A particle P P\,P has acceleration [(3t2−6)i+(2t2+6t)j] ms−2[(3t^2 - 6)\mathbf{i} + (2t^2 + 6t)\mathbf{j}] \text{ ms}^{-2}[(3t2−6)i+(2t2+6t)j] ms−2 where t t\,t is the time in seconds. When t=3t = 3t=3 the velocity of the particle is (14i+20j) ms−1(14\mathbf{i} + 20\mathbf{j}) \text{ ms}^{-1}(14i+20j) ms−1
Find an expression for the velocity of P P\,P in terms of ttt, giving your answer in exact form
Initially the particle is at the point with position vector (2i−5j)(2\mathbf{i} - 5\mathbf{j})(2i−5j). Find an expression for the displacement of P P\,P in terms of ttt
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.