| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Total distance with direction changes |
| Difficulty | Moderate -0.3 This is a straightforward kinematics question requiring students to (i) solve a quadratic equation by setting v=0, and (ii) integrate velocity to find displacement, being careful about direction changes. While it tests understanding of when distance differs from displacement, the techniques are standard M1 material with no conceptual surprises, making it slightly easier than average. |
| Spec | 3.02a Kinematics language: position, displacement, velocity, acceleration3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(4t^2 - 8t + 3 = 0\), \((2t-3)(2t-1)\) | M1 | Set \(v = 0\) and attempt to factorise or use quadratic formula or completing the square |
| \(t = 0.5\) and \(t = 1.5\) | A1 | Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(s = -\int(4t^2 - 8t + 3)\,dt\) | M1 | Integrating \(v\) to find \(s\). Allow minus sign omitted |
| \(-\left[\frac{4}{3}t^3 - 4t^2 + 3t\right]_{0.5}^{1.5}\) | M1 | Attempted integration with limits substituted and then subtracted but not necessarily fully evaluated. \([= -(0 - 2/3)]\). Allow first minus sign omitted |
| Distance travelled \(= \frac{2}{3}\) m | A1 | Must justify sign of answer. Total: 3 |
## Question 2(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $4t^2 - 8t + 3 = 0$, $(2t-3)(2t-1)$ | M1 | Set $v = 0$ and attempt to factorise or use quadratic formula or completing the square |
| $t = 0.5$ and $t = 1.5$ | A1 | **Total: 2** |
## Question 2(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $s = -\int(4t^2 - 8t + 3)\,dt$ | M1 | Integrating $v$ to find $s$. Allow minus sign omitted |
| $-\left[\frac{4}{3}t^3 - 4t^2 + 3t\right]_{0.5}^{1.5}$ | M1 | Attempted integration with limits substituted and then subtracted but not necessarily fully evaluated. $[= -(0 - 2/3)]$. Allow first minus sign omitted |
| Distance travelled $= \frac{2}{3}$ m | A1 | Must justify sign of answer. **Total: 3** |
---2 A particle $P$ moves in a straight line, starting from a point $O$. At time $t \mathrm {~s}$ after leaving $O$, the velocity of $P , v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, is given by $v = 4 t ^ { 2 } - 8 t + 3$.\\
(i) Find the two values of $t$ at which $P$ is at instantaneous rest.\\
(ii) Find the distance travelled by $P$ between these two times.
\hfill \mbox{\textit{CAIE M1 2016 Q2 [5]}}