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Further Kinematics

Further Kinematics

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Question 211

A boat B B\,B moves with a constant velocity. At noon, B B\,B is at the point with position vector (2i−5j) km(2\mathbf{i} - 5\mathbf{j}) \text{ km}(2i−5j) km with respect to a fixed origin OOO. The velocity of B B\,B is (−4i+4j) km h−1(-4\mathbf{i} + 4\mathbf{j}) \text{ km h}^{-1}(−4i+4j) km h−1.

a.

Find the position vector of B B\,B at 1430.

[3]
b.

Find the bearing of the velocity vector

[2]
c.

Find an expression, in terms of ttt, for the position of BBB t t\,t hours after noon.

[2]
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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