OCR M1 — Question 3 11 marks

Exam BoardOCR
ModuleM1 (Mechanics 1)
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTravel graphs
TypeDistance from velocity function using calculus
DifficultyModerate -0.3 This is a straightforward M1 kinematics question requiring standard techniques: differentiation to find acceleration, solving a linear equation, and integration to find displacement. The piecewise nature (accelerating then constant speed) is typical for M1, and all steps follow routine procedures with no novel problem-solving required. Slightly easier than average due to simple algebraic manipulation throughout.
Spec1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08d Evaluate definite integrals: between limits3.02f Non-uniform acceleration: using differentiation and integration

A motorcyclist starts from rest at a point \(O\) and travels in a straight line. His velocity after \(t\) seconds is \(v\) m s\(^{-1}\), for \(0 \leq t \leq T\), where \(v = 7.2t - 0.45t^2\). The motorcyclist's acceleration is zero when \(t = T\).
  1. Find the value of \(T\). [4]
  2. Show that \(v = 28.8\) when \(t = T\). [1]
For \(t \geq T\) the motorcyclist travels in the same direction as before, but with constant speed \(28.8\) m s\(^{-1}\).
  1. Find the displacement of the motorcyclist from \(O\) when \(t = 31\). [6]

A motorcyclist starts from rest at a point $O$ and travels in a straight line. His velocity after $t$ seconds is $v$ m s$^{-1}$, for $0 \leq t \leq T$, where $v = 7.2t - 0.45t^2$. The motorcyclist's acceleration is zero when $t = T$.

\begin{enumerate}[label=(\roman*)]
\item Find the value of $T$. [4]
\item Show that $v = 28.8$ when $t = T$. [1]
\end{enumerate}

For $t \geq T$ the motorcyclist travels in the same direction as before, but with constant speed $28.8$ m s$^{-1}$.

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Find the displacement of the motorcyclist from $O$ when $t = 31$. [6]
\end{enumerate}

\hfill \mbox{\textit{OCR M1  Q3 [11]}}