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Further Kinematics

Further Kinematics

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Question 302

At 6 a.m. a boat A A\,A has position vector (10i−8j)(10\mathbf{i} - 8\mathbf{j})(10i−8j) km relative to a fixed origin O O\,O and moves with constant velocity (6i−4j)(6\mathbf{i} - 4\mathbf{j})(6i−4j) km h−1\text{h}^{-1}h−1. Another boat B B\,B has position vector (31i−29j)(31\mathbf{i} - 29\mathbf{j})(31i−29j) km relative to a fixed origin O O\,O and moves with constant velocity (−8i+10j)(-8\mathbf{i} + 10\mathbf{j})(−8i+10j) km h−1\text{h}^{-1}h−1. If both boats maintain their course and speed, they would collide at 7:30 a.m.

At 7 a.m. boat A A\,A realises that a collision is imminent and changes course so that it now has velocity (−8i+12j)(-8\mathbf{i} + 12\mathbf{j})(−8i+12j) km h−1\text{h}^{-1}h−1. Find the distance between the two ships at the time when they would have collided.

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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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