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Further Kinematics

Further Kinematics

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Question 264

After t t\,t seconds, a particle P P\,P has velocity v\mathbf{v}v ms−1^{-1}−1 where v=3at12i−2tj\displaystyle \mathbf{v} = 3at^{\frac{1}{2}}\mathbf{i} - 2t\mathbf{j}v=3at21​i−2tj, where a>0a>0a>0. 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. Given that the exact distance AB AB\,AB is 421\sqrt{421}421​,

Find the value of aaa.

[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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