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Further Kinematics

Further Kinematics

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Question 118

At 09:00 hiker A A\,A has position vector (6i−2j)(6\mathbf{i} - 2\mathbf{j})(6i−2j) km and moves with constant velocity (−2i+4j)(-2\mathbf{i} + 4\mathbf{j})(−2i+4j) km h−1\text{h}^{-1}h−1. Hiker B B\,B has position vector (11i+3j)(11\mathbf{i} + 3\mathbf{j})(11i+3j) km and moves with constant velocity (2i+2j)(2\mathbf{i} + 2\mathbf{j})(2i+2j) km h−1\text{h}^{-1}h−1.

a.

Find the relative displacement of hiker A A\,A from hiker B B\,B after t t\,t hours.

[6]
b.

Find the time when A A\,A is due west of BBB.

[2]
c.

Find the time, after 09:00, when the hikers are exactly 290 \sqrt{290}\,290​ km apart.

[6]
Markscheme

Further Kinematics Questions

  1. A Level
  2. /Maths
  3. /Further Kinematics

363 exam-style questions on Edexcel A Level Maths Further Kinematics, covering 8.1 Vectors in Kinematics, 8.2 Vector Methods with Projectiles, 8.3 Variable Acceleration in One Dimension, 8.4 Differentiating Vectors, and 8.5 Integrating Vectors. Each one has a worked solution and a mark scheme showing where the marks go.

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