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Further Kinematics

Further Kinematics

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Question 24

After t t\,t seconds, a particle P P\,P has velocity v\mathbf{v}v ms−1^{-1}−1 where v=6ti−4t−12j\displaystyle \mathbf{v} = 6t\mathbf{i} - 4t^{-\frac{1}{2}}\mathbf{j}v=6ti−4t−21​j. When t=0t = 0t=0, the position vector of P P\,P is 8i+7j8\mathbf{i} + 7\mathbf{j}8i+7j m

a.

Find the acceleration of P P\,P when t=9t = 9t=9

[3]
b.

Find the position vector of P P\,P when t=9t = 9t=9

[3]
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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