$t$ graph with three straight lines, with positive, zero and negative gradients, continuous |
| 0, 3, 9, 13 shown on the $t$ axis | B1 | |
| $v = 2.7$ soi in either part | B1 | [3] |
### Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $[0.5 \times (6 + 13) \times 2.7]$ | M1 | Using area of trapezium |
| Total distance $= 25.65$ m | A1 | [2] Allow Distance $= 513/20$ m |
**Alternative method for 1(ii):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Stage 1: $s_1 = 0.5 \times 0.9 \times 3^2 = 4.05$; Stage 2: $s_2 = 2.7 \times 6 = 16.2$; Stage 3: $s_3 = 0.5 \times (2.7 + 0) \times 4 = 5.4$ | M1 | Complete method to find total distance using constant acceleration equations for all three stages |
| Total distance $= 25.65$ m | A1 | [2] |
---1 A lift moves upwards from rest and accelerates at $0.9 \mathrm {~ms} ^ { - 2 }$ for 3 s . The lift then travels for 6 s at constant speed and finally slows down, with a constant deceleration, stopping in a further 4 s .\\
(i) Sketch a velocity-time graph for the motion.\\
(ii) Find the total distance travelled by the lift.
\hfill \mbox{\textit{CAIE M1 2016 Q1 [5]}}