| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2015 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Displacement from velocity by integration |
| Difficulty | Standard +0.3 This is a standard M3 variable acceleration question requiring integration with initial conditions. Part (a) is a 'show that' involving one integration step with careful attention to direction/signs. Part (b) requires finding when v=0, then integrating velocity to find displacement. While it requires competence with fractional powers and sign conventions, it follows a predictable template for M3 questions with no novel problem-solving insight needed. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\frac{dv}{dt} = -2(t+4)^{-\frac{1}{2}}\) | M1 | Attempting expression for acceleration in form \(\frac{dv}{dt}\); minus may be omitted |
| \(v = -\int 2(t+4)^{-\frac{1}{2}}\,dt\) | ||
| \(v = -4(t+4)^{\frac{1}{2}}\ (+c)\) | DM1A1 | Attempting integration; correct integration (constant may be omitted) |
| \(t=0, v=8 \Rightarrow c = 16\) | M1 | Using initial conditions to obtain constant of integration |
| \(v = 16 - 4(t+4)^{\frac{1}{2}}\ \text{(m s}^{-1}\text{)}\) | A1cso | Substitute value of \(c\) and obtain given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(v=0:\ 16 = 4(t+4)^{\frac{1}{2}}\) | M1 | Setting given expression for \(v\) equal to 0 |
| \(16 = t+4 \Rightarrow t = 12\) | A1 | Solving to get \(t=12\) |
| \(x = 4\int\!\left(4-(t+4)^{\frac{1}{2}}\right)dt\) | M1 | Setting \(v = \frac{dx}{dt}\) and attempting integration wrt \(t\); at least one term clearly integrated |
| \(x = 4\!\left(4t - \frac{2}{3}(t+4)^{\frac{3}{2}}\right)(+d)\) | M1A1 | Correct integration; constant may be omitted |
| \(t=0,\, x=0:\ d = 4 \times \frac{2}{3} \times 4^{\frac{3}{2}} = \frac{64}{3}\) | A1 | Substituting \(t=0, x=0\) and obtaining correct value of \(d\) |
| \(t=12:\ x = 4\!\left(4\times12 - \frac{2}{3}\times16^{\frac{3}{2}}\right) + \frac{64}{3} = 42\frac{2}{3}\ \text{(m)}\) | DM1A1 | Substituting \(t=12\); correct final answer (oe, e.g. 43 or better); dependent on second M mark |
## Question 3(a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\frac{dv}{dt} = -2(t+4)^{-\frac{1}{2}}$ | M1 | Attempting expression for acceleration in form $\frac{dv}{dt}$; minus may be omitted |
| $v = -\int 2(t+4)^{-\frac{1}{2}}\,dt$ | | |
| $v = -4(t+4)^{\frac{1}{2}}\ (+c)$ | DM1A1 | Attempting integration; correct integration (constant may be omitted) |
| $t=0, v=8 \Rightarrow c = 16$ | M1 | Using initial conditions to obtain constant of integration |
| $v = 16 - 4(t+4)^{\frac{1}{2}}\ \text{(m s}^{-1}\text{)}$ | A1cso | Substitute value of $c$ and obtain given answer |
**Total: (5)**
## Question 3(b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $v=0:\ 16 = 4(t+4)^{\frac{1}{2}}$ | M1 | Setting given expression for $v$ equal to 0 |
| $16 = t+4 \Rightarrow t = 12$ | A1 | Solving to get $t=12$ |
| $x = 4\int\!\left(4-(t+4)^{\frac{1}{2}}\right)dt$ | M1 | Setting $v = \frac{dx}{dt}$ and attempting integration wrt $t$; at least one term clearly integrated |
| $x = 4\!\left(4t - \frac{2}{3}(t+4)^{\frac{3}{2}}\right)(+d)$ | M1A1 | Correct integration; constant may be omitted |
| $t=0,\, x=0:\ d = 4 \times \frac{2}{3} \times 4^{\frac{3}{2}} = \frac{64}{3}$ | A1 | Substituting $t=0, x=0$ and obtaining correct value of $d$ |
| $t=12:\ x = 4\!\left(4\times12 - \frac{2}{3}\times16^{\frac{3}{2}}\right) + \frac{64}{3} = 42\frac{2}{3}\ \text{(m)}$ | DM1A1 | Substituting $t=12$; correct final answer (oe, e.g. 43 or better); dependent on second M mark |
**Total: (7) [12]**
---\begin{enumerate}
\item At time $t = 0$, a particle $P$ is at the origin $O$, moving with speed $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in the positive $x$ direction. At time $t$ seconds, $t \geqslant 0$, the acceleration of $P$ has magnitude $2 ( t + 4 ) ^ { - \frac { 1 } { 2 } } \mathrm {~m} \mathrm {~s} ^ { - 2 }$ and is directed towards $O$.\\
(a) Show that, at time $t$ seconds, the velocity of $P$ is $16 - 4 ( t + 4 ) ^ { \frac { 1 } { 2 } } \mathrm {~ms} ^ { - 1 }$\\
(b) Find the distance of $P$ from $O$ when $P$ comes to instantaneous rest.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2015 Q3 [12]}}