WJEC Unit 4 2019 June — Question 6 9 marks

Exam BoardWJEC
ModuleUnit 4 (Unit 4)
Year2019
SessionJune
Marks9
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TopicVariable acceleration (1D)
TypeVector motion with components
DifficultyStandard +0.3 This is a straightforward mechanics question requiring standard differentiation and integration of trigonometric functions. Part (a) uses F=ma with differentiation, part (b) requires integration with initial conditions, and part (c) is simple substitution and magnitude calculation. All techniques are routine A-level mechanics with no problem-solving insight needed, making it slightly easier than average.
Spec1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)3.02g Two-dimensional variable acceleration3.03d Newton's second law: 2D vectors

A particle \(P\) of mass \(0.5\) kg moves on a horizontal plane such that its velocity vector \(\mathbf{v}\) ms\(^{-1}\) at time \(t\) seconds is given by $$\mathbf{v} = 12\cos(3t)\mathbf{i} - 5\sin(2t)\mathbf{j}.$$
  1. Find an expression for the force acting on \(P\) at time \(t\) s. [3]
  2. Given that when \(t = 0\), \(P\) has position vector \((\mathbf{4i} + \mathbf{7j})\) m relative to the origin \(O\), find an expression for the position vector of \(P\) at time \(t\) s. [4]
  3. Hence determine the distance of \(P\) from \(O\) at time \(t = \frac{\pi}{2}\). [2]

A particle $P$ of mass $0.5$ kg moves on a horizontal plane such that its velocity vector $\mathbf{v}$ ms$^{-1}$ at time $t$ seconds is given by
$$\mathbf{v} = 12\cos(3t)\mathbf{i} - 5\sin(2t)\mathbf{j}.$$

\begin{enumerate}[label=(\alph*)]
\item Find an expression for the force acting on $P$ at time $t$ s. [3]

\item Given that when $t = 0$, $P$ has position vector $(\mathbf{4i} + \mathbf{7j})$ m relative to the origin $O$, find an expression for the position vector of $P$ at time $t$ s. [4]

\item Hence determine the distance of $P$ from $O$ at time $t = \frac{\pi}{2}$. [2]
\end{enumerate}

\hfill \mbox{\textit{WJEC Unit 4 2019 Q6 [9]}}