| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Given velocity function find force |
| Difficulty | Standard +0.8 This M3 question requires applying the chain rule for acceleration (a = v dv/dx), integrating to find position as a function of time, and working with a rational velocity function. It demands multiple connected steps with careful algebraic manipulation, placing it moderately above average difficulty but within reach of well-prepared M3 students. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(v = \frac{50}{2x+3}\), \(\frac{dv}{dt} = \frac{dv}{dx} \times \frac{dx}{dt}\) | M1 | Uses chain rule of form \(\frac{dv}{dt} = \frac{dv}{dx} \times \frac{dx}{dt}\) or \(\frac{d(\frac{1}{2}v^2)}{dx}\). M0 for acc \(= \frac{1}{2}v^2\). |
| \(= \frac{-100}{(2x+3)^2} \times \frac{50}{2x+3} \left(= \frac{-5000}{(2x+3)^3}\right)\) | DM1A1 | DM1: differentiate \(v\) wrt \(x\). A1: correct differentiation. |
| \(x=12\): \(\frac{dv}{dt} = -\frac{5000}{27^3} = -0.2540... = -0.25\) or \(-0.254\) m s\(^{-2}\) | M1 | Sub \(x=12\) into their expression for acceleration. Must have attempted to differentiate. |
| deceleration \(= 0.25\) (m s\(^{-2}\)) or better | A1 (5) | Correct deceleration — must be positive. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(v = \frac{dx}{dt} = \frac{50}{2x+3}\) | M1 | Use \(v = \frac{dx}{dt}\) |
| \(\int(2x+3)\,dx = \int 50\,dt\) | M1 | Attempt at integration |
| \(x^2 + 3x = 50t \ (+c)\) | A1 | Correct integration but \(c\) may be missing |
| \(t=1, x=4 \Rightarrow 28 = 50 + c,\ c = -22\) | A1 | Use \(t=1\), \(x=4\) to obtain correct value of \(c\) for their correct integration |
| \(x=12 \Rightarrow 50t = 144 + 36 + 22,\ t = \frac{202}{50} = 4.04\) (accept 4.0) | A1 (5) [10] | Sub \(x=12\) to obtain correct value of \(t\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int_4^{12}(2x+3)\,dx = \int_1^T 50\,dt\) | M1 | |
| \(\left[x^2+3x\right]_4^{12} = \left[50t\right]_1^T\) | A1 | Correct integration |
| \(12^2+3(12)-4^2-3(4) = 50T - 50\) | A1 | Sub in limits correctly |
| Obtain correct value | A1 |
# Question 3:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $v = \frac{50}{2x+3}$, $\frac{dv}{dt} = \frac{dv}{dx} \times \frac{dx}{dt}$ | M1 | Uses chain rule of form $\frac{dv}{dt} = \frac{dv}{dx} \times \frac{dx}{dt}$ or $\frac{d(\frac{1}{2}v^2)}{dx}$. M0 for acc $= \frac{1}{2}v^2$. |
| $= \frac{-100}{(2x+3)^2} \times \frac{50}{2x+3} \left(= \frac{-5000}{(2x+3)^3}\right)$ | DM1A1 | DM1: differentiate $v$ wrt $x$. A1: correct differentiation. |
| $x=12$: $\frac{dv}{dt} = -\frac{5000}{27^3} = -0.2540... = -0.25$ or $-0.254$ m s$^{-2}$ | M1 | Sub $x=12$ into their expression for acceleration. Must have attempted to differentiate. |
| deceleration $= 0.25$ (m s$^{-2}$) or better | A1 (5) | Correct deceleration — must be positive. |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $v = \frac{dx}{dt} = \frac{50}{2x+3}$ | M1 | Use $v = \frac{dx}{dt}$ |
| $\int(2x+3)\,dx = \int 50\,dt$ | M1 | Attempt at integration |
| $x^2 + 3x = 50t \ (+c)$ | A1 | Correct integration but $c$ may be missing |
| $t=1, x=4 \Rightarrow 28 = 50 + c,\ c = -22$ | A1 | Use $t=1$, $x=4$ to obtain correct value of $c$ for their correct integration |
| $x=12 \Rightarrow 50t = 144 + 36 + 22,\ t = \frac{202}{50} = 4.04$ (accept 4.0) | A1 (5) **[10]** | Sub $x=12$ to obtain correct value of $t$ |
**ALT 3(b) — definite integration:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int_4^{12}(2x+3)\,dx = \int_1^T 50\,dt$ | M1 | |
| $\left[x^2+3x\right]_4^{12} = \left[50t\right]_1^T$ | A1 | Correct integration |
| $12^2+3(12)-4^2-3(4) = 50T - 50$ | A1 | Sub in limits correctly |
| Obtain correct value | A1 | |\begin{enumerate}
\item In this question you must show all stages of your working.
\end{enumerate}
\section*{Solutions relying entirely on calculator technology are not acceptable.}
A particle $P$ is moving along a straight line.
At time $t$ seconds, $P$ is a distance $x$ metres from a fixed point $O$ on the line and is moving away from $O$ with speed $\frac { 50 } { 2 x + 3 } \mathrm {~ms} ^ { - 1 }$\\
(a) Find the deceleration of $P$ when $x = 12$
Given that $x = 4$ when $t = 1$\\
(b) find the value of $t$ when $x = 12$
\hfill \mbox{\textit{Edexcel M3 2022 Q3 [10]}}