Edexcel M3 — Question 4 9 marks

Exam BoardEdexcel
ModuleM3 (Mechanics 3)
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable Force
TypeForce depends on time t
DifficultyStandard +0.3 This is a straightforward M3 kinematics question requiring integration of acceleration to find velocity, then applying initial conditions to find constants. The integration of 1/(1+t) is standard, and solving the resulting logarithmic equation in part (c) uses basic log laws. All steps are routine for M3 students with no novel problem-solving required, making it slightly easier than average.
Spec1.08k Separable differential equations: dy/dx = f(x)g(y)6.06a Variable force: dv/dt or v*dv/dx methods

The acceleration \(a\) ms\(^{-2}\) of a particle \(P\) moving in a straight line away from a fixed point \(O\) is given by \(a = \frac{k}{1+t}\), where \(t\) is the time that has elapsed since \(P\) left \(O\), and \(k\) is a constant.
  1. By solving a suitable differential equation, find an expression for the velocity \(v\) ms\(^{-1}\) of \(P\) in terms of \(t\), \(k\) and another constant \(c\). [3 marks]
Given that \(v = 0\) when \(t = 0\) and that \(v = 4\) when \(t = 2\),
  1. show that \(v \ln 3 = 4 \ln (1 + t)\). [3 marks]
  2. Calculate the time when \(P\) has a speed of 8 ms\(^{-1}\). [3 marks]

AnswerMarks
(a) \(\frac{dv}{dt} = \frac{-k}{1+t}\)B1 M1 A1
\(\int dv = k\int\frac{1}{1+t}dt\)
\(v = k\ln(1+t) + c\)
(b) \(t = 0, v = 0\), so \(c = 0\)M1 A1 A1
\(t = 2, v = 4\): \(k = \frac{4}{\ln 3}\); hence result
(c) When \(v = 8\), \(8 = \frac{4}{\ln 3}\ln(1+t)\)M1 A1 A1
\(\ln(1+t) = \ln 9\)
\(t = 8\)9 marks total
**(a)** $\frac{dv}{dt} = \frac{-k}{1+t}$ | B1 M1 A1
$\int dv = k\int\frac{1}{1+t}dt$ | 
$v = k\ln(1+t) + c$ | 

**(b)** $t = 0, v = 0$, so $c = 0$ | M1 A1 A1
$t = 2, v = 4$: $k = \frac{4}{\ln 3}$; hence result | 

**(c)** When $v = 8$, $8 = \frac{4}{\ln 3}\ln(1+t)$ | M1 A1 A1
$\ln(1+t) = \ln 9$ | 
$t = 8$ | 9 marks total
The acceleration $a$ ms$^{-2}$ of a particle $P$ moving in a straight line away from a fixed point $O$ is given by $a = \frac{k}{1+t}$, where $t$ is the time that has elapsed since $P$ left $O$, and $k$ is a constant.

\begin{enumerate}[label=(\alph*)]
\item By solving a suitable differential equation, find an expression for the velocity $v$ ms$^{-1}$ of $P$ in terms of $t$, $k$ and another constant $c$. [3 marks]
\end{enumerate}

Given that $v = 0$ when $t = 0$ and that $v = 4$ when $t = 2$,

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item show that $v \ln 3 = 4 \ln (1 + t)$. [3 marks]
\item Calculate the time when $P$ has a speed of 8 ms$^{-1}$. [3 marks]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M3  Q4 [9]}}