WJEC Unit 4 2018 June — Question 10 6 marks

Exam BoardWJEC
ModuleUnit 4 (Unit 4)
Year2018
SessionJune
Marks6
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Mark schemeDownload PDF ↗
TopicVectors Introduction & 2D
TypeNewton's second law with vector forces (find acceleration or force)
DifficultyModerate -0.3 This is a straightforward mechanics question requiring direct application of Newton's second law (F=ma) to find acceleration, then standard kinematics integration. Part (a) involves simple division and magnitude calculation; part (b) requires integrating acceleration twice with given initial conditions. All steps are routine with no problem-solving insight needed, making it slightly easier than average but not trivial due to 3D vectors and multiple steps.
Spec1.10b Vectors in 3D: i,j,k notation3.02g Two-dimensional variable acceleration3.03d Newton's second law: 2D vectors

A particle of mass 2 kg moves under the action of a constant force F N, where F is given by $$\mathbf{F} = -3\mathbf{i} + 4\mathbf{j} - 5\mathbf{k}.$$
  1. Find the magnitude of the acceleration of the particle. [3]
  2. Given that at time \(t = 0\) seconds, the position vector of the particle is \(2\mathbf{i} - 7\mathbf{j} + 9\mathbf{k}\) and it is moving with velocity \(3\mathbf{i} - 2\mathbf{j} + \mathbf{k}\), find the position vector of the particle when \(t = 2\) seconds. [3]

A particle of mass 2 kg moves under the action of a constant force F N, where F is given by

$$\mathbf{F} = -3\mathbf{i} + 4\mathbf{j} - 5\mathbf{k}.$$

\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of the acceleration of the particle. [3]

\item Given that at time $t = 0$ seconds, the position vector of the particle is $2\mathbf{i} - 7\mathbf{j} + 9\mathbf{k}$ and it is moving with velocity $3\mathbf{i} - 2\mathbf{j} + \mathbf{k}$, find the position vector of the particle when $t = 2$ seconds. [3]
\end{enumerate}

\hfill \mbox{\textit{WJEC Unit 4 2018 Q10 [6]}}