Edexcel M1 Specimen — Question 1 5 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
SessionSpecimen
Marks5
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Mark schemeDownload PDF ↗
TopicVectors Introduction & 2D
TypePosition vector at time t (constant velocity)
DifficultyModerate -0.8 This is a straightforward kinematics question requiring only the constant velocity equation r = r₀ + vt. Students must work backwards to find position at t=0, then forward to t=2, and finally calculate distance using Pythagoras. All steps are routine with no conceptual challenges beyond basic vector arithmetic.
Spec1.10a Vectors in 2D: i,j notation and column vectors1.10e Position vectors: and displacement1.10f Distance between points: using position vectors3.02a Kinematics language: position, displacement, velocity, acceleration

A particle \(P\) is moving with constant velocity \((-3\mathbf{i} + 2\mathbf{j})\) m s\(^{-1}\). At time \(t = 6\) s \(P\) is at the point with position vector \((-4\mathbf{i} - 7\mathbf{j})\) m. Find the distance of \(P\) from the origin at time \(t = 2\) s. [5]

A particle $P$ is moving with constant velocity $(-3\mathbf{i} + 2\mathbf{j})$ m s$^{-1}$. At time $t = 6$ s $P$ is at the point with position vector $(-4\mathbf{i} - 7\mathbf{j})$ m. Find the distance of $P$ from the origin at time $t = 2$ s.
[5]

\hfill \mbox{\textit{Edexcel M1  Q1 [5]}}