Edexcel M3 2015 June — Question 4 12 marks

Exam BoardEdexcel
ModuleM3 (Mechanics 3)
Year2015
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable Force
TypeForce depends on time t
DifficultyStandard +0.3 This is a standard M3 variable force question requiring Newton's second law with F=ma=m(dv/dt), integration to find v(t), finding constant k from given conditions, and applying trapezium rule. All techniques are routine for M3 students with clear signposting and no novel problem-solving required, making it slightly easier than average.
Spec1.09f Trapezium rule: numerical integration6.06a Variable force: dv/dt or v*dv/dx methods

  1. A vehicle of mass 900 kg moves along a straight horizontal road. At time \(t\) seconds the resultant force acting on the vehicle has magnitude \(\frac { 63000 } { k t ^ { 2 } } \mathrm {~N}\), where \(k\) is a positive constant. The force acts in the direction of motion of the vehicle. At time \(t\) seconds, \(t \geqslant 1\), the speed of the vehicle is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and the vehicle is a distance \(x\) metres from a fixed point \(O\) on the road. When \(t = 1\) the vehicle is at rest at \(O\) and when \(t = 4\) the speed of the vehicle is \(10.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
    1. Show that \(v = 14 - \frac { 14 } { t }\)
    2. Hence deduce that the speed of the vehicle never reaches \(14 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
    3. Use the trapezium rule, with 4 equal intervals, to estimate the value of \(x\) when \(v = 7\)

\begin{enumerate}
  \item A vehicle of mass 900 kg moves along a straight horizontal road. At time $t$ seconds the resultant force acting on the vehicle has magnitude $\frac { 63000 } { k t ^ { 2 } } \mathrm {~N}$, where $k$ is a positive constant. The force acts in the direction of motion of the vehicle. At time $t$ seconds, $t \geqslant 1$, the speed of the vehicle is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and the vehicle is a distance $x$ metres from a fixed point $O$ on the road. When $t = 1$ the vehicle is at rest at $O$ and when $t = 4$ the speed of the vehicle is $10.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(a) Show that $v = 14 - \frac { 14 } { t }$\\
(b) Hence deduce that the speed of the vehicle never reaches $14 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(c) Use the trapezium rule, with 4 equal intervals, to estimate the value of $x$ when $v = 7$
\end{enumerate}

\hfill \mbox{\textit{Edexcel M3 2015 Q4 [12]}}