| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2006 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Piecewise motion functions |
| Difficulty | Moderate -0.8 This is a straightforward M1 piecewise motion question requiring basic integration and differentiation. Part (i) uses constant velocity (trivial multiplication), part (ii) is routine integration with a constant of integration to find, and part (iii) involves differentiating to find acceleration then substituting back. All steps are standard textbook procedures with no problem-solving insight required. |
| Spec | 1.08d Evaluate definite integrals: between limits3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Displacement is 20 m | B1 | 1 \(20 + c\) (from integration) B0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For using \(s(t) = \int v(t)\,dt\) | |
| \(s(t) = 0.01t^3 - 0.15t^2 + 2t\) \((+A)\) | A1 | Can be awarded prior to cancelling |
| \(10 - 15 + 20 + A = 20\) | M1 | For using \(s(10) = \text{cv}(20)\) |
| Displacement is \(0.01t^3 - 0.15t^2 + 2t + 5\) | A1 | 4 AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a = 0.06t - 0.3\) | M1 | For using \(a(t) = dv/dt\) |
| A1 | ||
| \(0.06t - 0.3 = 0.6\) | DM1 | For starting solving \(a(t) = 0.6\); depends on previous M1 |
| \(t = 15\) | A1 | |
| Displacement is 35 m | B1 | 5 |
# Question 4:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Displacement is 20 m | B1 | **1** $20 + c$ (from integration) B0 |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using $s(t) = \int v(t)\,dt$ |
| $s(t) = 0.01t^3 - 0.15t^2 + 2t$ $(+A)$ | A1 | Can be awarded prior to cancelling |
| $10 - 15 + 20 + A = 20$ | M1 | For using $s(10) = \text{cv}(20)$ |
| Displacement is $0.01t^3 - 0.15t^2 + 2t + 5$ | A1 | **4** AG |
## Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = 0.06t - 0.3$ | M1 | For using $a(t) = dv/dt$ |
| | A1 | |
| $0.06t - 0.3 = 0.6$ | DM1 | For starting solving $a(t) = 0.6$; depends on previous M1 |
| $t = 15$ | A1 | |
| Displacement is 35 m | B1 | **5** |
---4 A cyclist travels along a straight road. Her velocity $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, at time $t$ seconds after starting from a point $O$, is given by
$$\begin{aligned}
& v = 2 \quad \text { for } 0 \leqslant t \leqslant 10 \\
& v = 0.03 t ^ { 2 } - 0.3 t + 2 \quad \text { for } t \geqslant 10 .
\end{aligned}$$
(i) Find the displacement of the cyclist from $O$ when $t = 10$.\\
(ii) Show that, for $t \geqslant 10$, the displacement of the cyclist from $O$ is given by the expression $0.01 t ^ { 3 } - 0.15 t ^ { 2 } + 2 t + 5$.\\
(iii) Find the time when the acceleration of the cyclist is $0.6 \mathrm {~m} \mathrm {~s} ^ { - 2 }$. Hence find the displacement of the cyclist from $O$ when her acceleration is $0.6 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
\hfill \mbox{\textit{OCR M1 2006 Q4 [10]}}