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Further Kinematics

Further Kinematics

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Question 190

At 10 a.m. plane A A\,A has position vector (2i−3j)(2\mathbf{i} - 3\mathbf{j})(2i−3j) km and moves with constant velocity (−i+4j)(-\mathbf{i} + 4\mathbf{j})(−i+4j) km h−1^{-1}−1. Another plane B B\,B has position vector (2i−8j)(2\mathbf{i} - 8\mathbf{j})(2i−8j) km and moves with constant velocity (i+5j)(\mathbf{i} + 5\mathbf{j})(i+5j) km h−1^{-1}−1.

a.

Show that the position vector of plane A A\,A after t t\,t hours is (2−t)i+(4t−3)j(2 - t)\mathbf{i} + (4t - 3)\mathbf{j}(2−t)i+(4t−3)j

[2]
b.

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

[6]
c.

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

[2]
d.

Find the time, after 10 a.m. when the planes are exactly 10 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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