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Further Kinematics

Further Kinematics

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Question 10

At time t t\,t seconds, where t≥0t \geq 0t≥0, a particle P P\,P moves so that its acceleration a ms−2\mathbf{a} \text{ ms}^{-2}a ms−2 is given by a=(2−4t)i+(6−3t2)j\mathbf{a} = (2 - 4t)\mathbf{i} + (6 - 3t^2)\mathbf{j}a=(2−4t)i+(6−3t2)j. At the instant when t=0t = 0t=0, the velocity of P P\,P is 12i ms−112\mathbf{i} \text{ ms}^{-1}12i ms−1

a.

Find the velocity of P P\,P when t=3t = 3t=3

[3]
b.

Find the value of t t\,t at the instant when P P\,P is moving in a direction perpendicular to i\mathbf{i}i

[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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