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Further Kinematics

Further Kinematics

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Question 256

After t t\,t seconds, a particle P P\,P has acceleration a\mathbf{a}a ms−2^{-2}−2 where a=4i+t−32j\displaystyle \mathbf{a} = 4\mathbf{i} + t^{-\frac{3}{2}}\mathbf{j}a=4i+t−23​j. When t=4t = 4t=4, the velocity of P P\,P is 16i−j16\mathbf{i} - \mathbf{j}16i−j ms−1^{-1}−1 and its position vector is 42i−3j42\mathbf{i} - 3\mathbf{j}42i−3j m

a.

Find the velocity of PPP

[3]
b.

Find the position vector of P P\,P when t=0t = 0t=0

[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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