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Further Kinematics

Further Kinematics

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Question 152

After t t\,t seconds, a particle P P\,P has velocity v\mathbf{v}v ms−1^{-1}−1 where v=6t12i−6tj\displaystyle \mathbf{v} = 6t^{\frac{1}{2}}\mathbf{i} - 6t\mathbf{j}v=6t21​i−6tj. When t=1t = 1t=1, P P\,P is at the point A A\,A and when t=4t = 4t=4, P P\,P is at the point BBB.

Find the exact distance ABABAB.

[6]
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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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