| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Two vehicles: overtaking or meeting (algebraic) |
| Difficulty | Standard +0.3 This is a standard two-stage SUVAT problem with a straightforward overtaking scenario. Part (i) requires applying SUVAT equations across three phases of motion (acceleration, constant speed, deceleration) with clear given values. Part (ii) involves setting up a simple equation where distances are equal, with the constraint already specified (overtaking during constant speed phase). The problem requires multiple steps but uses routine techniques with no conceptual challenges beyond careful bookkeeping of the motion phases. |
| Spec | 3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(36 = 0 + 0.5 \times 0.5t^2\) | ||
| \(t = 12\) | B1 | |
| \(v^2 = 0 + 2 \times 0.5 \times 36\) | ||
| \(v = 6\) | B1 | |
| \(s = 6 \times 25\) | ||
| remaining distance \(= 210 - 36 - 150 = 24\) | B1 | |
| \(24 = (6 + 0)/2 \times t\) | M1 | Using \(s = \frac{(u+v)t}{2}\) |
| \(t = 8\) | ||
| Total Time \(= 12 + 25 + 8 = 45\) s | A1 | [5 marks] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Distance travelled by cyclist \(= 36 + 6(t - 12)\) | M1 | For attempting distance travelled by cyclist for \(t > 12\) |
| Distance travelled by car \(= 0.5 \times 4 \times (t-24)^2\) | M1 | For attempting distance travelled by car |
| \(2t^2 - 96t + 1152 = 36 + 6t - 72\) | M1 | Equating expressions and attempting to solve a three term quadratic equation |
| \([t^2 - 51t + 594 = 0]\) | ||
| \(t = 33\) or \(t = 18\) | A1 | |
| Time \(= 33\) s | B1 | Choosing the correct solution [5 marks] |
## Question 7:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $36 = 0 + 0.5 \times 0.5t^2$ | | |
| $t = 12$ | B1 | |
| $v^2 = 0 + 2 \times 0.5 \times 36$ | | |
| $v = 6$ | B1 | |
| $s = 6 \times 25$ | | |
| remaining distance $= 210 - 36 - 150 = 24$ | B1 | |
| $24 = (6 + 0)/2 \times t$ | M1 | Using $s = \frac{(u+v)t}{2}$ |
| $t = 8$ | | |
| Total Time $= 12 + 25 + 8 = 45$ s | A1 | **[5 marks]** |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Distance travelled by cyclist $= 36 + 6(t - 12)$ | M1 | For attempting distance travelled by cyclist for $t > 12$ |
| Distance travelled by car $= 0.5 \times 4 \times (t-24)^2$ | M1 | For attempting distance travelled by car |
| $2t^2 - 96t + 1152 = 36 + 6t - 72$ | M1 | Equating expressions and attempting to solve a three term quadratic equation |
| $[t^2 - 51t + 594 = 0]$ | | |
| $t = 33$ or $t = 18$ | A1 | |
| Time $= 33$ s | B1 | Choosing the correct solution **[5 marks]** |7 A cyclist starts from rest at point $A$ and moves in a straight line with acceleration $0.5 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ for a distance of 36 m . The cyclist then travels at constant speed for 25 s before slowing down, with constant deceleration, to come to rest at point $B$. The distance $A B$ is 210 m .\\
(i) Find the total time that the cyclist takes to travel from $A$ to $B$.
24 s after the cyclist leaves point $A$, a car starts from rest from point $A$, with constant acceleration $4 \mathrm {~m} \mathrm {~s} ^ { - 2 }$, towards $B$. It is given that the car overtakes the cyclist while the cyclist is moving with constant speed.\\
(ii) Find the time that it takes from when the cyclist starts until the car overtakes her.
\hfill \mbox{\textit{CAIE M1 2015 Q7 [10]}}