OCR MEI M1 2013 June — Question 4 6 marks

Exam BoardOCR MEI
ModuleM1 (Mechanics 1)
Year2013
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable acceleration (vectors)
TypeBearing or compass direction of motion
DifficultyStandard +0.3 This question requires understanding that bearing 045° means equal i and j components, leading to a simple quadratic equation. The multi-step nature (equating components, solving quadratic, finding speed) and vector context add slight complexity beyond routine problems, but the mathematical techniques are standard M1 material with no novel insight required.
Spec1.10a Vectors in 2D: i,j notation and column vectors3.02a Kinematics language: position, displacement, velocity, acceleration

4 The directions of the unit vectors \(\mathbf { i }\) and \(\mathbf { j }\) are east and north.
The velocity of a particle, \(\mathrm { vm } \mathrm { s } ^ { - 1 }\), at time \(t \mathrm {~s}\) is given by $$\mathbf { v } = \left( 16 - t ^ { 2 } \right) \mathbf { i } + ( 31 - 8 t ) \mathbf { j } .$$ Find the time at which the particle is travelling on a bearing of \(045 ^ { \circ }\) and the speed of the particle at this time.

Question 4:
AnswerMarks Guidance
AnswerMarks Guidance
Bearing 045° means equal east and north components, both positiveM1
\(16 - t^2 = 31 - 8t\)M1 Setting i and j components equal
\(t^2 - 8t + 15 = 0 \Rightarrow (t-3)(t-5)=0\)M1
\(t = 3\) or \(t = 5\); check both components positive: \(t=3\): east \(=7\), north \(=7\) ✓; \(t=5\): east \(=-9\) ✗A1 \(t=3\) only
Speed \(= \sqrt{7^2+7^2} = 7\sqrt{2} \approx 9.90\) m s\(^{-1}\)M1 A1
# Question 4:

| Answer | Marks | Guidance |
|--------|-------|----------|
| Bearing 045° means equal east and north components, both positive | M1 | |
| $16 - t^2 = 31 - 8t$ | M1 | Setting i and j components equal |
| $t^2 - 8t + 15 = 0 \Rightarrow (t-3)(t-5)=0$ | M1 | |
| $t = 3$ or $t = 5$; check both components positive: $t=3$: east $=7$, north $=7$ ✓; $t=5$: east $=-9$ ✗ | A1 | $t=3$ only |
| Speed $= \sqrt{7^2+7^2} = 7\sqrt{2} \approx 9.90$ m s$^{-1}$ | M1 A1 | |

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4 The directions of the unit vectors $\mathbf { i }$ and $\mathbf { j }$ are east and north.\\
The velocity of a particle, $\mathrm { vm } \mathrm { s } ^ { - 1 }$, at time $t \mathrm {~s}$ is given by

$$\mathbf { v } = \left( 16 - t ^ { 2 } \right) \mathbf { i } + ( 31 - 8 t ) \mathbf { j } .$$

Find the time at which the particle is travelling on a bearing of $045 ^ { \circ }$ and the speed of the particle at this time.

\hfill \mbox{\textit{OCR MEI M1 2013 Q4 [6]}}