A particle has an initial velocity of (i−5j) ms−1(\mathbf{i} - 5\mathbf{j}) \text{ ms}^{-1}(i−5j) ms−1 and is accelerating uniformly in the direction (i+2j)(\mathbf{i} + 2\mathbf{j})(i+2j) where i\mathbf{i}i and j\mathbf{j}j are perpendicular unit vectors. Given that the magnitude of the acceleration is 35 ms−23\sqrt{5} \text{ ms}^{-2}35 ms−2,
show that, after t t\,t seconds, the velocity vector of the particle is [(3t+1)i+(6t−5)j] ms−1[(3t + 1)\mathbf{i} + (6t - 5)\mathbf{j}] \text{ ms}^{-1}[(3t+1)i+(6t−5)j] ms−1.
Using your answer to part (a), or otherwise, find the exact value of t t\,t for which the speed of the particle is at its minimum.
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.