| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Sketch velocity-time graph |
| Difficulty | Standard +0.3 This is a standard two-vehicle SUVAT problem requiring systematic application of kinematic equations across multiple phases of motion. Part (i) is straightforward calculation, part (ii) involves setting up equations for the motorcycle's motion with constraints, and part (iii) is routine graph sketching. While multi-step, it follows predictable patterns without requiring novel insight—slightly easier than average. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v = 3 \times 10 = 30 \text{ ms}^{-1}\) | B1 | Velocity after 10 seconds |
| \(s = \frac{1}{2}(30 + 40) \times 30\) or equivalent complete method | M1 | For determining distance travelled in first 40 seconds |
| Total distance \(= 1050 \text{ m}\) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Distance \(= 450 \text{ m}\), Time taken \(= 450/15 = 30 \text{ s}\) | M1 | For finding distance covered in deceleration stage and time taken for this stage |
| Total time of motion for car \(= 70 \text{ s}\) | A1 | May be implied by time for motorcycle \(= 50 \text{ s}\) |
| Motorcycle takes 50 s to travel 1500 m: \(1500 = \frac{1}{2}(30 + 50) \times V\) or \(1500 = 30V + 0.5 \times 20V\) | M1 | For setting up an equation for distance travelled by M/C \((v-t\) graph or other) involving \(V\) or \(a\) and up to one other variable |
| \(V = 37.5 \text{ ms}^{-1}\) | A1 | |
| 20 s is split between 5 s accelerating and 15 s decelerating | M1 | For finding time taken to accelerate to speed \(V\) |
| \(a = 37.5/5 = 7.5 \text{ ms}^{-2}\) | A1 [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Displacement-time graph | B1 | Two of the three graph stages correct with correct curvature |
| B1 | All three stages of the graph correct with correct curvature | |
| B1 [3] | Correct graph, fully labelled \(t = 10, 40, 70 \text{ s}\); \(s = 150, 1050, 1500\) |
## Question 7:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v = 3 \times 10 = 30 \text{ ms}^{-1}$ | B1 | Velocity after 10 seconds |
| $s = \frac{1}{2}(30 + 40) \times 30$ or equivalent complete method | M1 | For determining distance travelled in first 40 seconds |
| Total distance $= 1050 \text{ m}$ | A1 [3] | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Distance $= 450 \text{ m}$, Time taken $= 450/15 = 30 \text{ s}$ | M1 | For finding distance covered in deceleration stage and time taken for this stage |
| Total time of motion for car $= 70 \text{ s}$ | A1 | May be implied by time for motorcycle $= 50 \text{ s}$ |
| Motorcycle takes 50 s to travel 1500 m: $1500 = \frac{1}{2}(30 + 50) \times V$ or $1500 = 30V + 0.5 \times 20V$ | M1 | For setting up an equation for distance travelled by M/C $(v-t$ graph or other) involving $V$ or $a$ and up to one other variable |
| $V = 37.5 \text{ ms}^{-1}$ | A1 | |
| 20 s is split between 5 s accelerating and 15 s decelerating | M1 | For finding time taken to accelerate to speed $V$ |
| $a = 37.5/5 = 7.5 \text{ ms}^{-2}$ | A1 [6] | |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Displacement-time graph | B1 | Two of the three graph stages correct with correct curvature |
| | B1 | All three stages of the graph correct with correct curvature |
| | B1 [3] | Correct graph, fully labelled $t = 10, 40, 70 \text{ s}$; $s = 150, 1050, 1500$ |7 A car starts from rest and moves in a straight line from point $A$ with constant acceleration $3 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ for 10 s . The car then travels at constant speed for 30 s before decelerating uniformly, coming to rest at point $B$. The distance $A B$ is 1.5 km .\\
(i) Find the total distance travelled in the first 40 s of motion.
When the car has been moving for 20 s , a motorcycle starts from rest and accelerates uniformly in a straight line from point $A$ to a speed $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$. It then maintains this speed for 30 s before decelerating uniformly to rest at point $B$. The motorcycle comes to rest at the same time as the car.\\
(ii) Given that the magnitude of the acceleration $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$ of the motorcycle is three times the magnitude of its deceleration, find the value of $a$.\\
(iii) Sketch the displacement-time graph for the motion of the car.
\hfill \mbox{\textit{CAIE M1 2016 Q7 [12]}}