| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2005 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Given acceleration function find velocity |
| Difficulty | Standard +0.8 This M3 variable force question requires integration of a non-standard function (involving square roots), careful attention to direction/signs, and a second integration to find distance. Part (a) is a structured 'show that' requiring integration of t^(-1/2) with initial conditions. Part (b) requires finding when v=0, then integrating velocity (involving both linear and t^(3/2) terms) between appropriate limits. While the techniques are standard M3 content, the algebraic manipulation and multi-step nature with non-trivial functions makes this moderately challenging, above average difficulty. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Content | Marks | Notes |
| \(\frac{dv}{dt} = -\frac{3}{\sqrt{t+4}}\) | M1 | |
| \(v = -3\int(t+4)^{-\frac{1}{2}}dt\) | M1 | |
| \(v = -6(t+4)^{\frac{1}{2}} + C\) | A1 | |
| Using \(v=18\) when \(t=8\): \(18 = -6\sqrt{12} + C \Rightarrow C = 30\) | M1 | |
| \(v = 30 - 6\sqrt{t+4}\) | A1 esc. | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Content | Marks | Notes |
| \(x = \int 30 - 6(t+4)^{\frac{1}{2}}dt\) | M1 | |
| \(= 30t - 4(t+4)^{\frac{3}{2}} + D\) | M1 | |
| Using \(t=0, x=0\): \(0 = 0 - 4 \times 8 + D \Rightarrow D = 32\) | M1 | |
| When \(t = 21\): \(x = 30 \times 21 - 4 \times 5^2 + 32\) | A1 | |
| \(= 162(u)\) | A1 | (7), (12) |
## Part (a)
| Content | Marks | Notes |
|---------|-------|-------|
| $\frac{dv}{dt} = -\frac{3}{\sqrt{t+4}}$ | M1 | |
| $v = -3\int(t+4)^{-\frac{1}{2}}dt$ | M1 | |
| $v = -6(t+4)^{\frac{1}{2}} + C$ | A1 | |
| Using $v=18$ when $t=8$: $18 = -6\sqrt{12} + C \Rightarrow C = 30$ | M1 | |
| $v = 30 - 6\sqrt{t+4}$ | A1 esc. | (5) |
## Part (b)
| Content | Marks | Notes |
|---------|-------|-------|
| $x = \int 30 - 6(t+4)^{\frac{1}{2}}dt$ | M1 | |
| $= 30t - 4(t+4)^{\frac{3}{2}} + D$ | M1 | |
| Using $t=0, x=0$: $0 = 0 - 4 \times 8 + D \Rightarrow D = 32$ | M1 | |
| When $t = 21$: $x = 30 \times 21 - 4 \times 5^2 + 32$ | A1 | |
| $= 162(u)$ | A1 | (7), (12) |
---At time $t = 0$, a particle $P$ is at the origin $O$, moving with speed 18 m s$^{-1}$ along the $x$-axis, in the positive $x$-direction. At time $t$ seconds ($t > 0$) the acceleration of $P$ has magnitude $\frac{3}{\sqrt{(t + 4)}}$ m s$^{-2}$ and is directed towards $O$.
\begin{enumerate}[label=(\alph*)]
\item Show that, at time $t$ seconds, the velocity of $P$ is $[30 - 6\sqrt{(t + 4)}]$ m s$^{-1}$. [5]
\item Find the distance of $P$ from $O$ when $P$ comes to instantaneous rest. [7]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2005 Q5 [12]}}