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Further Kinematics

Further Kinematics

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Question 140

A particle P P\,P has acceleration [(5t+2)i+(3t−1)j] ms−2[(5t + 2)\mathbf{i} + (3t - 1)\mathbf{j}] \text{ ms}^{-2}[(5t+2)i+(3t−1)j] ms−2 where t t\,t is the time in seconds. Initially P P\,P has velocity (−2i+2j) ms−1(-2\mathbf{i} + 2\mathbf{j}) \text{ ms}^{-1}(−2i+2j) ms−1

a.

Find an expression for the velocity of P P\,P in terms of ttt

[3]
b.

Find the velocities at the two times P P\,P is moving parallel to the vector (2i+j)(2\mathbf{i} + \mathbf{j})(2i+j)

[5]
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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