| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2018 |
| Session | Specimen |
| 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 of a given acceleration function with initial conditions, then a second integration to find displacement. The algebraic manipulation is straightforward (power of 1/2), and the 'show that' part guides students through the first step. While it requires calculus and careful sign handling, it follows a predictable template for this topic with no novel problem-solving required. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration6.06a Variable force: dv/dt or v*dv/dx methods |
| VIIIV SIHI NI JIIIM ION OC | VIIV SIHI NI JAHM ION OO | VI4V SIHI NI JIIIM I ON OO |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{dv}{dt} = -2(t+4)^{-\frac{1}{2}}\) | M1 | Attempting expression for acceleration in form \(\frac{dv}{dt}\); minus may be omitted |
| \(v = -\int 2(t+4)^{-\frac{1}{2}}\,dt\) | — | Setting up integral |
| \(v = -4(t+4)^{\frac{1}{2}} \quad (+c)\) | dM1, A1 | Attempting integration; correct integration, constant may be omitted (no ft) |
| \(t = 0, v = 8 \Rightarrow c = 16\) | M1 | Using initial conditions to obtain value for constant of integration |
| \(v = 16 - 4(t+4)^{\frac{1}{2}} \quad (\text{m s}^{-1})\) | A1 cso | Substituting value of \(c\) and obtaining the given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(v = 0 \Rightarrow 16 = 4(t+4)^{\frac{1}{2}}\) | M1 | Setting the 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 must clearly be integrated |
| \(x = 4\!\left(4t - \frac{2}{3}(t+4)^{\frac{3}{2}}\right) \quad (+d)\) | M1, A1 | Correct integration, constant may be omitted |
| \(t = 0,\; x = 0 \Rightarrow d = 4 \times \frac{2}{3} \times 4^2 = \frac{64}{3}\) | A1 | Substituting \(t=0, x=0\) and obtaining correct value of \(d\); any equivalent number including decimals |
| \(t = 12:\; x = 4\!\left(4\times12 - \frac{2}{3}\times16^{\frac{3}{2}}\right) + \frac{64}{3} = 42\frac{2}{3}\) (m) | dM1, A1 | Substituting their \(t\) and obtaining value for required distance; correct final answer, any equivalent form |
## Question 3:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{dv}{dt} = -2(t+4)^{-\frac{1}{2}}$ | M1 | Attempting expression for acceleration in form $\frac{dv}{dt}$; minus may be omitted |
| $v = -\int 2(t+4)^{-\frac{1}{2}}\,dt$ | — | Setting up integral |
| $v = -4(t+4)^{\frac{1}{2}} \quad (+c)$ | dM1, A1 | Attempting integration; correct integration, constant may be omitted (no ft) |
| $t = 0, v = 8 \Rightarrow c = 16$ | M1 | Using initial conditions to obtain value for constant of integration |
| $v = 16 - 4(t+4)^{\frac{1}{2}} \quad (\text{m s}^{-1})$ | A1 cso | Substituting value of $c$ and obtaining the given answer |
**(5 marks)**
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $v = 0 \Rightarrow 16 = 4(t+4)^{\frac{1}{2}}$ | M1 | Setting the 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 must clearly be integrated |
| $x = 4\!\left(4t - \frac{2}{3}(t+4)^{\frac{3}{2}}\right) \quad (+d)$ | M1, A1 | Correct integration, constant may be omitted |
| $t = 0,\; x = 0 \Rightarrow d = 4 \times \frac{2}{3} \times 4^2 = \frac{64}{3}$ | A1 | Substituting $t=0, x=0$ and obtaining correct value of $d$; any equivalent number including decimals |
| $t = 12:\; x = 4\!\left(4\times12 - \frac{2}{3}\times16^{\frac{3}{2}}\right) + \frac{64}{3} = 42\frac{2}{3}$ (m) | dM1, A1 | Substituting their $t$ and obtaining value for required distance; correct final answer, any equivalent form |
**(7 marks) — Total: 12 marks**
---3. 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 $2 ( t + 4 ) ^ { - \frac { 1 } { 2 } } \mathrm {~ms} ^ { - 2 }$ and is directed towards $O$.
\begin{enumerate}[label=(\alph*)]
\item Show that, at time $t$ seconds, the velocity of $P$ is $16 - 4 ( t + 4 ) ^ { \frac { 1 } { 2 } } \mathrm {~ms} ^ { - 1 }$
\item Find the distance of $P$ from $O$ when $P$ comes to instantaneous rest.
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VIIIV SIHI NI JIIIM ION OC & VIIV SIHI NI JAHM ION OO & VI4V SIHI NI JIIIM I ON OO \\
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\hfill \mbox{\textit{Edexcel M3 2018 Q3 [12]}}